Chapter 1: Sobolev Spaces Introduction In many problems of mathematical physics and variational calculus it is not sufficient to deal with the classical solutions of differential equations. It is necessary to introduce the notion of weak derivatives and to work in the so called Sobolev spaces. Let us consider the simplest example — the Dirichlet problem for the Laplace equation in a bounded domain Ω ⊂ Rn: u = 0, x ∈ Ω u(x) = ϕ(x), x ∈ ∂Ω, (∗) where ϕ(x) is a given function on the boundary ∂Ω. It is known that the Laplace equation is the Euler equation for the functional l(u) = Ω n j=1 ∂u ∂xj 2 dx. We can consider (∗) as a variational problem: to find the minimum of l(u) on the set of functions satisfying condition u|∂Ω = ϕ. It is much easier to minimize this functional not in C1(Ω), but in a larger class. Namely, in the Sobolev class W1 2 (Ω). W1 2 (Ω) consists of all functions u ∈ L2(Ω), having the weak derivatives ∂ju ∈ L2(Ω), j = 1, . . . , n. If the boundary ∂Ω is smooth, then the trace of u(x) on ∂Ω is well defined and relation u|∂Ω = ϕ makes sense. (This follows from the so called ” boundary trace theorem“ for Sobolev spaces.) If we consider l(u) on W1 2 (Ω), it is easy to prove the existence and uniqueness of solution of our variational problem. The function u ∈ W1 2 (Ω), that gives minimum to l(u) under the condition u|∂Ω = ϕ, is called the weak solution of the Dirichlet problem (∗). We’ll study the Sobolev spaces, the extension theorems, the boundary trace theorems and the embedding theorems. Next, we’ll apply this theory to elliptic boundary value problems. 1 §1: Preliminaries Let us recall some definitions and notation. Definition An open connected set Ω ⊂ Rn is called a domain. By Ω we denote the closure of Ω; ∂Ω is the boundary. Definition We say that a domain Ω ⊂ Ω ⊂ Rn is a strictly interior subdomain of Ω and write Ω ⊂⊂ Ω, if Ω ⊂ Ω. If Ω is bounded and Ω ⊂⊂ Ω, then dist {Ω , ∂Ω} > 0. We use the following notation: x = (x1, x2, . . . , xn) ∈ Rn , ∂ju = ∂u ∂xj , α = (α1, α2, . . . , αn) ∈ Zn + is a multi–index |α| = α1 + α2 + . . . + αn, ∂α u = ∂|α|u ∂xα1 1 ∂xα2 2 ...∂xαn n Next, u = (∂1u, . . . , ∂nu) , | u| =   n j=1 |∂ju|2   1/2 Definition Lq(Ω), 1 ≤ q < ∞ , is the set of all measurable functions u(x) in Ω such that the norm u q,Ω = Ω |u(x)|q dx 1/q is finite. Lq(Ω) is a Banach space. We’ll use the following property: Let u ∈ Lq(Ω), 1 ≤ q < ∞. We denote Jρ(u; Lq) = sup |z|≤ρ  n |u(x + z) − u(x)|q dx 1/q . Here u(x) is extended by zero on Rn\Ω. Jρ(u; Lq) is called the modulus of continuity of a function u in Lq(Ω). Then Jρ(u; Lq) → 0 as ρ → 0. 2 Definition Lq,loc(Ω), 1 ≤ q < ∞, is the set of all measurable functions u(x) in Ω such that Ω |u(x)|p dx < ∞ for any bounded strictly interior subdomain Ω ⊂⊂ Ω. Lq,loc(Ω) is a topological space (but not a Banach space). We say that uk k→∞ −→ u in Lq,loc(Ω), if uk − u q,Ω k→∞ −→ 0 for any bounded Ω ⊂⊂ Ω Definition L∞(Ω) is the set of all bounded measurable functions in Ω; the norm is defined by u ∞,Ω = ess sup x∈Ω |u(x)| Definition Cl(Ω) is the Banach space of all functions in Ω such that u(x) and ∂αu(x) with |α| ≤ l are uniformly continuous in Ω and the norm u Cl(Ω) = |α|≤l sup x∈Ω |∂α u(x)| is finite. If l = 0, we denote C0(Ω) = C(Ω). Remark If Ω is bounded, then u Cl(Ω) < ∞ follows from the uniform continuity of u, ∂αu, |α| ≤ l Definition Cl(Ω) is the class of functions in Ω such that u(x) and ∂αu, |α| ≤ l, are continuous in Ω. Remark Even if Ω is bounded, a function u ∈ Cl(Ω) may be not bounded; it may grow near the boundary. Definition C∞ 0 (Ω) is the class of the functions u(x) in Ω such that a) u(x) is infinitely smooth, which means that ∂αu is uniformly continuous in Ω, ∀α; b) u(x) is compactly supported: supp u is a compact subset of Ω. 3 §2: Mollification of functions 1. Definition of mollification The procedure of mollification allows us to approximate function u ∈ Lq(Ω) by smooth functions. Let ω(x), x ∈ Rn, be a function such that ω ∈ C∞ 0 (Rn ), ω(x) ≥ 0, ω(x) = 0 if |x| ≥ 1, and   ω(x)dx = 1. (1) For example, we may take ω(x) = c exp − 1 1−|x|2 if |x| < 1 0 if |x| ≥ 1 where constant c is chosen so that condition (1) is satisfied. For ρ > 0 we put ωρ(x) = ρ−n ω x ρ , x ∈ Rn . (2) Then ωρ ∈ C∞ 0 (Rn), ωρ(x) ≥ 0, ωρ(x) = 0 if |x| ≥ ρ, (3)  n ωρ(x)dx = 1. (4) Definition wρ is called a mollifier. Let Ω ⊂ Rn be a domain, and let u ∈ Lq(Ω) with some 1 ≤ q ≤ ∞. We extend u(x) by zero on Rn\Ω and consider the convolution ωρ ∗ u =: uρ uρ(x) =  n ωρ(x − y)u(y)dy. (5) In fact, the integral is over Ω ∩ {y : |x − y| < ρ}. Definition uρ(x) is called a mollification or regularization of u(x). 4 2. Properties of mollification 1) uρ ∈ C∞(Rn), and ∂αuρ(x) =  n ∂α x ωρ(x − y)u(y)dy. This follows from ωρ ∈ C∞. 2) uρ(x) = 0 if dist {x; Ω} ≥ ρ, since ωρ(x − y) = 0, y ∈ Ω. 3) Let u ∈ Lq(Ω) with some q ∈ [1, ∞]. Then uρ q,  n ≤ u q,Ω . (6) In other words, the operator Yρ : u → uρ is a linear continuous operator from Lq(Ω) to Lq(Rn) and Yρ Lq(Ω)→Lq(  n) ≤ 1. Proof: Case 1 : 1 < q < ∞. Let 1 q + 1 q = 1. By the H¨older inequality and (4), we have |uρ(x)| =  n ωρ(x − y)1/q ωρ(x − y)1/q u(y)dy ≤  n ωρ(x − y) 1/q =1  n ωρ(x − y)|u(y)|q dy 1/q ⇒ |uρ(x)|q ≤  n ωρ(x − y)|u(y)|q dy By (4), we obtain  n |uρ(x)|q dx ≤  n dx  n ωρ(x − y) |u(y)|q dy =  n dy |u(y)|q  n ωρ(x − y)dx =1 =  n |u(y)|q dy Case 2 : q = ∞. We have |uρ(x)| ≤  n ωρ(x − y)|u(y)|dy ≤ u ∞  n ωρ(x − y)dy =1 5 ⇒ uρ ∞ ≤ u ∞ Case 3 : q = 1. We integrate the inequality |uρ(x)| ≤  n ωρ(x − y)|u(y)|dy and obtain:  n |uρ(x)| dx ≤  n dx  n ωρ(x − y)|u(y)|dy =  n |u(y)|dy 4) Let u ∈ Lq(Ω), 1 ≤ q < ∞. Then uρ − u q,  n → 0 as ρ → 0. (7) Consequently, uρ − u q,Ω → 0 as ρ → 0. Proof The proof is based on the following property: if u ∈ Lq(Ω) (and u(x) is extended by 0), then sup |z|≤ρ  n |u(x + z) − u(x)|q dx 1/q =: Jρ(u; Lq) → 0 as ρ → 0. (Jρ(u; Lq) is called the modulus of continuity of u in Lq.) Case 1 : 1 < q < ∞. By (4) and (5) we have |uρ(x) − u(x)| ≤  n ωρ(x − y) (u(y) − u(x)) dy =  n ωρ(x − y)1/q ωρ(x − y)1/q (u(y) − u(x)) dy Then, by the H¨older inequality, it follows that |uρ(x) − u(x)| ≤  n ωρ(x − y)dy 1/q =1  n ωρ(x − y) |u(y) − u(x)|q dy 1/q 6 Hence,  n |uρ(x) − u(x)|q dx ≤  n dx  n ωρ(x − y) |u(y) − u(x)|q dy x−y=z = |z|<ρ dz ωρ(z)  n |u(y + z) − u(y)|q dy ≤ sup |z|≤ρ  n |u(y + z) − u(y)|q dy |z|<ρ dz ωρ(z) =1 = (Jρ(u; Lq))q . ⇒ uρ − u q,  n ≤ Jρ(u; Lq) → 0 as ρ → 0. Case 2 : q = 1. We have |uρ(x) − u(x)| ≤  n ωρ(x − y) |u(y) − u(x)| dy ⇒  n |uρ(x) − u(x)| dx ≤  n dx  n ωρ(x − y) |u(y) − u(x)| dy x−y=z = |z|<ρ dz ωρ(z)  n |u(y + z) − u(y)| dy ≤ Jρ(u; L1) → 0 as ρ → 0. Remark If q = ∞, there is NO such property, since L∞–limit of smooth functions uρ(x) must be a continuous function. If u ∈ C(Ω) and we extend u(x) by zero, then we may loose continuity. In general, uρ − u C(Ω) 0 as ρ → 0. However, we have the following property: 5) If u ∈ C(Ω) , Ω ⊂⊂ Ω and Ω is bounded, then uρ − u C(Ω ) → 0 as ρ → 0 7 Proof Let ρ < dist {Ω ; ∂Ω}. Then uρ(x) − u(x) =  n ωρ(x − y) (u(y) − u(x)) dy x−y=z =  n ωρ(z) (u(x − z) − u(x)) dz ⇒ sup x∈Ω |uρ(x) − u(x)| ≤ sup x∈Ω sup |z|≤ρ |u(x − z) − u(x)| → 0 as ρ → 0 (since u(x) is continuous in Ω). 8 §3: Class C∞ 0 (Ω) By C∞ 0 (Ω) we denote the class of infinitely smooth functions in Ω with compact support: u ∈ C∞ 0 (Ω) ⇔ u ∈ C∞ (Ω) and supp u ⊂ Ω. Theorem 1 C∞ 0 (Ω) is dense in Lq(Ω), 1 ≤ q < ∞ Proof Let u ∈ Lq(Ω) and ε > 0. Let Ω be a bounded domain, Ω ⊂⊂ Ω, and u q,Ω\Ω ≤ ε 2 . We put u(ε) (x) = u(x) if x ∈ Ω 0 if x ∈ Ω\Ω Then u − u(ε) q,Ω ≤ ε 2. Let u (ε) ρ (x) be the mollification of u(ε)(x). By property 4) of mollification, u (ε) ρ − u(ε) q,Ω ≤ ε 2 for sufficiently small ρ. Hence, u (ε) ρ − u q,Ω ≤ ε for sufficiently small ρ. Note that u (ε) ρ ∈ C∞ 0 (Ω) if ρ < dist {Ω , ∂Ω} . Theorem 2 Let u ∈ L1,loc(Ω), and suppose that Ω u(x)η(x)dx = 0, ∀η ∈ C∞ 0 (Ω). (8) Then u(x) = 0, a. e. x ∈ Ω. Theorem 2 is an analog of the Main Lemma of variational calculus. Proof 1) First, let us prove that Ω u(x)η(x)dx = 0 9 for any η ∈ L∞(Ω) with compact support supp η ⊂ Ω. Suppose that supp η ⊂ Ω , where Ω is a bounded domain and Ω ⊂⊂ Ω. Then ηρ ∈ C∞ 0 (Ω) if ρ < dist Ω ; ∂Ω =: 2ρ0. Let Ωρ0 = {x : dist {x; Ω } < ρ0}, and let χρ0 (x) = 1 if x ∈ Ωρ0 0 otherwise By (8), Ω u(x)ηρ(x)dx = 0, ρ < ρ0. (9) Since η ∈ L1(Ω), by property 4) of mollification, ηρ − η 1,Ω → 0 as ρ → 0. Then there exists a sequence {ρk}k∈  , ρk → 0, ρk < ρ0, such that ηρk (x) k→∞ −→ η(x) for almost every x ∈ Ω. Then also ηρk (x)u(x) k→∞ −→ η(x)u(x) for a. e. x ∈ Ω. Using property 3) (that ηρ ∞ ≤ η ∞), we have |u(x)ηρk (x)| ≤ χρ0 (x)|u(x)| η ∞ , (10) and the right-hand side in (10) belongs to L1(Ω). Then, by the Lebesgue Theorem, Ω u(x)ηρk (x)dx k→∞ −→ Ω u(x)η(x)dx. By (9), the left-hand side is equal to zero. Hence, Ω u(x)η(x)dx = 0. 2) Now, let Ω be a bounded domain such that Ω ⊂⊂ Ω. We put η(x) = u(x) |u(x)| , if u(x) = 0, x ∈ Ω 0 , otherwise Then u(x)η(x) = |u(x)| , x ∈ Ω 0 , x ∈ Ω\Ω Since η(x) is L∞–function with compact support supp η ⊂ Ω ⊂ Ω, then, by part 1), 0 = Ω u(x)η(x)dx = Ω |u(x)|dx. It follows that u(x) = 0 for a. e. x ∈ Ω . Since Ω is an arbitrary bounded domain such that Ω ⊂⊂ Ω, then u(x) = 0, a.e. x ∈ Ω 10 §4: Weak derivatives 1. Definition and properties of weak derivatives Definition 1 Let α be a multi–index. Suppose that u, v ∈ L1,loc(Ω), and Ω u(x)∂α η(x)dx = (−1)|α| Ω v(x)η(x)dx, ∀η ∈ C∞ 0 (Ω). (11) Then v is called the weak ( or distributional ) partial derivative of u in Ω, and is denoted by ∂αu. If u(x) is sufficiently smooth to have continuous derivative ∂αu, we can integrate by parts: Ω u(x)∂α η(x)dx = Ω (−1)|α| ∂α u(x)η(x)dx. Hence, the classical derivative ∂αu is also the weak derivative. Of course, ∂αu may exist in the weak sense wihout existing in the classical sense. Remark 1) To define the weak derivative ∂αu, we don’t need the existence of derivatives of the smaller order (like in the classical definition). 2) The weak derivative is defined as an element of L1,loc(Ω), so we can change it on some set of measure zero. Properties of ∂α u 1) Uniqueness Proof Uniqueness of the weak derivative follows from Theorem 2. Suppose that u ∈ L1,loc(Ω) and v, w ∈ L1,loc(Ω) are both weak derivatives of u. Then, by (11), Ω (v(x) − w(x)) η(x)dx = 0, ∀η ∈ C∞ 0 (Ω). By Theorem 2, v(x) = w(x), a.e. x ∈ Ω. 11 2) Linearity If u1, u2 ∈ L1,loc(Ω) and there exist weak derivatives v1 = ∂αu1, v2 = ∂αu2 ∈ L1,loc(Ω), then there exists ∂α (c1u1 + c2u2) and ∂α (c1u1 + c2u2) = c1∂α u1 + c2∂α u2, c1, c2 ∈ C. Proof Obviously, Ω (c1u1 + c2u2) ∂α ηdx = c1 Ω u1∂α ηdx + c2 Ω u2∂α ηdx = (−1)|α| c1 Ω v1ηdx + (−1)|α| c2 Ω v2ηdx = (−1)|α| Ω (c1v1 + c2v2) =∂α(c1u1+c2u2) ηdx. 3) If v = ∂αu in Ω , then v = ∂αu in Ω for any Ω ⊂ Ω. Obvious 4) Mollification of the weak derivative ” Derivative of mollification is equal to mollification of derivative“. This is true in any bounded strictly interior domain Ω ⊂⊂ Ω. Suppose that u, v ∈ L1,loc(Ω) and v = ∂αu. Then vρ(x) = ∂α uρ(x) if ρ < dist {x, ∂Ω} . (12) The functions uρ and vρ are smooth; the derivative ∂αuρ in (12) is understood in the classical sense. Proof Let ρ < dist {x, ∂Ω}. We have uρ(x) = Ω ωρ(x − y)u(y)dy Then ∂αuρ(x) = Ω ∂α x ωρ(x − y)u(y)dy. Note that ∂α x ωρ(x − y) = (−1)|α|∂α y ωρ(x − y). Hence, ∂α uρ(x) = (−1)|α| Ω ∂α y ωρ(x − y)u(y)dy. Since ρ < dist {x, ∂Ω}, then for η(y) := ωρ(x−y) we have η ∈ C∞ 0 (Ω). By definition of the weak derivative ∂αu = v, we obtain ∂α uρ(x) = Ω ωρ(x − y)v(y)dy = vρ(x). 12 5) Suppose that u ∈ L1,loc(Ω) and there exists the weak derivative ∂αu such that ∂αu ∈ Lq(Ω), 1 ≤ q < ∞. Then ∂αuρ − ∂αu q,Ω → 0 as ρ → 0, for any bounded strictly interior domain Ω ⊂⊂ Ω. Proof This follows from property 4) of mollification and property 4) of weak derivatives: ∂α u = v ∈ Lq(Ω); ∂α uρ = vρ in Ω ( for sufficiently small ρ) ; vρ − v q,Ω → 0 as ρ → 0. Remark If we extend u(x) by zero on Rn\Ω, then, in general, the weak derivative ∂αu in Rn does not exist. Hence, we have convergence ∂αuρ ρ→0 −→ ∂αu in Lq(Ω ) only for bounded strictly interior domain Ω . Exclusion: if u(x) = 0, if dist {x; ∂Ω} < ρo, and ∂αu ∈ Lq(Ω), then ∂αuρ − ∂αu q,Ω ρ→0 −→ 0. 2. Another definition of the weak derivative Definition 2 Suppose that u, v ∈ L1,loc(Ω) and there exists a sequence um ∈ Cl(Ω), m ∈ N, such that um m→∞ −→ u and ∂αum m→∞ −→ v in L1,loc(Ω). Here α is a multi–index and |α| = l. Then v is called the weak derivative of u in Ω : ∂αu = v. Definition 1 ⇔ Definition 2 Proof 1) Definition 1 ⇐ Definition 2. Since um ∈ Cl(Ω), then Ω um∂α ηdx = (−1)|α| Ω ∂α umηdx, ∀η ∈ C∞ 0 (Ω). (13) 13 For η fixed, the left–hand side of (13) tends to Ω u∂αηdx as m → ∞: Ω (um − u) ∂α ηdx ≤ max |∂α η| supp η |um − u| dx m→∞ −→ 0. Similarly, the right–hand side of (13) tends to (−1)|α| Ω vηdx. Conse- quently, Ω u∂α ηdx = (−1)|α| Ω vηdx, ∀η ∈ C∞ 0 (Ω). It means that v = ∂αu in the sense of Definition 1. 2) Definition 1 ⇒ Definition 2. Let u, v ∈ L1,loc(Ω), and let v = ∂αu in the sense of Definition 1. We want to find a sequence um ∈ C∞(Ω) such that um m→∞ −→ u and ∂αum m→∞ −→ v in L1,loc(Ω). Let {Ωm} , m ∈ N, be a sequence of bounded domains such that Ωm ⊂⊂ Ω, Ωm ⊂ Ωm+1 and m∈  Ωm = Ω. We put u(m) (x) = u(x) if x ∈ Ωm 0 otherwise Then u(m) ∈ L1(Ω). Consider the mollification of u(m) : u (m) ρ ∈ C∞(Ω). Let {ρm}m∈  be a sequence of positive numbers such that ρm → 0 as m → ∞. We put um(x) = u(m) ρm (x), x ∈ Ω. Then um ∈ C∞(Ω) and um m→∞ −→ u in L1,loc(Ω). Prove this yourself, using property 4) of mollification. Next, by property 5) of ∂αu, prove that ∂αum m→∞ −→ v in L1,loc(Ω). Thus, v = ∂αu in the sense of Definition 2. 14 Theorem 3 Let um ∈ L1,loc(Ω) and um m→∞ −→ u in L1,loc(Ω). Suppose that there exist weak derivatives ∂αum ∈ L1,loc(Ω) and ∂αum m→∞ −→ v in L1,loc(Ω). Then v = ∂αu. In other words, the operator ∂α is closed. Proof By Definition 1, for ∂αum we have Ω um∂α ηdx = (−1)|α| Ω ∂α umηdx, ∀η ∈ C∞ 0 (Ω) ↓ m → ∞ ↓ m → ∞ Ω u∂α ηdx = (−1)|α| Ω vηdx, ∀η ∈ C∞ 0 (Ω) ⇒ v = ∂αu in the sense of Definition 1. Remark The conclusion of Theorem 3 remains true under weaker assumptions that Ω umηdx m→∞ −→ Ω uηdx and Ω ∂α umηdx m→∞ −→ Ω vηdx, ∀η ∈ C∞ 0 (Ω). (It means that um → u and ∂αum → v in D (Ω).) 3. Weak derivatives of the product of functions Proposition If u, ∂ju ∈ Lq,loc(Ω), and v, ∂jv ∈ Lq ,loc(Ω) with some 1 < q < ∞, 1 q + 1 q = 1 or if u, ∂ju ∈ L1,loc(Ω) and v, ∂jv ∈ C(Ω), then ∂j (uv) = (∂ju) v + u (∂jv) . Proof 1) Case 1: 1 < q < ∞ Let us fix η ∈ C∞ 0 (Ω). Let Ω be a bounded domain such that supp η ⊂ Ω ⊂⊂ Ω. We put ˜u(x) = u(x) , x ∈ Ω 0 otherwise ˜v(x) = v(x) , x ∈ Ω 0 otherwise 15 Then ˜u ∈ Lq(Ω ), ˜v ∈ Lq (Ω ). By property 4) of mollifications, ˜uρ − ˜u q,Ω → 0, ˜vρ − ˜v q ,Ω → 0 as ρ → 0 Next, ∂j ˜u = ∂ju in Ω , ∂j ˜v = ∂jv in Ω (it is clear from Definition 1). So, ∂j ˜u ∈ Lq(Ω ), ∂j ˜v ∈ Lq (Ω ). By property 5) of weak derivatives, ∂j ˜uρ − ∂j ˜u q,suppη → 0, as ρ → 0, ∂j ˜vρ − ∂j ˜v q ,supp η → 0, as ρ → 0. Since ˜uρ, ˜vρ are smooth functions, we have Ω ˜uρ˜vρ∂jηdx = − Ω ∂j (˜uρ˜vρ) ηdx = − Ω (∂j ˜uρ) ˜vρηdx − Ω ˜uρ (∂j ˜vρ) ηdx. (14) Let us show that Ω ˜uρ˜vρ∂jηdx ρ→0 −→ Ω ˜u˜v∂jηdx = Ω uv∂jηdx (15) We have Ω (˜uρ˜vρ − ˜u˜v) ∂jηdx ≤ Ω (˜uρ − ˜u) ˜vρ∂jηdx + Ω ˜u (˜vρ − ˜v) ∂jηdx ≤ ˜uρ − ˜u q,Ω →0 ˜vρ q ,Ω bounded max |∂jη| + + ˜u q,Ω ˜vρ − ˜v q ,Ω →0 max |∂jη| → 0 as ρ → 0 Similarly, we can show that Ω ((∂j ˜uρ) ˜vρ + ˜uρ (∂j ˜vρ)) ηdx ρ→0 → Ω ((∂j ˜u) ˜v + ˜u (∂j ˜v)) ηdx = Ω ((∂ju) v + u (∂jv)) ηdx (16) From (14) - (16) it follows that Ω uv∂jηdx = − Ω ((∂ju) v + u (∂jv)) ηdx This identity is proved for any η ∈ C∞ 0 (Ω). It means (by Definition 1) that there exists the weak derivative ∂j(uv) and ∂j(uv) = (∂ju) v + u (∂jv) 2) Case q=1. Prove yourself 16 4. Change of variables Suppose that u ∈ L1,loc(Ω) and there exist weak derivatives ∂ju ∈ L1,loc(Ω), j = 1, . . . , n. Let y = f(x) be a diffeomorphism of class C1 and f(Ω) = ˜Ω. We put ˜u(y) = u(f−1(y)). Then ˜u ∈ L1,loc(˜Ω). Let us show that there exist weak derivatives ∂˜u ∂yκ , κ = 1, . . . , n, and ∂˜u ∂yk = n j=1 ∂u ∂xj ∂xj ∂yκ Proof Since there exist weak derivatives ∂u ∂xj ∈ L1,loc(Ω), j = 1, . . . , n, then there exists a sequence um ∈ C1(Ω) such that um m→∞ −→ u and ∂um ∂xj m→∞ −→ ∂u ∂xj in L1,loc(Ω) for all j = 1, . . . , n. (We can construct this sequence like in the proof, that Definition 1 and Definition 2 are equivalent ). We denote ˜um(y) = um(f−1(y)). Then ˜um ∈ C1(˜Ω), and, by usual rule (for classical derivatives), ∂˜um ∂yk = n j=1 ∂um ∂xj ∂xj ∂yκ Let us check that ˜um m→∞ −→ ˜u in L1,loc( ˜Ω). Indeed, for every bounded domain ˜Ω ⊂⊂ ˜Ω we have ˜Ω |˜um(y) − ˜u(y)| dy = ˜Ω um(f−1 (y)) − u(f−1 (y)) dy = Ω |um(x) − u(x)| |J(x)| dx → 0 as m → ∞ Here Ω = f−1 (˜Ω ) and J(x) is the Jacobian of the transformation f(x) J(x) = det ∂y ∂x . Here the right-hand side tends to zero, since |J(x)| is bounded in Ω ; Ω is a bounded domain such that Ω ⊂⊂ Ω; and um m→∞ −→ u in L1,loc(Ω). Similarly, using that ∂um ∂xj m→∞ −→ ∂u ∂xj in L1,loc(Ω), one can show that ∂˜um ∂yk = n j=1 ∂um ∂xj ∂xj ∂yκ m→∞ −→ n j=1 ∂u ∂xj ∂xj ∂yκ in L1,loc(˜Ω) Then, by Definition 2, there exist weak derivatives ∂˜u ∂yk and ∂˜u ∂yk = n j=1 ∂u ∂xj ∂xj ∂yk 17 Thus, for weak derivatives we have the usual rule of change of variables. The same is true for derivatives of higher order. 5. For ordinary derivatives we have the following property: if ∂u ∂xj = 0 in Ω, j = 1, . . . , n, then u = const. The same is true for weak derivatives. Theorem 4 Suppose that u ∈ L1,loc(Ω) and there exist weak derivatives ∂αu for any multi-index α such that |α| = l (l ∈ N) and ∂αu = 0 in Ω, |α| = l. Then u(x) is a polynomial of order ≤ l − 1 in Ω. Proof 1) Let Ω be a bounded domain such that Ω ⊂⊂ Ω. Let Ω be another bounded domain such that Ω ⊂⊂ Ω ⊂⊂ Ω. We put ˜u(x) = u(x) , x ∈ Ω 0 otherwise Then ˜u ∈ L1(Ω ) , and ∂α ˜u = ∂αu = 0, |α| = l, in Ω . Consider the mollification ˜uρ(x). If ρ < dist {Ω , ∂Ω }, then, by property 4) of ∂αu, ∂α ˜uρ(x) = (∂α ˜u)ρ (x), x ∈ Ω , |α| = l. Hence, ∂α ˜uρ = 0 in Ω . Thus, ˜uρ(x) is a smooth function in Ω and all its derivatives of order l are equal to zero. It follows that ˜uρ(x) = P (ρ) l−1(x), x ∈ Ω , where P (ρ) l−1 is a polynomial of order ≤ l−1. By property 4) of mollification, ˜uρ − u 1,Ω → 0 as ρ → 0, i. e. , P (ρ) l−1 ρ→0 −→ u in L1(Ω ). The set of all polynomials in Ω of order ≤ l − 1 is a finite-dimensional (and, so, closed!) subspace in L1(Ω ). Therefore, the limit u(x) must be also a polynomial of order ≤ l − 1: u(x) = Pl−1(x), x ∈ Ω . 2) Now it is easy to complete the proof by the standard procedure. Let {Ωk}k∈  be a sequence of bounded domains such that Ωk ⊂⊂ Ω, Ωk ⊂ Ωk+1, and k∈  Ωk = Ω. 18 We have poved that for each domain Ωk u(x) = P (k) l−1(x), x ∈ Ωk Then P (k+1) l−1 (x) is continuation of P (k) l−1(x), but continuation of a polynomial is unique. ⇒ There exists a polynomial Pl−1(x) such that u(x) = Pl−1(x), x ∈ Ω. 6. Absolute continuity property The existence of the weak derivative is related to the absolute continuity property. Recall the definition of absolute continuity for function of one variable. Definition Function u : [a, b] → R is called absolutely continuous, if for any ε > 0 there exists δ > 0 such that for any finite set of disjoint intervals x1, x1 , x2, x2 , . . . , xm, xm (⊂ [ a, b ]) with m j=1 xj − xj < δ, one has m j=1 u(xj) − u(xj) < ε. We’ll use the following facts: 1) u : [a, b] → R is absolutely continuous if and only if there exists a function v ∈ L1(a, b) such that u(x) = u(a) + x a v(t)dt, x ∈ [a, b] . 2) If u : [a, b] → R is absolutely continuous, then there exists derivative du dx for almost every x ∈ (a, b) and du dx = v (∈ L1(a, b)). Theorem 5 Let n = 1. A measurable function u(x) is absolutely continuous on [a, b] if and only if there exists the weak derivative du dx ∈ L1(a, b). The weak derivative coincides with the classical derivative almost everywhere. 19 Remark When we speak about measurable functions, we mean not just one function but a class of functions, that are equal to each other almost everywhere. So, when we say that a measurable function u(x) is absolutely continuous, it means that in the class of functions equivalent to u, there exists an absolutely continuous representative. Proof 1) u(x) is absolutely continuous. ⇒ ∃ weak derivative du dx ∈ L1. If u(x) is a. c., then there exists the classical derivative du dx = v almost everywhere and v ∈ L1(a, b). Next, let η ∈ C∞ 0 (a, b). Then the product η u is also absolutely continuous. There exists the classical derivative d(η u) dx for almost every x ∈ (a, b). We have the usual rule: d (η u) dx = vη + u dη dx . Integrate this identity over (a, b). Then b a d(η u) dx dx = 0. (Since η(x) = 0 near a and b). Hence, b a vη + u dη dx dx = 0. The obtained identity b a u dη dx = − b a vηdx, ∀η ∈ C∞ 0 (a, b), by Definition 1, means that v is the weak derivative du dx 2) ∃ weak derivative v = du dx ∈ L1(a, b) ⇒ u is a. c. Consider w(x) = x a v(t)dt. Then w(x) is absolutely continuous. There exists classical derivative dw dx = v, a.e. x ∈ (a, b). By statement 1) (already proved), there exists the weak derivative dw dx which coincides with the classical one and with v. Thus, du dx = dw dx , i. e. d (u − w) dx = 0. (the weak derivative is equal to zero.) By Theorem 4, u − w = const. Since w(x) is absolutely consinuous, then u = c + w is also absolutely continuous. If x = a, we have u(a) = c + w(a) =0 = c. Thus, u(x) = u(a) + x a v(t)dt. 20 u(x) is absolutely continuous; it has classical derivative for a. e. x ∈ (a, b); classical derivative = weak derivative = v ∈ L1(a, b). Theorem 6 Let Ω ⊂ Rn, n > 1. We denote x = (x1, . . . , xj−1, xj+1, . . . , xn) and write x = {x , xj}. Suppose that [a(x ), b(x )] are some intervals such that {x } × [a(x ), b(x )] ⊂ Ω. Let u ∈ L1,loc(Ω) and there exists the weak derivative ∂u ∂xj ∈ L1,loc(Ω). Then for almost every x the function u(x , xj) is absolutely continuous on interval [a(x ), b(x )] (as a function of one variable xj). Exercise: Prove Theorem 6. 7. Examples 1) Let Ω = (0, 1)2 and u(x1, x2) = ϕ(x1) + ψ(x2), where ϕ and ψ are not absolutely continuous on [0, 1], but ϕ, ψ ∈ L1(0, 1). Then, by Theorem 6, u(x1, x2) does not have weak derivatives ∂u ∂x1 , ∂u ∂x2 in Ω. (Since, if they exist, then u(x) must be also absolutely continuous in x1 for x2 fixed, and in x2 for x1 fixed.) However, there exists the weak derivative ∂2u ∂x1∂x2 = 0. Indeed, for ∀ η ∈ C∞ 0 (Ω), Ω u ∂2 η ∂x1∂x2 dx = 1 0 1 0 ϕ(x1) ∂2 η ∂x1∂x2 dx1dx2 + 1 0 1 0 ψ(x2) ∂2 η ∂x1∂x2 dx1dx2 = 1 0 dx1ϕ(x1) 1 0 ∂2 η ∂x1∂x2 dx2 =0 + 1 0 dx2ψ(x2) 1 0 ∂2 η ∂x1∂x2 dx1 =0 = 0. By Definition 1 of weak derivative, it means that there exists weak derivative ∂2u ∂x1∂x2 and ∂2u ∂x1∂x2 =0. This example shows that functions may have derivative of higher order, not having derivatives of lower order. 2) Suppose that the domain Ω ⊂ Rn is devided by a smooth (n-1)–dimensional surface Γ into two parts Ω1 and Ω2. So, Ω = Ω1 ∪ 21 Ω2 ∪ Γ. Let u1 ∈ C1(Ω1), u2 ∈ C1(Ω2), u(x) = u1(x) , x ∈ Ω1 u2(x) , x ∈ Ω2 If u1|Γ = u2|Γ, then, in general, weak derivatives do not exist. Let n(x) be the unit normal vector to Γ exterior with respect to Ω1. Since uk(x), k = 1, 2, is a C1–function in Ωk, we can integrate by parts in Ωk: for η ∈ C∞ 0 (Ω) we have: Ω u ∂η ∂xj dx = Ω1 u1 ∂η ∂xj dx + Ω2 u2 ∂η ∂xj dx = − Ω1 ∂u1 ∂xj η dx − Ω2 ∂u2 ∂xj η dx + + Γ (u1(x) − u2(x)) η cos (∠(n, 0xj)) dS(x) If u1 = u2 on Γ (we have NO jump on Γ), then the integral over Γ is equal to zero. In this case, there exists the weak derivative ∂u ∂xj and ∂u ∂xj = ∂u1 ∂xj in Ω1 ∂u2 ∂xj in Ω2 Also, if cos (∠(n, 0xj)) = 0, then there exists ∂u ∂xj . For example, if Γ is parallel to the axis 0xj, then cos (∠(n, 0xj)) = 0. ⇒ Even if u1|Γ = u2|Γ, the tangential derivative exists. If Γ (u1(x) − u2(x)) η cos (∠(n, 0xj)) dS(x) = 0, then ∂u ∂xj does not exist. Exercise Let Ω = {x ∈ Rn : |x| < 1}, and let u(x) = |x|α , α > −n + 1. Prove that there exist the weak derivatives ∂u ∂xj and ∂u ∂xj = αxj |x|α−2 , j = 1, . . . , n. For this consider continuous functions u(δ) (x) = |x|α , |x| > δ δα , |x| ≤ δ From the previous example we know that ∃ ∂u(δ) ∂xj = αxj |x|α−2 , |x| > δ 0 , |x| ≤ δ Check that u(δ) L1(Ω) −→ u and ∂u(δ) ∂xj L1(Ω) −→ vj := αxj |x|α−2 . Then, by Theorem 3, it follows that ∃ ∂u ∂xj = vj 22 §5: The Sobolev spaces Wl p(Ω) and ◦ Wl p(Ω) 1. Definition of Wl p(Ω) (1 ≤ p < ∞, l ∈ Z+) Definition Suppose that u ∈ Lp(Ω) and there exist weak derivatives ∂αu for any α with |α| ≤ l (all derivatives up to order l), such that ∂α u ∈ Lp(Ω), |α| ≤ l. Then we say that u ∈ Wl p(Ω). We introduce the (standard) norm in Wl p(Ω): u W l p(Ω) =   Ω |α|≤l |∂α u|p dx   1/p . Remark 1) The norm |α|≤l ∂αu p,Ω is equivalent to the standard norm. 2) W0 p (Ω) = Lp(Ω). Proposition Wl p(Ω) is complete. In other words, Wl p(Ω) is a Banach space. Proof Let {um} be a fundamental sequence in Wl p(Ω). It is equivalent to the fact that all sequences {∂αum} for |α| ≤ l are fundamental sequences in Lp(Ω). Since the space Lp(Ω) is complete , there exist functions u, vα ∈ Lp(Ω) such that um Lp(Ω) −→ u, ∂α um Lp(Ω) −→ vα as m → ∞. Then also um → u, ∂αum → vα in L1,loc(Ω). By Theorem 3, vα = ∂αu. Hence, um W l p(Ω) −→ u as m → ∞. If p = 2, the space Wl 2(Ω) is a Hilbert space with the inner product (u, v)W l 2(Ω) = Ω |α|≤l ∂α u(x)∂αv(x)dx. 23 For Wl 2(Ω) another notation Hl(Ω) is often used: Wl 2(Ω) = Hl(Ω). Using the properties of weak derivatives (see section 4 ” Change of variables“ in § 4), we can show that the class Wl p(Ω) is invariant with respect to smooth (Cl-class) change of variables. Theorem 7 Let f : Ω → ˜Ω be a diffeomorphism of class Cl, so that f ∈ Cl(Ω), f−1 ∈ Cl(˜Ω). Then, if u ∈ Wl p(Ω) , then ˜u = u ◦ f−1 ∈ Wl p(˜Ω), and c1 u W l p(Ω) ≤ ˜u W l p(˜Ω) ≤ c2 u W l p(Ω) . (17) The constants c1, c2 do not depend on u; they depend only on f Cl(Ω) and f−1 Cl(˜Ω) . Proof: For simplicity, let us prove Theorem 7 in the case l = 1. We have u ∈ W1 p (Ω), ˜u(y) = u(f−1 (y)). By section 4 in §4, there exist the weak derivatives ∂˜u ∂yk = n j=1 ∂u ∂xj ∂xj ∂yk . Let us check that ∂˜u ∂yk ∈ Lp(˜Ω): ˜Ω ∂˜u ∂yk p dy 1/p =   Ω n j=1 ∂u ∂xj ∂xj ∂yk p |J(x)| dx   1/p ≤ n j=1 max x∈Ω ∂xj ∂yk |J(x)|1/p Ω ∂u ∂xj p dx 1/p ≤ c n j=1 Ω ∂u ∂xj p dx 1/p . Here J(x) = det f (x) and the constant c := maxj,k maxx∈Ω ∂xj ∂yk |J(x)|1/p depends only on the norms f C1(Ω) and f−1 C1(˜Ω) . Also we have ˜Ω |˜u(y)|p dy = Ω |u(x)|p |J(x)| dx ≤ (max |J(x)|) Ω |u(x)|p dx. Thus, ˜u W 1 p (˜Ω) ≤ c2 u W 1 p (Ω) with the constant c2 depending only on f C1 and f−1 C1 . Prove the lower estimate in (17) yourself (for this change the roles of u and ˜u in the argument). 24 2. Definition of ◦ Wl p(Ω) Definition The closure of C∞ 0 (Ω) in the norm of Wl p(Ω) is denoted by ◦ Wl p(Ω). So, ◦ Wl p(Ω) is a subspace in the space Wl p(Ω). Proposition Let u ∈ ◦ Wl p(Ω), and let ˜u(x) = u(x) x ∈ Ω 0 x ∈ Rn\Ω. Then ˜u ∈ Wl p(Ω1) for any Ω1 such that Ω ⊂ Ω1. In particular, ˜u ∈ Wl p(Rn). Proof By definition of ◦ Wl p(Ω), there exists a sequence um ∈ C∞ 0 (Ω) such that um W l p(Ω) −→ u as m → ∞. We put ˜um(x) = um(x) x ∈ Ω 0 otherwise . Then ˜um ∈ C∞ 0 (Ω1) and ˜um W l p(Ω1) −→ ˜u as m → ∞ (since ˜um − ˜u W l p(Ω1) = um − u W l p(Ω)). Hence, ˜u ∈ ◦ Wl p(Ω1). Theorem 8 Let u ∈ ◦ Wl p(Ω) and let ˜u(x) = u(x) x ∈ Ω 0 x ∈ Rn\Ω. Then for mollifications uρ(x) we have uρ ρ→0 −→ u in Wl p(Ω). 25 Proof We have already proved that ˜u ∈ Wl p(Rn). Then ∂α˜u ∈ Lp(Rn), |α| ≤ l. By property 4) and 5) of ∂αu (mollification of the weak derivative), ∂α ˜uρ ρ→0 −→ ∂α ˜u in Lp(Ω), |α| ≤ l. It means that ˜uρ ρ→0 −→ ˜u in Wl p(Ω). But, by definition of ˜u and definition of mollification, ˜u = u in Ω, and ˜uρ = uρ. So, uρ ρ→0 −→ u in Wl p(Ω). Remark If u(x) is an arbitrary function in Wl p(Ω), and ˜u(x) is defined as above, then, in general, ˜u(x) does not have weak derivatives in Rn. (See example 2 in Section 7 of §4). So, in general, ◦ Wl p(Ω) = Wl p(Ω). 3. Integration by parts Proposition Let u ∈ Wl p(Ω) and v ∈ ◦ Wl p(Ω), where 1 p + 1 p = 1. Then Ω ∂α uvdx = (−1)|α| Ω u ∂α vdx, |α| ≤ l. (18) Proof Let vm ∈ C∞ 0 (Ω) and vm → v as m → ∞ in ◦ Wl p(Ω). By Definition 1 of the weak derivative ∂αu, we have Ω ∂α uvmdx = (−1)|α| Ω u ∂α vmdx. (19) Let us show that Ω ∂α u vm dx m→∞ −→ Ω ∂α u v dx, Ω u ∂α vm dx m→∞ −→ Ω u ∂α v dx. We have Ω ∂α u (vm − v) dx ≤ Ω |∂α u|p dx 1/p Ω |vm − v|p dx 1/p ≤ u W l p(Ω) vm − v W l p (Ω) → 0 as m → ∞; 26 Ω u (∂α vm − ∂α v) dx ≤ Ω |u|p dx 1/p Ω |∂α vm − ∂α v|p dx 1/p ≤ u W l p(Ω) vm − v W l p (Ω) → 0 as m → ∞. Tending to the limit in (19) as m → ∞, we obtain (18). 4. Separability By V l p (Ω) we denote the linear space of all vector–valued functions v = {vα}|α|≤l such that vα ∈ Lp(Ω), |α| ≤ l. We introduce the norm in V l p (Ω): v V l p (Ω) = |α|≤l vα p,Ω . Then V l p (Ω) is the direct product of a finite number (equal to the number of multi–indices α with |α| ≤ l) of Lp(Ω). We know that Lp(Ω) is a separable Banach space if 1 ≤ p < ∞. Then so is V l p (Ω). Now, consider the transformation J from Wl p(Ω) (equipped with the norm ||u||W l p(Ω) = |α|≤l ∂αu p,Ω, which is equivalent to the standard norm) to V l p (Ω): J : Wl p(Ω) → V l p (Ω), Ju = {∂α u}|α|≤l . Then J is a linear operator; it preserves the norm: Ju V l p (Ω) = ||u||W l p(Ω); and J is injective. Such an operator is called isometric. The range Ran J = ˜V l p (Ω) is a linear set in V l p (Ω) consisting of vector–valued functions v of the form v = {∂αu}|α|≤l , u ∈ Wl p(Ω). From Theorem 3 it follows that ˜V l p (Ω) is a closed subspace of V l p (Ω). Hence, ˜V l p (Ω) is separable together with V l p (Ω). (Since any subspace of some separable space is also separable.) Since J is isometric, we can identify Wl p(Ω) with ˜V l p (Ω). It follows that Wl p(Ω) is separable if 1 ≤ p < ∞. 5. The space Wl p(Rn ) Proposition ◦ Wl p(Rn) = Wl p(Rn) . In other words, C∞ 0 (Rn) is dense in Wl p(Rn). 27 Proof Let ζ ∈ C∞(R+) be such that 0 ≤ ζ(t) ≤ 1, ζ(t) = 1 if 0 ≤ t ≤ 1, ζ(t) = 0 if t ≥ 2. Let u ∈ Wl p(Rn). We put u(R)(x) = u(x)ζ |x| R . Then u(R) (x) = u(x) if |x| ≤ R, u(R) (x) = 0 if |x| ≥ 2R. Note that derivatives ∂β x ζ |x| R are uniformly bounded with respect to R ≥ 1. Calculating the derivatives of u(R)(x), we obtain the inequality ∂α u(R) (x) ≤ c |β|≤|α| ∂β u(x) , a.e. x ∈ Rn . Then for |α| ≤ l we have ∂α u(R) − ∂α u p,  n =      n ∂α u(R) (x) − ∂α u(x) p =0 for |x|≤R dx     1/p = |x|>R ∂α u(R) (x) − ∂α u(x) p dx 1/p ≤ c |β|≤|α| |x|>R ∂β u(x) p dx 1/p → 0 as R → ∞ This expression tends to zero as R → ∞, since u ∈ Wl p(Rn), and so, ∂βu p ∈ L1. Thus, u(R) → u as R → ∞ in Wl p(Rn). Now, we consider mollification u (R) ρ of u(R). Then u (R) ρ ∈ C∞ 0 (Rn) and u (R) ρ → u(R) as ρ → 0 in Wl p(Rn). It follows that C∞ 0 (Rn) is dense in Wl p(Rn). Indeed, let u ∈ Wl p(Rn) and let ε > 0. We find R so large that u(R) − u W l p(  n) < ε 2. Next, we find ρ so small that u (R) ρ − u(R) W l p(  n) < ε 2. Then u (R) ρ − u W l p(  n) < ε. 28 6. The Friedrichs inequality Theorem 9 If Ω is a bounded domain in Rn, then for any function u ∈ ◦ Wl p(Ω) we have u p,Ω ≤ (diamΩ)l —u—p,l,Ω. (20) Here —u—p,l,Ω =   |α|=l ∂α u p p,Ω   1/p . (21) Proof Since C∞ 0 (Ω) is dense in ◦ Wl p(Ω), it suffices to prove (20) for u ∈ C∞ 0 (Ω). 1) So, let u ∈ C∞ 0 (Ω). Let Q be a cube with the edge d = diamΩ, such that Ω ⊂ Q. We extend u(x) by zero to Q\Ω. We can choose the coordinate system so that Q = {x : 0 < xj < d, j = 1, . . . n}. Obviously, u(x) = xn 0 ∂u ∂xn (x , y)dy, x ∈ Q. Here x = (x1, . . . , xn−1 x , xn) = (x , xn). Then, by the H¨older inequality, |u(x)|p ≤ xn 0 ∂u ∂xn (x , y) p dy xn 0 1dy p/p ≤ dp/p ≤ dp/p d 0 ∂u(x , xn) ∂xn p dxn Here 1 p + 1 p = 1. We integrate both sides of this inequality: Ω |u|p dx = Q |u|p dx ≤ dp/p d 0 dxn Q ∂u ∂xn p dx p p +1=p = dp Ω ∂u ∂xn p dx. We have proved that u p,Ω ≤ (diamΩ) Ω ∂u ∂xn p dx 1/p ≤ (diamΩ)—u—p,1,Ω. (22) This is inequality (20) for l = 1. 29 2) In order to prove (20) with l > 1, we iterate (22): Ω ∂u ∂xn p dx ≤ dp Ω ∂2u ∂x2 n p dx, etc. ⇒ Ω |u|p dx ≤ dlp Ω ∂lu ∂xl n p dx ≤ dlp —u—p p,l,Ω. Remark Inequality (20) is not valid for all u ∈ Wl p(Ω). Example If Ω is a bounded domain and u(x) = Pl−1(x)(= 0) is a polynomial of order ≤ l − 1, then —u—p,l,Ω = 0, but u p,Ω = 0. 30 §6. Domains of star type A natural question: Can we approximate functions in Wl p(Ω) by smooth functions? The answer depends on domain Ω. We’ll consider the class of domains for which the answer is ” YES, we can“. Definition We say that a bounded domain Ω is of star type with respect to a point 0, if any half–line starting at point 0 intersects ∂Ω only in one point. Theorem 10 Let Ω be a bounded domain of star type with respect to a point 0. Then C∞(Ω) is dense in Wl p(Ω). Proof Let us use the coordinate system with origin 0. Consider a sequence of domains Ωk = x : k−1 k x ∈ Ω , k ∈ N. Then Ωk+1 ⊂ Ωk and Ω ⊂ Ωk. Let u ∈ Wl p(Ω). We put uk(x) = u(k−1 k x). Clearly, uk ∈ Wl p(Ωk). Let us show that uk − u W l p(Ω) → 0 as k → ∞. We have uk − u p,Ω = Ω u k − 1 k x − u(x) p dx 1/p −→ 0 as k → ∞. This follows from the property of Lp–functions: if u ∈ Lp(Ω), then sup |z(x)|≤ c k Ω |u(x + z(x)) − u(x)|p dx k→∞ −→ 0. (In our case z(x) = −x k and |x| ≤ diamΩ = d ⇒ |z(x)| ≤ d k .) Let α be a multi–index with |α| ≤ l. Then ∂α uk − ∂α u p,Ω = Ω k − 1 k |α| ∂α u k − 1 k x − ∂α u(x) dx 1/p ≤ 1 − k − 1 k |α| →0 as k→∞ Ω ∂α u k − 1 k x p dx 1/p ≤c u W l p(Ω) + + Ω ∂α u k − 1 k x − ∂α u(x) p dx 1/p →0 as k→∞ . 31 Hence, uk k→∞ −→ u in Wl p(Ω). Consider mollifications uk,ρ(x). Then uk,ρ ∈ C∞(Ω) and uk,ρ ρ→0 −→ uk in Wl p(Ω) (since Ω is bounded and Ω ⊂⊂ Ωk). We can choose a sequence {ρk}, so that ρk → 0 as k → ∞, and a sequence ˜uk(x) := uk,ρk (x) tends to u(x) in Wl p(Ω): ˜uk ∈ C∞(Ω) and ˜uk − u W l p(Ω) k→∞ −→ 0. Remark Let Ω = {x : |x| < 1, xn > 0} be a half–ball. Ω is of star type with respect to any interior point 0 . Suppose that u ∈ Wl p(Ω) and u(x) = 0 if |x| > 1 − ε. Then ˜uk ∈ C∞(Ω) and ˜uk(x) = 0 if |x| > 1 − ε 2 for sufficiently large k. 32 §7: Extension theorems We can always extend a function u ∈ ◦ Wl p(Ω) by zero and the extended function ∈ Wl p(˜Ω) in ˜Ω (⊃ Ω). It is a natural question if we can extend functions of class Wl p(Ω). We start with the case l = 1. Theorem 11 Suppose that Ω ⊂ Rn is a bounded domain such that Ω is a compact manifold of class C1. Let ˜Ω be a domain in Rn such that Ω ⊂ ˜Ω. Then there exists a linear bounded extension operator Π : W1 p (Ω) → ◦ W1 p (˜Ω) such that (Πu) (x) = u(x), x ∈ Ω. Proof We proceed in three steps: Step 1 Let Ω = K+ = {x : |x| < 1, xn > 0} be a half–ball, and let u ∈ W1 p (K+) and u(x) = 0 near Σ+ = {x ∈ ∂K+ : |x| = 1}. We extend u to the left half–ball K− = {x : |x| < 1, xn < 0} as follows: v(x) = u(x) x ∈ K+ u(x , −xn) x ∈ K−. Let us show that v ∈ W1 p (K) and v W 1 p (K) = 21/p u W 1 p (K+) . (23) Here K = {x : |x| < 1}. Using construction of Theorem 10 (and Remark after Theorem 10), we can find a sequence um(x) such that um ∈ C∞(K+), um(x) = 0 near Σ+, and um − u W 1 p (K+) → 0 as m → ∞. We put vm(x) = um(x) x ∈ K+ um(x , −xn) x ∈ K−. Then vm ∈ C(K), vm(x) = 0 near ∂K, vm ∈ C∞(K+), vm ∈ C∞(K−). It follows that vm ∈ ◦ W1 p (K) (see §4, Subsection 7, Example 2). For the norm of vm we have: vm p W 1 p (K) = K (|vm(x)|p + | vm(x)|p ) dx = 2 K+ (|um(x)|p + | um(x)|p ) dx ⇒ vm W 1 p (K) = 21/p um W 1 p (K+) . (24) 33 Next, ∂vm(x) ∂xj = ∂um(x) ∂xj x ∈ K+ ∂ ∂xj (um(x , −xn)) x ∈ K−. Since um m→∞ −→ u in W1 p (K+), it follows that vm m→∞ −→ v in Lp(K) and ∂vm ∂xj m→∞ −→ wj in Lp(K), where wj(x) = ∂u(x) ∂xj x ∈ K+ ∂u ∂xj (x , −xn) x ∈ K− , j = 1, . . . , n − 1; wn(x) = ∂u(x) ∂xn x ∈ K+ − ∂u ∂xn (x , −xn) x ∈ K−. By Theorem 3, there exist weak derivatives ∂v ∂xj in K and ∂v ∂xj = wj. Thus, vm m→∞ −→ v in W1 p (K). Relation (23) follows from (24) by the limit procedure (as m → ∞). Step 2 Suppose that u ∈ W1 p (Ω) and suppu ⊂ U, where U is a neighbourhood of x0 ∈ ∂Ω, such that U ⊂ ˜Ω and ∃ diffeomorphism f : U → K, f ∈ C1(U), f−1 ∈ C1(K), f(U) = K, f(U ∩ Ω) = K+, f(U ∩ ∂Ω) = ∂K+\Σ+. We consider the function ˜u(y) = u(f−1(y)), y ∈ K+. Then ˜u ∈ W1 p (K+) and ˜u(y) = 0 near Σ+. We extend ˜u(y) on K− like in step 1: ˜v(y) = ˜u(y) y ∈ K+ ˜u(y , −yn) y ∈ K−. As it was proved in step 1, ˜v ∈ ◦ W1 p (K), and ˜v W 1 p (K) = 21/p ˜u W 1 p (K+) . Consider the function v(x) = ˜v(f(x)), x ∈ U. Then v ∈ ◦ W1 p (U). We extend v(x) by zero on ˜Ω\U. Then v ∈ ◦ W1 p (˜Ω), v|Ω = u, and v W 1 p (˜Ω) = v W 1 p (U) ≤ c1 ˜v W 1 p (K) ≤ c121/p ˜u W 1 p (K+) ≤ c2c121/p =c u W 1 p (Ω) . The constant c depends on f C1 , f−1 C1 and on p. Step 3 (general case) Let Ω be a bounded domain such that Ω is a compact manifold of class 34 C1 with boundary ∂Ω. Then (by definition of such manifolds) there exists a finite number of open sets U1, U2, . . . , UN such that either Uj ⊂ Ω or Uj is a neighbourhood of some point x(j) ∈ ∂Ω, and ∃ a diffeomorphism fj ∈ C1(Uj), f−1 j ∈ C1(K), fj(Uj) = K, fj(Uj ∩ Ω) = K+, fj(Uj ∩ ∂Ω) = ∂K+\Σ+. Finally, Ω ⊂ N j=1 Uj. We can choose the sets U1, U2, . . . , UN so that N j=1 Uj ⊂ ˜Ω. There exists a partition of unity {ζj(x)}j=1,...,N such that ζj ∈ C∞ 0 (Rn), supp ζj ⊂ Uj, N j=1 ζj(x) = 1, x ∈ Ω. Let u ∈ W1 p (Ω). We represent u(x) as u(x) = N j=1 uj(x), where uj(x) = ζj(x)u(x). If Uj ⊂ Ω, then uj ∈ ◦ W1 p (Ω) (since supp ζj ⊂ Uj) and we can extend uj(x) by zero to ˜Ω\Ω : vj(x) = uj(x) x ∈ Ω 0 x ∈ ˜Ω\Ω. If U ∩ ∂Ω = ∅, then uj(x) satisfies assumptions of step 2. By the result of step 2, we can extend uj(x) to some function vj ∈ ◦ W1 p (˜Ω) such that vj(x) = uj(x), x ∈ Ω, and vj W 1 p (˜Ω) ≤ cj uj W 1 p (Ω). The constant cj depends on fj C1 , f−1 j C1 and on p. We put Πu = v = N j=1 vj. Then v ∈ ◦ W1 p (˜Ω), v(x) = N j=1 vj(x) = N j=1 uj(x) = u(x), x ∈ Ω, and v W 1 p (˜Ω) ≤ N j=1 vj W 1 p (˜Ω) ≤ N j=1 cj uj W 1 p (Ω) ≤ c N j=1 uj W 1 p (Ω) , where c = max1≤j≤N {cj}. Finally, uj W 1 p (Ω) = ζju W 1 p (Ω) ≤ ˆcj u W 1 p (Ω) . The constant ˆcj depends on ζj C1 . Hence, v W 1 p (˜Ω) ≤ cˆc u W 1 p (Ω) , ˆc = N j=1 ˆcj. Thus, we constructed the linear continuous extension operator Π : W1 p (Ω) → W1 p (˜Ω). 35 Remark It is clear from the proof that for the constructed extension operator Π we have v Lp(˜Ω) ≤ c u Lp(Ω) , v = Πu. The constant c depends on p, Ω and ˜Ω. 2. Similar extension theorem is true for unbounded domain Ω ⊂ Rn satisfying the following condition. Suppose that there exist bounded open sets {Uj} , j ∈ N, such that Ω ⊂ ∞ j=1 Uj. Here either Uj ⊂ Ω or Uj is a neighbourhood of a point x(j) ∈ ∂Ω and ∃ a diffeomorphism fj ∈ C1(Uj), f−1 j ∈ C1(K), fj(Uj) = K, fj(Ω ∩ Uj) = K+, fj(∂Ω ∩ Uj) = ∂K+\Σ+. Moreover, suppose that the norms fj C1(Uj ) and f−1 j C1(K) are uniformly bounded for all j ∈ N. Suppose also that each point x ∈ Ω belongs only to a finite number N(x) of sets Uj, and that N(x) ≤ N < ∞, ∀x ∈ Ω. (This means that the multiplicity of covering is finite.) Theorem 12 Under the above conditions on Ω ⊂ Rn, let ˜Ω be a domain in Rn such that ∞ j=1 Uj ⊂ ˜Ω. Then there exists a linear bounded extension operator Π : W1 p (Ω) → ◦ W1 p (˜Ω) such that (Πu) (x) = u(x), x ∈ Ω. We omit the proof. 3. Now we consider the case l > 1 Theorem 13 Suppose that Ω ⊂ Rn is a bounded domain such that Ω is a compact manifold of class Cl. Let ˜Ω be a domain in Rn such that Ω ⊂ ˜Ω. Then there exists a linear bounded extension operator Π : Wl p(Ω) → ◦ Wl p(˜Ω), i. e., (Πu) (x) = u(x), for x ∈ Ω and Πu W l p(˜Ω) ≤ c1 u W l p(Ω). Besides, Πu Lp(˜Ω) ≤ c2 u Lp(Ω). The constants c1, c2 depend on l, p, Ω and ˜Ω. 36 Proof Like in the proof of Theorem 11, the question reduces to the case, where Ω = K+ and u(x) = 0 near Σ+. Moreover, it suffices to consider smooth functions u ∈ C∞(K+). So, let u ∈ C∞(K+) and u(x) = 0 near Σ+. We extend u(x) by zero to Rn +\K+. We put v(x) = u(x) x ∈ K+ l−1 j=0 cj u(x , −2jxn) x ∈ K−. The constants cj, j = 0, . . . , l − 1, are chosen so that ∂mv ∂xm n (x , +0) = ∂mv ∂xm n (x , −0), m = 0, . . . , l − 1. These conditions are equivalent to the following system of linear equations for c0, c1, . . . , cl−1: l−1 j=0 −2j m cj = 1, m = 0, 1, . . . , l − 1. The determinant of this system is not zero. Hence, such constants c0, . . . , cl−1 exist. It is easy to check that v ∈ Wl p(K), and ∂α v p,K ≤ cα ∂α u p,K+ , |α| ≤ l. (We use that v ∈ C∞(K+), v ∈ C∞(K−) and v ∈ Cl−1(K). Then v ∈ Wl p(K).) Hence , v W l p(K) ≤ c1 u W l p(K+). Obviously, v(x) = 0 near ∂K. So, v ∈ ◦ Wl p(K). Next, for arbitrary domain Ω, we use the covering Ω ⊂ N j=1 Uj and the partition of unity. The argument is the same as in proof of Theorem 11. The only difference is that we consider diffeomorphisms of class Cl. Remark 1) The conclusion of Theorem 13 remains true under weaker assumptions on the domain Ω. It suffices to assume that Ω is domain of class C1 (for arbitrary l!) or, even that Ω is Lipschitz domain (it means that diffeomorphisms fj, f−1 j ∈ Lip1). 2) Extension theorems allow us to reduce the study of functions in Wl p(Ω) to the study of functions in ◦ Wl p(˜Ω). In particular, from the fact that C∞ 0 (˜Ω) is dense in ◦ Wl p(˜Ω) it follows that C∞(Ω) is dense in Wl p(Ω), if domain Ω satisfies conditions of Theorem 13. 37 Chapter 2: Embedding Theorems Introduction Embedding theorems give relations between different functional spaces. Definition Let B1 and B2 be two Banach spaces. We say that B1 is embedded into B2 and write B1 → B2, if for any u ∈ B1 we have u ∈ B2 and u B2 ≤ c u B1 , where the constant c does not depend on u ∈ B1. We define the embedding operator J : B1 → B2, which takes u ∈ B1 into the same element u considered as an element of B2. The fact that B1 → B2 is equivalent to the fact that the embedding operator J : B1 → B2 is continuous linear operator. If u B2 ≤ c u B1 , ∀u ∈ B1, then J B1→B2 ≤ c. Definition If B1 → B2 and the embedding operator J : B1 → B2 is a compact operator, then we say that B1 is compactly embedded into B2. The compactness of operator J is equivalent to the fact that any bounded set in B1 is a compact set in B2. Some embeddings are obvious. For example, it is obvious that Wl1 p (Ω) → Wl2 p (Ω), if l1 > l2. In particular, Wl p(Ω) → Lp(Ω), l > 0. But the fact that for bounded domain Ω, these embeddings are compact, is non–trivial. (This is the Rellich embedding theo- rem.) More general is the Sobolev embedding theorem : Wl p(Ω) → Wr q (Ω) under some conditions on p, l, q, r (with q > p and r < l). Another embedding theorem is that, if pl > n, then a function u ∈ Wl p(Ω) is continuous (precisely, u(x) coincides with a continuous function for a. e. x ∈ Ω). The trace embedding theorems show that functions in Wl p(Ω) have traces on some surfaces of lower dimension. The embedding theorems are very important for the modern analysis and boundary value problems. 38 §1: Integral operators in Lp(Ω) In order to prove embedding theorems, we need some auxiliary material about integral operators. 1. Let Ω ⊂ Rn and D ⊂ Rm be some bounded domains. We consider the integral operator (Ku) (x) = v(x) = Ω K(x, y)u(y)dy, x ∈ D, u ∈ Lp(Ω) (1 ≤ p < ∞). We’ll show that under some conditions on the kernel K(x, y), the operator K is continuous or, even, compact from Lp(Ω) to Lq(D), or from Lp(Ω) to C(D). We always assume that K(x, y) is a measurable function on D × Ω, and K satisfies one or several of the following conditions: a) Ω |K(x, y)|t dy ≤ M for a. e. x ∈ D, where t ≥ 1. b) D |K(x, y)|s dx ≤ N for a. e. y ∈ Ω, where s > 0. c) ess sup x∈D,y∈Ω |K(x, y)| ≤ L < ∞ (K is bounded). d) sup x,z∈D |x−y|≤ρ sup y∈Ω |K(x, y) − K(z, y)| ≤ ε(ρ) → 0 as ρ → 0 (K is continuous in x). Lemma 1 If K(x, y) satisfies conditions c) and d) , then K : Lp(Ω) → C(D) is compact operator. Here 1 ≤ p < ∞. Proof Let u ∈ Lp(Ω) and v(x) = (Ku) (x). Then, by condition c), |v(x)| ≤ L Ω |u(y)| dy ≤ L Ω |u(y)|p dy 1/p Ω 1p dy 1/p = L|Ω|1/p u p,Ω , (1) 39 where 1 p + 1 p = 1. (If p = 1 then (1) is also true with p = ∞, |Ω|1/p = 1.) Next, if |x − z| ≤ ρ (x, z ∈ D), then, by condition d), |v(x) − v(z)| = Ω (K(x, y) − K(z, y)) u(y)dy ≤ ε(ρ) Ω |u(y)| dy ≤ ε(ρ)|Ω|1/p u p,Ω . (2) From (1) and (2) it follows that, if u belongs to some bounded set in Lp(Ω): u p,Ω ≤ c, then the set of functions {v} is uniformly bounded ( v C(D) ≤ L|Ω|1/p c) and equicontinuous (|v(x) − v(z)| ≤ ε(ρ)|Ω|1/p c, if |x − z| ≤ ρ). By the Arzela Theorem, this set is compact in C(D). It means that the operator K : Lp(Ω) → C(D) is compact. Lemma 2 1) If p > 1, 1 p + 1 p = 1, and K(x, y) satisfies conditions a) and b) with some t < p and s p + t p ≥ 1, then v = Ku ∈ Lq(D) (for u ∈ Lp(Ω)), where q ≥ p is defined from the relation s q + t p = 1. We have v q,D ≤ M1/p N1/q u p,Ω , u ∈ Lp(Ω). (3) 2) If p = 1, and K(x, y) satisfies condition b) with s = q ≥ 1. Then v = Ku ∈ Lq(D) and v q,D ≤ N1/q u 1,Ω , u ∈ L1(Ω). (4) 3) If p > 1, and K(x, y) satisfies condition a) with t = p , then v = Ku ∈ L∞(D) and v ∞,D ≤ M1/p u p,Ω , u ∈ Lp(Ω). (5) 4) If p = 1, and K(x, y) satisfies condition c), then v = Ku ∈ L∞(D) and v ∞,D ≤ L u 1,Ω , u ∈ L1(Ω). (6) 40 Proof 1) Let p > 1. Using that s q + t p = 1, we obtain: |K(x, y)u(y)| = |K(x, y)|s/q |u(y)|p/q |u(y)|1− p q |K(x, y)|t/p We apply the H¨older inequality for the product of three functions: Ω |f1(y)f2(y)f3(y)| dy ≤ Ω |f1|p1 dy 1 p1 Ω |f2|p2 dy 1 p2 Ω |f3|p3 dy 1 p3 with 1 p1 + 1 p2 + 1 p3 = 1. We take p1 = q, p2 = pq q−p, p3 = p . Then |v(x)| ≤ Ω |K(x, y)u(y)| dy ≤ Ω |K(x, y)|s |u(y)|p dy 1 q Ω |u(y)|p dy 1 p − 1 q Ω |K(x, y)|t dy 1 p ≤M1/p ( by cond. a) ) ≤ M 1 p u 1− p q p,Ω Ω |K(x, y)|s |u(y)|p dy 1 q . Note that in the case q = p, we simply apply the ordinary H¨older inequality and obtain the same result. We have |v(x)|q ≤ Mq/p u q−p p,Ω Ω |K(x, y)|s |u(y)|p dy ⇒ D |v(x)|q dx ≤ Mq/p u q−p p,Ω D dx Ω |K(x, y)|s |u(y)|p dy = Mq/p u q−p p,Ω Ω |u(y)|p dy D |K(x, y)|s dx ≤N( by cond b )) ≤ NMq/p u q p,Ω . This gives estimate (3). 2) Let p = 1 and s = q ≥ 1. If q > 1, we have |K(x, y)u(y)| = (|K(x, y)|q |u(y)|)1/q |u(y)|1/q , 1 q + 1 q = 1. Then, by the H¨older inequality, |v(x)| ≤ Ω |K(x, y)u(y)| dy ≤ Ω |K(x, y)|q |u(y)| dy 1/q Ω |u(y)| dy 1/q 41 ⇒ D |v(x)|q dx ≤ D dx Ω |K(x, y)q | |u(y)| dy u q/q 1,Ω = Ω |u(y)| dy D |K(x, y)|q dx ≤N( by cond b) with s=q) u q/q 1,Ω ≤ N u q 1,Ω . This implies (4) (in the case q > 1). If q = 1, then |v(x)| ≤ Ω |K(x, y)| |u(y)| dy ⇒ D |v(x)| dx ≤ D dx Ω |K(x, y)| |u(y)| dy = Ω |u(y)| dy D |K(x, y)| dx ≤N ≤ N u 1,Ω . This implies (4) (in the case q = 1). 3) Let p > 1, and condition a) is satisfied with t = p . Then, by the H¨older inequality, |v(x)| ≤ Ω |K(x, y)u(y)| dy ≤ Ω |K(x, y)|p dy 1/p ≤M1/p Ω |u(y)|p 1/p ≤ M1/p u p,Ω . This yields (5). 4) Let p = 1 and K satisfies condition c). Then |v(x)| ≤ Ω |K(x, y)| |u(y)| dy ≤ L Ω |u(y)| dy = L u 1,Ω , which gives (6). 42 Remark Statements 1) and 2) mean that the operator K : Lp(Ω) → Lq(D) is continuous, and K Lp(Ω)→Lq(D) ≤ M1/p N1/q , p > 1; (7) K Lp(Ω)→Lq(D) ≤ N1/q , p = 1. (8) Under conditions of 3) and 4) the operator K : Lp(Ω) → L∞(D) is continuous and K Lp(Ω)→L∞(D) ≤ M1/p , p > 1; (9) K Lp(Ω)→L∞(D) ≤ L, p = 1. (10) 2. Now, we’ll show that under some additional assumptions on K(x, y), the operator K is compact. We’ll assume that K(x, y) can be approximated by Kh(x, y) (as h → 0) and Kh(x, y) are bounded and continuous in x. Lemma 3 Suppose that Kh(x, y), 0 < h < h0, satisfies conditions c) and d) (where L = L(h) and ε(ρ) = ε(ρ; h) depend on h). 1) Suppose that K(x, y) and Kh(x, y), 0 < h < h0, satisfy conditions of Lemma 2(1) with common t, s, M, N, and Ω |Kh(x, y) − K(x, y)|t dy ≤ mh h→0 −→ 0, for a. e. x ∈ D; (11) D |Kh(x, y) − K(x, y)|s dx ≤ nh h→0 −→ 0, for a. e. y ∈ Ω. (12) Then the operator K : Lp(Ω) → Lq(D) is compact. 2) Suppose that K(x, y), Kh(x, y), 0 < h < h0, satisfy conditions of Lemma 2(2) with common s = q, N, and D |Kh(x, y) − K(x, y)|q dx ≤ nh h→0 −→ 0, for a. e. y ∈ Ω. (13) Then the operator K : L1(Ω) → Lq(D) is compact. 3) Suppose that K(x, y), Kh(x, y) satisfy conditions of Lemma 2(3) with common t = p , M, and condition (11) is satisfied with t = p . Then the operator K : Lp(Ω) → C(D) is compact. (Here p > 1.) 43 Proof We denote (Khu) (x) = vh(x) = Ω Kh(x, y)u(y)dy. By Lemma 1, the operator Kh : Lp(Ω) → C(D) is compact. Obviously, the embedding C(D) → Lq(D) (for a bounded domain D) is continuous. Hence, the operator Kh : Lp(Ω) → Lq(D) is also compact. 1) From conditions (11), (12) and the estimate (7) it follows that Kh − K Lp(Ω)→Lq(D) ≤ m 1/p h n 1/q h → 0 as h → 0. Thus, K is the limit in the operator norm of compact operators Kh. Hence, K : Lp(Ω) → Lq(D) is compact. 2) Similarly, if p = 1, from condition (13) and estimate (8) it follows that Kh − K L1(Ω)→Lq(D) ≤ n 1/q h → 0 as h → 0. It follows that K : L1(Ω) → Lq(D) is compact. 3) From condition (11) with t = p and estimate (5), it follows that vh − v ∞,D ≤ m 1/p h u p,Ω , u ∈ Lp(Ω). Hence, vh − v ∞,D → 0 as h → 0. Since (Khu) (x) = vh(x) is uniformly continuous, then v(x) is also uniformly continuous: v ∈ C(D). Thus, the operator K maps Lp(Ω) into C(D), and Kh − K Lp(Ω)→C(D) ≤ m 1/p h → 0 as h → 0. Since Kh : Lp(Ω) → C(D) is compact and Kh h→0 −→ K in the operator norm, then K : Lp(Ω) → C(D) is also compact operator. 3. Now we apply Lemmas 1–3 to the study of the operator (Kju) (x) = Ω xj − yj |x − y|n u(y)dy, j = 1, . . . , n. Here x ∈ Ω or x ∈ Ωm, where Ωm is some section of Ω by m–dimensional hyper–plane (m < n). So, either D = Ω or D = Ωm. If m = n, we agree that Ωn ≡ Ω. 44 Lemma 4 1) Suppose that 1 ≤ p ≤ n, n − p < m ≤ n, q ≥ 1 and 1 − n p + m q > 0. Then the operator Kj : Lp(Ω) → Lq(Ωm) is compact. (If m = n, then Ωm = Ω; if m < n, then Ωm is arbitrary section of Ω by m–dimensional hyper–plane.) 2) If p > n, then the operator Kj : Lp(Ω) → C(Ωm) is compact. In particular, Kj : Lp(Ω) → C(Ω) is compact. Proof The proof is based on Lemmas 2 and 3. 1) Case 1 < p ≤ n Suppose that conditions 1) are satisfied and, moreover, that q ≥ p > 1. We put θ = 1 − n p + m q , t = n n − 1 + θ = n n p + m q ; s = m n − 1 + θ = m n p + m q . Then s q + t p = 1. Since q ≥ p, then s p + t p ≥ 1. Clearly, t < p . Since m ≤ n, q ≥ p, then θ ≤ 1. Hence, t ≥ 1. Thus, the numbers t and s satisfy conditions of Lemma 2(1). Let us check that Kj(x, y) = xj−yj |x−y|n satisfy conditions a) and b) with these t and s. Note that t(n − 1) < n (since t(n − 1) < t(n − 1 + θ) = n) and s(n − 1) < m (since s(n − 1) < s(n − 1 + θ) = m). We have Ω |Kj(x, y)|t dy = Ω |xj − yj|t |x − y|tn dy ≤ Ω dy |x − y|t(n−1) . This integral converges since t(n − 1) < n. Let d = diamΩ, B(x) = {y ∈ Rn : |x − y| ≤ d}. Obviously, Ω ⊂ B(x). Then Ω |Kj(x, y)|t dy ≤ B(x) dy |x − y|t(n−1) y=x+rξ ξ∈ n−1 = κn d 0 rn−1dr rt(n−1) = κndn−t(n−1) n − t(n − 1) , where κn is the square of the unit sphere Sn−1 in Rn. Thus, condition a) is satisfied with M = κndn−t(n−1) n − t(n − 1) < ∞. 45 Let us check condition b): Ωm |Kj(x, y)|s dx ≤ Ωm dx |x − y|s(n−1) ≤ Ωm dx |x − y |s(n−1) , where y is the projection of point y onto the hyper–plane Πm (which contains Ωm : Ωm ⊂ Πm ). The integral is finite, since s(n − 1) < m. Consider the m–dimensional ball Bm(y ) = {x ∈ Πm : |x − y | ≤ d}. Clearly, Ωm ⊂ Bm(y ). Then Ωm |Kj(x, y)|s dx ≤ Bm(y ) dx |x − y |s(n−1) = κmdm−s(n−1) m − s(n − 1) . Hence, condition b) is satisfied with N = κmdm−s(n−1) m − s(n − 1) < ∞. The constant N depends on m and on d = diamΩ, but it does not depend on Ωm (it is one and the same for all sections Ωm of dimension m). Thus, conditions of Lemma 2(1) are satisfied and, therefore, Kj : Lp(Ω) → Lq(Ωm) is continuous. We want to prove that this operator is compact. For this, we want to find the operators Kjh satisfying conditions of Lemma 3. Let Ψ(r), r ∈ [0, ∞), be a smooth function such that Ψ ∈ C∞([0, ∞)), Ψ(r) = 0 if 0 ≤ r ≤ 1 2, Ψ(r) = 1 if r ≥ 1, and 0 ≤ Ψ(r) ≤ 1, ∀r. We put Ψh(r) = Ψ(r h ). Then Ψh(r) = 0 if 0 ≤ r ≤ h 2 , Ψh(r) = 1 if r ≥ h. Consider the kernels Kjh(x, y) = xj − yj |x − y|n Ψh(|x − y|). Obviously, |Kjh(x, y)| ≤ |Kj(x, y)| , ∀x, y. Hence, Kjh satisfy conditions a) and b) together with K with the same constants t, s, M and N. Clearly, Kjh(x, y) are bounded: |Kjh(x, y)| ≤ Ψh(|x − y|) |x − y|n−1 ≤ 1 h 2 n−1 = L(h). So, Kjh satisfy condition c). And, finally, Kjh is uniformly continuous in both variables. So, condition d) for Kjh is also satisfied. Let us check 46 condition (11): Ω |Kjh(x, y) − Kj(x, y)|t dy = |x−y| 1). 2) If 1 ≤ q < p, then we apply the result that Kj : Lp(Ω) → Lp(Ωm) is compact (i. e. , we apply 1) with p = q; condition 1 − n p + m p > 0 is true, since m > n − p). Since Ωm is a bounded domain, then Lp(Ωm) → Lq(Ωm) (if q < p), and any compact set in Lp(Ωm) is also compact in Lq(Ωm). It follows that the operator Kj : Lp(Ω) → Lq(Ωm) is compact. 3) Case p=1 Condition n − p < m ≤ n means that n − 1 < m ≤ n. Then m = n, so, now Ωm = Ω. Next, condition 1 − n p + m q > 0 means that 1 − n + n q > 0 ⇒ 1 ≤ q < n n−1. Let us check that condition b) with s = q is true: Ω |Kj(x, y)|q dx = Ω |xj − yj|q |x − y|nq dx ≤ Ω dx |x − y|(n−1)q . The integral is finite since (n − 1)q < n ⇒ Ω |Kj(x, y)|q dx ≤ κndn−q(n−1) n − q(n − 1) . 47 Also, condition (13) is satisfied: Ω |Kjh(x, y) − Kj(x, y)|q dx ≤ κnhn−q(n−1) n − q(n − 1) → 0 as h → 0. By Lemma 3(2), the operator Kj : L1(Ω) → Lq(Ω) is compact. 4) Case p ¿ n Let us check that conditions of Lemma 3(3) are satisfied. Indeed, the kernels Kj(x, y) (and Kjh(x, y) with it) satisfy condition a) with t = p : Ω |Kj(x, y)|p dy ≤ Ω dy |x − y|p (n−1) ≤ κndn−p (n−1) n − p (n − 1) < ∞. Since p > n, then 1 p < 1 n , 1 p = 1 − 1 p > 1 − 1 n = n−1 n . Hence n > p (n − 1). Next, Ω |Kjh(x, y) − Kj(x, y)|p dy ≤ |x−y|≤h dy |x − y|p (n−1) ≤ κnhn−p (n−1) n − p (n − 1) → 0 as h → 0. Thus, conditions of Lemma 3(3) are satisfied. It follows that the operator Kj : Lp(Ω) → C(Ω) is compact. Lemma 5 If 1 < p < n, n−p < m ≤ n, then the operator Kj : Lp(Ω) → Lq∗ (Ωm) is continuous (but not compact), where 1 − n p + m q∗ = 0 ⇔ q∗ = mp n − p . Without proof. 48 §2: Embedding theorems for W1 p (Ω) 1. The integral representation for functions in ◦ W1 p (Ω) Lemma 6 Let Ω ⊂ Rn be a bounded domain. Let u ∈ ◦ W1 p (Ω). Then u(x) = 1 κn n j=1 Ω xj − yj |x − y|n ∂u ∂yj dy, for a. e. x ∈ Ω. (14) Proof 1) First, assume that u ∈ C∞ 0 (Ω). Consider the fundamental solution of the Poisson equation E(z) = δ(z): E(z) = − 1 κn(n−2)|z|n−2 n > 2 1 κ2 ln |z| n = 2. Then for any u ∈ C∞ 0 (Ω) we have u(x) = Ω E(x − y)( u)(y)dy. The function E(z) has weak derivatives ∂E(z) ∂zj = 1 κn zj |z|n . Then, ∂ ∂yj E(x − y) = − 1 κn xj −yj |x−y|n , j = 1, . . . , n. By Definition 1 of weak derivatives, we have Ω E(x − y)( u)(y)dy = − n j=1 Ω ∂E(x − y) ∂yj ∂u ∂yj dy. Then, u(x) = 1 κn n j=1 Ω xj − yj |x − y|n ∂u ∂yj dy, ∀u ∈ C∞ 0 (Ω). 2) Now, let u ∈ ◦ W1 p (Ω), and let uk ∈ C∞ 0 (Ω), uk k→∞ −→ u in W1 p (Ω). For uk we have uk(x) = 1 κn n j=1 Ω xj − yj |x − y|n ∂uk ∂yj dy. 49 Thus, uk = 1 κn n j=1 Kj ∂uk ∂yj . By Lemma 4, each operator Kj is compact from Lp(Ω) to Lp(Ω). We know that uk k→∞ −→ u in Lp(Ω) and ∂uk ∂yj k→∞ −→ ∂u ∂yj in Lp(Ω). Then Kj(∂uk ∂yj ) k→∞ −→ Kj( ∂u ∂yj ) in Lp(Ω). (since Kj : Lp(Ω) → Lp(Ω) is continuous operator.) Hence, by the limit procedure as k → ∞ we obtain: u(x) = 1 κn n j=1 Ω xj − yj |x − y|n ∂u ∂yj dy, ∀u ∈ ◦ W1 p (Ω). 2. Embedding theorems for ◦ W1 p (Ω) Theorem 1 Let Ω ⊂ Rn be a bounded domain. 1) If 1 ≤ p ≤ n, m > n−p, q < ∞ and 1− n p + m q ≥ 0, then ◦ W1 p (Ω) is embedded into Lq(Ωm), where Ωm = Ω (if m = n ) and Ωm is any section of Ω by m–dimensional plane (if m < n). In the case 1 − n p + m q > 0, this embedding is compact. 2) If p > n, then ◦ W1 p (Ω) is compactly embedded into C(Ω). Comments 1) Let us distinguish the case m = n (Ωm = Ω): If 1 ≤ p ≤ n, q < ∞ and q ≤ np n−p = q∗, then ◦ W1 p (Ω) → Lq(Ω). If q < q∗, then this embedding is compact. 2) What does it mean that ◦ W1 p (Ω) → Lq(Ωm) in the case m < n ? A function u ∈ ◦ W1 p (Ω) is a measurable function in Ω; it can be changed on any set of measure zero; Ωm is a set of measure zero. First we consider u ∈ C∞ 0 (Ω), and put Tu = u|Ωm . Then T : C∞ 0 (Ω) → C∞ 0 (Ωm) is a linear operator. This linear operator can be extended by continuity to a continuous operator T : ◦ W1 p (Ω) → Lq(Ωm). We have the estimate Tu q,Ωm ≤ c u W 1 p (Ω) , ∀u ∈ C∞ 0 (Ω). 50 Let u ∈ ◦ W1 p (Ω). Then ∃ uk ∈ C∞ 0 (Ω), uk − u W 1 p (Ω) k→∞ −→ 0. Then Tuk − Tuj q,Ωm ≤ c uk − uj W 1 p (Ω) k,j→∞ −→ 0. Hence {Tuk} is a Cauchy sequence in Lq(Ωm). There exists limit Tuk k→∞ −→ w in Lq(Ωm). By definition, w = Tu. Proof of Theorem 1 By Dj we denote operators Dju = ∂u ∂xj . Then Dj : ◦ W1 p (Ω) → Lp(Ω) is continuous operator, j = 1, . . . , n. Then representation (14) can be written as u = 1 κn n j=1 KjDju. (15) 1) Suppose that 1 ≤ p ≤ n, m > n − p, q < ∞ and 1 − n p + m q > 0. Then conditions of Lemma 4(1) are satisfied. So, operator Kj : Lp(Ω) → Lq(Ωm) is compact. Hence, the embedding operator J = κ−1 n n j=1 KjDj : ◦ W1 p (Ω) → Lq(Ωm) is compact. (We use the fact that if A1 : B1 → B2 is continuous operator and A2 : B2 → B3 is compact operator, then A2A1 : B1 → B3 is compact. Here B1, B2, B3 are Banach spaces.) If p > 1 and 1 − n p + m q = 0 (i. e., q = q∗), then, by Lemma 5, the operator Kj : Lp(Ω) → Lq(Ωm) is continuous. Hence, the embedding operator J = κ−1 n n j=1 KjDj : ◦ W1 p (Ω) → Lq(Ωm) is continuous. For p = 1 – without proof. 2) Let p > n. Then, by Lemma 4(2), operators Kj : Lp(Ω) → C(Ω) are compact. Hence, the embedding operator J = κ−1 n n j=1 KjDj : ◦ W1 p (Ω) → C(Ω) is compact. 51 Remark 1) Under conditions of Theorem 1(1), we have the estimate u q,Ωm ≤ κ−1 n n j=1 Kj Lp(Ω)→Lq(Ωm) Dju Lp(Ω) ≤ c n j=1 ∂ju p,Ω ≤ c u W 1 p (Ω) , u ∈ ◦ W1 p (Ω). (16) 2) Under conditions of Theorem 1(2), we have u C(Ω) ≤ κ−1 n n j=1 Kj Lp(Ω)→C(Ω) ∂ju p,Ω ≤ c n j=1 ∂ju p,Ω ≤ c u W 1 p (Ω) , u ∈ ◦ W1 p (Ω). (17) Using the estimates from Lemma 4 it is easy to see that the constants in estimates (16), (17) depend only on diamΩ, n, m, p, q, but they do not depend on Ωm (they are one and the same for any section Ωm). 3. Embedding theorems for W1 p (Ω) Theorem 2 Let Ω ⊂ Rn be a bounded domain of class C1. Then both statements of Theorem 1 are true for W1 p (Ω). 1) If 1 ≤ p ≤ n, m > n − p, q < ∞ and 1 − n p + m q ≥ 0, then W1 p (Ω) is embedded into Lq(Ωm). In the case 1− n p + m q > 0, this embedding is compact. 2) If p > n, then W1 p (Ω) is compactly embedded into C(Ω). Proof Let ˜Ω ⊂ Rn be a bounded domain such that Ω ⊂ ˜Ω. (For example, ˜Ω is a ball of sufficiently large diameter.) By Theorem 11 (Chapter 1), there exists a linear continuous extension operator Π : W1 p (Ω) → ◦ W1 p (˜Ω). If u ∈ W1 p (Ω), then v = Πu ∈ ◦ W1 p (˜Ω), and v W 1 p (˜Ω) ≤ c u W 1 p (Ω) , c = Π . 52 1) Under conditions of part 1), by Theorem 1, ◦ W1 p (˜Ω) → Lq(˜Ωm); if 1 − n p + m q > 0, this embedding is compact. (Here Ωm ⊂ Πm, where Πm is m–dimensional plane, and ˜Ωm is the section of ˜Ω by the same Πm.) a) Let 1 − n p + m q > 0. Let   be some bounded set in W1 p (Ω). Then Π   = {v = Πu : u ∈   } is a bounded set in ◦ W1 p (˜Ω). Then, by Theorem 1, this set is compact in Lq(˜Ωm). Then   is compact in Lq(Ωm) (because functions in   are restrictions of functions in Π   back to Ω). Hence, W1 p (Ω) compactly embedded into Lq(Ωm). b) Let 1 − n p + m q = 0. In this case embedding ◦ W1 p (˜Ω) → Lq(˜Ωm) is continuous (but not compact). By similar arguments, we show that embedding W1 p (Ω) → Lq(Ωm) is also continuous. 2) Under condition p > n, by Theorem 1(2), ◦ W1 p (˜Ω) is compactly embedded into C(˜Ω). If   is a bounded set in W1 p (Ω), then Π   is bounded set in ◦ W1 p (˜Ω); Π   is compact in C(˜Ω). Hence,   is compact in C(Ω). Comments 1) Under conditions of Theorem2(1), let JΩ : W1 p (Ω) → Lq(Ωm) be the embedding operator and let J˜Ω : ◦ W1 p (˜Ω) → Lq(˜Ωm) be the embedding operator; Π : W1 p (Ω) → ◦ W1 p (˜Ω) is the extension operator; R : Lq(˜Ωm) → Lq(Ωm) is the restriction operator. Then JΩ = RJ˜ΩΠ. We have the estimate for all u ∈ W1 p (Ω). u q,Ωm = RJ˜ΩΠu q,Ω ≤ J˜ΩΠu q,˜Ωm ≤ J˜Ω ◦ W 1 p (˜Ω)→Lq(˜Ωm) =c1 Π W 1 p (Ω)→ ◦ W 1 p (˜Ω) =c2 u W 1 p (Ω) ⇒ u q,Ωm ≤ c u W 1 p (Ω) , ∀u ∈ W1 p (Ω). (18) Compare (18) with estimate (16): in the case u ∈ ◦ W1 p (Ω) we can estimate u q,Ωm by the norms of derivatives n j=1 ∂ju p,Ω. Now it is impossible. (It is clear for u = const = 0 : u q,Ωm = 0, but ∂ju ≡ 0.) 53 2) Similarly, under conditions of Theorem 2(2), we have the estimate u C(Ω) ≤ C u W 1 p (Ω) , ∀u ∈ W1 p (Ω). (19) The constants in estimates (18), (19) depend on Π and, so, on the properties of ∂Ω. (While constants in estimates (16), (17) depend only on diamΩ and on p, q, m, n.) Let us formulate the analog of Theorem 2 for unbounded domain. Theorem 3 Suppose that Ω ⊂ Rn is unbounded domain satisfying conditions of Theorem 12 (Chapter 1). Then 1) If p ≥ 1, m > n − p, p ≤ q < ∞ and 1 − n p + m q ≥ 0, then W1 p (Ω) → Lq(Ωm). 2) If p > n, then W1 p (Ω) → C(Ω). Remark 1) In Theorem 3 embeddings are continuous, but not compact. 2) In part 1) we have condition q ≥ p (we don’t need this condition in Theorem 2.). 3) If Ω is bounded and p > n, then 1) follows from 2). Now 1) does not follow from 2). 4. Comments. Examples. All conditions in Theorems 2, 3 are precise. 1) If 1 − n p + m q < 0, then W1 p (Ω) → Lq(Ωm). Example. Let Ω = {x ∈ Rn : |x| < 1}. Let u(x) = |x|λ with 1 − n p < λ < −m q . Then u ∈ W1 p (Ω), but u /∈ Lq(Ω). Indeed, | u| ≤ c |x|λ−1 , Ω | u|p dx ≤ c Ω |x|p(λ−1) dx = cκn 1 0 rn−1+p(λ−1) dr < ∞, since n − 1 + p(λ − 1) > −1 ⇔ λ > 1 − n p . 54 Also, Ω |u|pdx < ∞. However, Ωm |u(y)|q dy = Ωm |y|qλ dy = κm 1 0 rm−1+qλ dr = ∞, since m − 1 + qλ < −1(⇔ λ < − m q ). Here Ωm is a section of Ω by some m–dimensional plane Πm such that point 0 ∈ Πm. 2) For unbounded domains, if p < q, then W1 p (Ω) → Lq(Ω). Example Let Ω = {x ∈ Rn : |x| > 1} , u(x) = |x|λ. Let −n q < λ < −n p . Then u ∈ W1 p (Ω), but u /∈ Lq(Ω). Check yourself. 3) The ” critical exponent“ q∗ is defined by the relation 1 − n p + m q∗ = 0. (q∗ = mp n−p). Here p < n. We have q∗ > p, since m > n − p. W1 p (Ω) → Lq(Ωm) for q ≤ q∗, but not for q > q∗. If p ≥ n, then W1 p (Ω) → Lq(Ωm) for all q < ∞ (if Ω is bounded) and all p ≤ q < ∞ (if Ω is unbounded). If p > n, then W1 p (Ω) → C(Ω). But for p = n > 1, W1 n(Ω) → C(Ω) and even W1 n(Ω) → L∞(Ω) . (Here q∗ = ∞.) Example Let Ω = x ∈ Rn : |x| < 1 e . Consider u(x) = ln | ln |x||. Then u ∈ W1 n(Ω), but u /∈ L∞(Ω). Indeed, | u(x)| ≤ 1 |x|| ln |x||. Then Ω | u(x)|n dx ≤ Ω dx |x|n| ln |x||n = κn 1/e 0 rn−1dr rn| ln r|n = κn 1/e 0 dr r| ln r|n < ∞. Also, Ω |u(x)|ndx < ∞. Then u ∈ W1 n(Ω). 4) If p = n = 1, Ω = (a, b), then any function u ∈ W1 1 (Ω) is absolutely continuous. This follows from Theorem 5 (Chapter 1). 5) For unbounded domains embeddings from Theorem 3 are not compact. Example Let u ∈ C∞ 0 (Rn) and let x(k) be a sequence of points x(k) ∈ Rn such 55 that x(k) → ∞ as k → ∞. We put uk(x) = u(x − x(k)). Then the set {uk} is bounded in W1 p (Rn). Here p > n. (Obviously, uk W 1 p (  n) = u W 1 p (  n) = const.) But the set {uk} is not compact in C(Rn). Indeed, suppose that there exists a subsequence ukj such that ukj j→∞ −→ u0 in C(Rn). Since ukj j→∞ −→ 0 in C(Ω) for any bounded domain Ω (simply ukj ≡ 0 in Ω for sufficiently large j), then u0(x) ≡ 0. But ukj C(  n) = u C(  n) = 0. Contradication. Example Let u ∈ C∞ 0 (Rn), and vk(x) = k − n p u x k . Then vk ∈ W1 p (Rn) and {vk} is bounded in W1 p (Rn). But {vk} is not compact in Lp(Rn). Thus, the embedding W1 p (Rn) → Lp(Rn) is not compact. 6) For bounded domains Ω and q = q∗ embedding W1 p (Ω) → Lq∗ (Ω) is not compact. Example Ω = {x : |x| < 1}, u ∈ C∞ 0 (Rn), wk(x) = k n p −1 u(kx), p < n. Then {wk} is bounded in W1 p (Ω), but {wk} is not compact in Lq∗ (Ω). Check this yourself. 5. Embeddings on submanifolds Instead of the section of Ω by m–dimensional planes we can consider sections of Ω by some m–dimensional manifolds. Theorem 4 Let Ω ⊂ Rn be a bounded domain of class C1. Let 1 ≤ p ≤ n, m > n − p, 1 ≤ q < ∞ and 1 − n p + m q ≥ 0. Let Γ ⊂ Rn be a manifold of class C1, dimΓ = m. Let ΩΓ = Γ ∩ Ω. Then W1 p (Ω) → Lq(ΩΓ). If 1 − n p + m q > 0, then this embedding is compact. Without proof (The proof is based on Theorem 2 and using of covering Uj, diffemorphisms fj and patition of unity.) Important case Γ = ∂Ω (then also ΩΓ = ∂Ω). dimΓ = n − 1. Conditions: m = n−1 > n−p ⇒ p > 1, 1− n p + n−1 q ≥ 0 ⇔ q ≤ (n−1)p n−p = q∗. If q∗ < ∞ (1 < p < n), then W1 p (Ω) → Lq(∂Ω), ∀q ≤ q∗. For q < q∗ this embedding is compact. If n = p > 1, then q∗ = ∞, W1 n(Ω) → Lq(∂Ω), ∀q < ∞. 56 §3: Embedding theorems for Wl p(Ω) Theorem 5 Let Ω ⊂ Rn be a bounded domain of class C1. 1) If p ≥ 1, 1 ≤ q < ∞, 0 ≤ r < l, l − r − n p + n q ≥ 0, then Wl p(Ω) → Wr q (Ω). If l − r − n p + n q > 0, then this embedding is compact. 2) If p(l − r) > n, then Wl p(Ω) → Cr(Ω) and this embedding is compact. Proof 1) We put s = l − r and fix the numbers q0, q1, . . . , qs such that qj ≥ 1, q0 = p, qs = q and 1 − n qj + n qj+1 ≥ 0. Such numbers exist due to condition s− n p + n q ≥ 0. If l−r− n p + n q = 0, then q0, . . . , qs are defined uniquely from the equations 1 − n qj + n qj+1 = 0, j = 0, . . . , s − 1. If θ = s − n p + n q > 0, such numbers exist (but they are not unique). By Theorem 2(1), W1 qj (Ω) → Lqj+1 (Ω). It follows that Wl−j qj (Ω) → Wl−j−1 qj+1 (Ω). Indeed, let u ∈ Wl−j qj (Ω). Then ∂αu ∈ W1 qj (Ω) for |α| ≤ l − j − 1. Since W1 qj (Ω) → Lqj+1 (Ω), then ∂αu ∈ Lqj+1(Ω), |α| ≤ l − j − 1, and ∂α u qj+1,Ω ≤ c ∂α u W 1 qj (Ω) ≤ ˜c u W l−j qj (Ω) for all α with |α| ≤ l − j − 1. ⇒ u ∈ Wl−j−1 qj+1 (Ω) and u W l−j−1 qj+1 (Ω) ≤ c u W l−j qj (Ω) . We denote the embedding operator by Jj, Jj : Wl−j qj (Ω) → Wl−j−1 qj+1 (Ω), j = 0, 1, . . . , s − 1. Jj is a continuous operator. We have: Wl p(Ω) = Wl q0 (Ω) J0 → Wl−1 q1 (Ω) J1 → Wl−2 q2 (Ω) J3 → . . . Js−1 → Wl−s qs (Ω) = Wr q (Ω). ⇒ The embedding operator J : Wl p(Ω) → Wr q (Ω) is represented as J = Js−1 . . . J1J0. Each operator Jj is continuous , then J is also continuous. If θ > 0, then at least one of Jj is compact (at least for one index j we have 1 − n qj + n qj+1 > 0). In this case J is also compact. 57 2) Let p(l − r) > n ⇔ l − r − n p > 0 Case a) r = l − 1 l − l + 1 − n p > 0 ⇔ p > n. By Theorem 2(2), the embedding W1 p (Ω) → C(Ω) is compact. It follows that Wl p(Ω) → Cl−1(Ω) and this embedding is compact. (If u ∈ Wl p(Ω), then ∂αu ∈ W1 p (Ω) → C(Ω) for |α| ≤ l − 1.) Case b) r < l − 1 Then there exists a number q such that q > n and l−(r+1)−n p +n q > 0. (Indeed, l − r − n p =: ε > 0. We can find q > n such that 1 − n q < ε, i. e., n < q < n 1−ε .) Then we can represent the embedding operator J : Wl p(Ω) → Cr(Ω) as J = J2J1, where J1 : Wl p(Ω) → Wr+1 q (Ω) (J1 is compact by part 1) of Theorem 5) and J2 : Wr+1 q → Cr(Ω) (J2 is compact by case a), since q > n). Hence, J : Wl p(Ω) → Cr(Ω) is compact. Particular cases 1) Let r = 0, pl < n. The critical exponent q∗ is defined from the condition l− n p + n q∗ = 0 ⇔ q∗ = np n−lp . Since pl < n, then q∗ < ∞. Embedding Wl p(Ω) → Lq(Ω) is compact for q < q∗, and continuous for q = q∗. 2) If pl = n, then q∗ = ∞. In this case Wl p(Ω) → Lq(Ω) ∀q < ∞ (and this embedding is compact). But Wl p(Ω) → L∞(Ω). 3) If pl > n, then Wl p(Ω) → C(Ω) and this embedding is compact. 4) Let q = p, r < l. Then embedding Wl p(Ω) → Wr p (Ω) is compact. In particular, embedding Wl p(Ω) → Lp(Ω) (for l ≥ 1) is compact. Remarks 1) The embedding theorem for Ωm with m < n (W1 p (Ω) → Lq(Ωm)) can be also generalized for Wl p(Ω). However, for the proof we need another integral representation for u ∈ ◦ Wl p(Ω) (including derivatives of higher order). 2) The embedding theorems for Wl p(Ω) can be also generalized for the case of unbounded domains. 58 Equivalent norms in Sobolev spaces Wl p(Ω) (lecture by prof. M. Birman) 1. Finitedimensional linear spaces and norms in these spaces Let X be a linear space, dim X = N < ∞. It means that there exists a system of linear independent elements x1, . . . , xN ∈ X, such that any x ∈ X can be represented as a linear combination of x1, . . . , xN : x = N k=1 ξk xk, , ξk ∈ C , k = 1, . . . N. (1) There is a one–to–one correspondence of elements x ∈ X and coordinates ξ = ξk N k=1 . We denote x = N k=1 ξk 2 1/2 . Check yourself, that this functional has all properties of the norm. X is a Banach space with respect to this norm (i. e., the space X with this norm is complete). The mapping x → ξ is an isometric isomorphism of X and CN (with the standard norm). Proposition Any other norm x on X is equivalent to x . Therefore, all norms on X are equivalent to each other. Proof From (1) it follows that x ≤ N k=1 ξk xk ≤ N k=1 ξk 2 1/2 N k=1 xk 2 1/2 , i. e. , x ≤ γ x , γ = N k=1 xk 2 1/2 > 0. (2) Now, let us prove the opposite inequality. Let us check that the function x is continuous on X with respect to x . From (2) and from the triangle inequality it follows that x − x ≤ x − x ≤ γ x − x . Now we restrict the continuous function x to the unit sphere x = 1. Then x is a continuous function of ξ on the closed bounded set ξ ∈ CN : |ξ| = 1 in CN . Since x > 0, then by the Weierstrass Theorem, x ≥ β > 0 for x = 1. Then y = y y y ≥ β y , ∀ y ∈ X. Thus, x x , ∀ x ∈ X. (3) 59 2. ” Trivial“ equivalent norms in Wl p(Ω). The standard norm in Wl p(Ω), l ∈ N, 1 ≤ p < ∞, is u W l p(Ω) =   |α|≤l ∂α u p Lp(Ω)   1/p , Ω ⊆ Rd . (4) Let N be the number of all multi–indices α with |α| ≤ l. In CN we introduce the norm of lp–type by the formula η p  N = N s=1 |ηs |p , η ∈ CN . (5) Then we can rewrite (4) as u p W l p(Ω) = η p  n , (6) where η = ∂αu Lp(Ω) , |α| ≤ l. If we replace the norm (5) in relation (6) by any other (equivalent!) norm of vector η in CN , then (6) will automatically define some norm in Wl p(Ω), which is equivalent to the standard one. Such new norms in Wl p(Ω) are trivial. Example The norm u Lp(Ω) + max1≤|α|≤l ∂αu Lp(Ω) is equivalent to the standard norm in Wl p(Ω). Give yourself several examples of new ” trivial“ norms in Wl p(Ω). 3. The notion of seminorm. Definition A functional ϕ on a linear space X is called a seminorm on X, if 1) 0 ≤ ϕ(x) < ∞, ∀ x ∈ X, 2) ϕ(cx) = |c|ϕ(x), ∀ x ∈ X, ∀ c ∈ C, 3) ϕ(x1 + x2) ≤ ϕ(x1) + ϕ(x2). Thus, a seminorm ϕ has all properties of the norm besides one: from ϕ(x) = 0 it does not follow x = 0. Example X = Wl p(Ω), ϕ(x) = Ω u(x)dx . This functional is equal to zero for any u ∈ Wl p(Ω) with zero mean value. 60 4. General theorem about equivalent norms in Wl p(Ω). Assume that Ω ⊂ Rn is bounded and ∂Ω ∈ C1. By Pl we denote the class of all polynomials in Rn of order ≤ l − 1. Let ϕ be a seminorm on Wl p(Ω) which satisfies properties: 4) ϕ(u) ≤ c u W l p(Ω) (It means that ϕ is bounded, and, therefore, continuous in Wl p(Ω).) 5) If u ∈ Pl and ϕ(u) = 0, then u = 0 (ϕ is non–degenerate on the subspace Pl ⊂ Wl 2(Ω)). Theorem Let ϕ be a functional on Wl p(Ω) satisfying conditions 1) – 5). Then the functional —u—W l p(Ω) =   |α|=l ∂α u p Lp(Ω) + ϕ(u)p   1/p (7) defines the norm in Wl p(Ω), which is equivalent to the standard norm. Proof Obviously, functional (7) is homogeneous and satisfies the triangle inequality. Next, if —u—W l p(Ω) = 0, then ∂αu = 0 for ∀ α with |α| = l. Then it follows that u ∈ Pl. Besides, ϕ(u) = 0, and, by property 5), u=0. Thus, functional (7) is a norm on Wl p(Ω). Taking account of property 4), it suffices to check that u W l p(Ω) ≤ C—u—W l p(Ω), u ∈ Wl p(Ω). (8) Suppose the opposite. Then for any C > 0, (8) is not true. Then there exists a sequence {um}, um ∈ Wl p(Ω) such that m—um—W l p(Ω) ≤ um W l p(Ω) . (9) We put vm = um um W l p(Ω) . Then, by (9), vm W l p(Ω) = 1, (10) —vm—W l p(Ω) ≤ 1 m → 0 as m → ∞. (11) Since the embedding Wl p(Ω) → Wl−1 p (Ω) is compact it follows from (10) that there exists a subsequence vmj , which converges in Wl−1 p (Ω) to some v0 ∈ Wl−1 p (Ω): vmj − v0 W l−1 p (Ω) → 0 as j → ∞ (12) 61 From (11) it follows that ∂α vmj Lp(Ω) j→∞ −→ 0 for ∀ α with |α| = l. (13) Since the operator ∂α is closed in Lp(Ω), ∂αv0 = 0 for ∀ α with |α| = l. Then by (12) and (13), we have vmj W l p(Ω) −→ v0 as j → ∞, v0 ∈ Pl. (14) From (11) it follows that ϕ(vmj ) → 0 as j → ∞. By (14) and property 4), ϕ(vmj ) → ϕ(v0) as j → ∞. Thus, ϕ(v0) = 0, v0 ∈ Pl. By property 5), v0 = 0. Together with (14) this contradicts to (10). Mention that, in the proof of inequality (8), we did not use any explicit construction and we did not obtain any upper bound for the constant C. However, we have proved rather general theorem, which in particular cases implies a number of concrete inequalities (proved before by special tricks). Control question Where did we use that Ω is bounded and ∂Ω ∈ C1? 5. Examples. Additions. 1. Let l ≥ 2 and ϕ(u) = u Lp(Ω). Conditions 1) - 2) are obviously satisfies. By Theorem, the norm —u—W l p(Ω) =   |α|=l ∂α u p Lp(Ω) + u p Lp(Ω)   1/p (15) is equivalent to the standard one. It follows that ∂αu Lp(Ω), 0 < |α| < l, is estimated by the norm (15). Exercise In the case p = 2, l = 2, prove this estimate using Fourier transform. 2. Let l = 1, ω ⊆ Ω, ω is a measurable set such that mesnω > 0. Now Pl consists of constants. Let ϕ(u) = ω u(x)dx . Clearly, conditions 1) – 5) are satisfied. Then the Theorem implies that u p Lp(Ω) ≤ C Ω | u|p dx + ω u(x)dx p . For ω = Ω and p = 2 this is the classical Poincare inequality. 62 3. Let l = 1, Γ ⊂ ∂Ω, mesd−1Γ > 0. We put ϕ(u) = Γ udS . Properties 1) – 5) are satisfied. Condition 4) follows from the estimate ∂Ω |u|p dS ≤ C u p W l p(Ω) , i. e. , from the trace embedding theorem. By Theorem (on equivalent norms) we obtain u p Lp(Ω) ≤ C Ω | u|p dx + Γ udS p . This generalizes and strengthens the Friedrichs inequality Ω |u|2 dx ≤ C Ω | u|2 dx + ∂Ω |u|2 dS . 4. Let l = 2. P2 consists of linear functions, i. e. , of linear combinations of the basis funcrtions 1, x1, . . . , xn. Let ω ⊆ Ω be a measurable set, mesdω > 0. We put ϕ(u) = ω u(x)dx + n k=1 ω xk u(x)dx . (16) We have to check condition 5). Consider P2 as a finite–dimensional subspace in L2(ω). If ϕ(u) = 0, then u is orthogonal in L2(ω) to the basis in P2. Then, if u ∈ P2, it follows that u = 0. Thus, the norm (7) with l = 2 and such ϕ(u) is equivalent to the standard norm in W2 p (Ω). 5. Let l = 2, Γ ⊂ ∂Ω, mesd−1Γ > 0. We put ϕ(u) = Γ |u|dS. (17) Condition 4) follows from the trace embedding theorem. Let us check 5): ϕ(u) = 0 ⇔ u|Γ = 0. If u ∈ P2 (u(x) is a linear function), then condition u|Γ = 0 and u = 0 is equivalent to the fact that Γ is a plane part of the boundary, and u(x) = 0 is equation of this plane. In the case where Γ does not lie in some plane, from u ∈ P2, u|Γ = 0, it follows that u = 0. Then the norm (7) with l = 2 and ϕ(u) given by (17) is equivalent to the standard norm in W2 p (Ω). In particular, it is always so, if Γ = ∂Ω. 63 6. In conclusion, we discuss one example, which does not follow from Theorem. The norm —u—W l p(Ω) = n k=1 ∂lu ∂(xk)l p Lp(Ω) + u p Lp(Ω) 1/p is equivalent to the standard one. For example, in W2 2 (Rd) = H2(Rd), this fact follows from the inequality 2|ξjξk| ≤ |ξj|2 + |ξk|2. 64 Chapter 3: Sobolev spaces Hs (Rn ) §1: Classes S(Rn ) and S (Rn ). Fourier transform. Definition S(Rn) is a class of functions ϕ ∈ C∞(Rn) such that for any multi–index α and any k ∈ N, sup x∈  n (1 + |x|)k |∂α ϕ(x)| < ∞. S(Rn) is called the Schwartz class. For ϕ ∈ S(Rn) all derivatives ∂αϕ(x) are rapidly decreasing as |x| → ∞. We can introduce topology in S(Rn). Definition We say that ϕm m→∞ −→ ϕ in S(Rn), if sup x∈  n (1 + |x|)k |∂α ϕm(x) − ∂α ϕ(x)| m→∞ −→ 0, ∀α, ∀k. S(Rn) is a topological space, but not Banach space. Definition Let f ∈ S(Rn). We define the transformation F : f → f, f(ξ) = (2π) −n 2  n f(x)e−ixξ dx. F is called the Fourier transformation. It is known that f ∈ S(Rn), if f ∈ S(Rn). So, F : S(Rn) → S(Rn) is a linear operator. The inverse transformation F−1 is given by the formula f(x) = (2π) −n 2  n f(ξ)eixξ dξ, F−1 : S(Rn) → S(Rn). It is known that the Fourier transform F can be extended by continuity to L2(Rn), and F is unitary operator in L2(Rn): F : L2(Rn ) → L2(Rn ),  n |f(x)|2 dx =  n f(ξ) 2 dξ, f ∈ L2(Rn ). 65 Definition By S (Rn) we denote the dual space to S(Rn) (i. e., the space of linear continuous functionals on S(Rn)). Sometimes, S (Rn) is called the space of slowly increasing distributions. If v ∈ S (Rn), ϕ ∈ S(Rn), by v, ϕ we denote the meaning of functional v on function ϕ. The Fourier transformation is extended to the class S (Rn). Definition Let f ∈ S (Rn). A functional f ∈ S (Rn) is called the Fourier image of f, if f, ϕ := f, ϕ , ∀ϕ ∈ S(Rn ). It is known that F : S (Rn) onto −→ S (Rn), F−1 : S (Rn) → S (Rn). 66 §2: Spaces Hs (Rn ) 1. Definition of Hs (Rn ) We know that the spaces Wl 2(Ω) (l ∈ N) are Hilbert spaces : Wl 2(Ω) = Hl(Ω). Let Ω = Rn. We can use the Fourier transform and express the norm in Wl 2(Rn) = Hl(Rn) in terms of the Fourier image. Let u ∈ Hl(Rn). Consider the Fourier image u(ξ) = (2π) −n 2  n u(x)e−ixξ dx. Then u(x) = (2π) −n 2  n u(ξ)eixξ dξ. For the derivatives ∂αu(x), we have ∂αu(ξ) = (iξ)α u(ξ) = i|α| ξα u(ξ). Then u 2 Hl(  n) = |α|≤l  n |∂α u|2 dx =  n   |α|≤l |ξα |2   |u(ξ)|2 dξ. Since c1(1 + |ξ|2)l ≤ |α|≤l |ξα|2 ≤ c2(1 + |ξ|2)l (prove this!), then c1  n (1 + |ξ|2 )l |u(ξ)|2 dξ ≤ u 2 Hl(  n) ≤ c2  n (1 + |ξ|2 )l |u(ξ)|2 dξ. Thus, the norm  n (1 + |ξ|2)l |u(ξ)|2 dξ 1/2 is equivalent to the standard norm in Wl 2(Rn). We introduce the space with this norm; now we consider arbitrary l (not only l ∈ N). Definition Hs(Rn) = u ∈ S (Rn) :  n (1 + |ξ|2)s |u(ξ)|2 dξ < ∞ , s ∈ Rn. The inner product in Hs(Rn) is defined by (u, v)Hs(  n) =  n (1 + |ξ|2 )s u(ξ)v(ξ)dξ. Theorem 1 Hs(Rn) is the closure of C∞ 0 (Rn) with respect to the norm u Hs . 67 Proof 1) Let us show that any u ∈ Hs(Rn) can be approximated by functions in C∞ 0 (Rn). If u ∈ Hs(Rn), then u∗(ξ) = u(ξ)(1 + |ξ|2)s/2 ∈ L2(Rn). Since C∞ 0 (Rn) is dense in L2(Rn), there exists a sequence vk ∈ C∞ 0 (Rn) such that vk(ξ) k→∞ −→ u∗(ξ) in L2(Rn). We put vk(ξ)(1 + |ξ|2)−s/2 = wk(ξ). Then wk ∈ C∞ 0 (Rn) and wk(ξ)(1 + |ξ|2)s/2 → u∗(ξ) in L2(Rn). Obviously, wk ∈ S(Rn). We put uk = F−1wk. Then also uk ∈ S(Rn) and wk = uk. Since uk(ξ)(1 + |ξ|2)s/2 k→∞ −→ u∗(ξ) = u(ξ)(1 + |ξ|2)s/2 in L2(Rn), then uk k→∞ −→ u in Hs(Rn). It remains to approximate functions uk ∈ S(Rn) by functions ukj ∈ C∞ 0 (Rn) (in the Hs–norm). For this, we fix h ∈ C∞ 0 (Rn) such that h(x) = 1 for |x| ≤ 1. We put ukj(x) = uk(x)h x j . Then ukj ∈ C∞ 0 (Rn) and ukj − uk 2 Hs =  n |ukj(ξ) − uk(ξ)|2 1 + |ξ|2 s dξ l≥s,l∈  ≤  n |ukj(ξ) − uk(ξ)|2 1 + |ξ|2 l dξ = ukj − uk 2 Hl . For l ∈ N we can use another norm (which is equivalent to the standard one): ukj − uk 2 Hl ≤ c |α|≤l  n ∂α uk(x) 1 − h x j 2 dx ≤ ˜c |β|≤l |x|>j |∂α uk(x)|2 dx since h x j = 1 for |x| ≤ j. → 0 as j → ∞. It follows that ukj j→∞ −→ uk in Hs(Rn). 2) Let us show that each element of the closure of C∞ 0 (Rn) in Hs– norm belongs to Hs(Rn). Suppose that um ∈ C∞ 0 (Rn) and {um} is the Cauchy sequence in Hs(Rn), i. e. , um − ul Hs(  n) → 0 as m, l → ∞. It means that um(ξ)(1 + |ξ|2)s/2 =: u∗ m(ξ) is a fundamental sequence in L2(Rn). Since L2(Rn) is complete, there exists a limit u∗ m(ξ) m→∞ −→ u∗(ξ) in L2(Rn). We put w(ξ) = u∗(ξ)(1 + |ξ|2)−s/2. Then w ∈ S (Rn), and, therefore, F−1w = u ∈ S (Rn). We have: w(ξ) = u(ξ), u∗(ξ) = u(ξ)(1 + |ξ|2 )s/2 ∈ L2(Rn ), um(ξ)(1 + |ξ|2 )s/2 m→∞ −→ u(ξ)(1 + |ξ|2 )s/2 in L2(Rn ). 68 It means that um m→∞ −→ u in Hs(Rn). Thus, each element of the closure of C∞ 0 (Rn) in · Hs belongs to Hs(Rn). 2. Duality of Hs and H−s . Theorem 2 Let u ∈ Hs(Rn), v ∈ H−s(Rn), and let uj, vj ∈ C∞ 0 (Rn), uj j→∞ −→ u in Hs(Rn), vj j→∞ −→ v in H−s(Rn). Then there exists the limit lim j→∞  n uj(x)vj(x)dx. We denote this limit by  n u(x)v(x)dx. We have  n uvdx ≤ u Hs v H−s . Proof We have  n uj(x)vj(x)dx =  n uj(ξ)vj(ξ)dξ =  n uj(ξ)(1 + |ξ|2 )s/2 · vj(ξ)(1 + |ξ|2 )−s/2 dξ. (1) Since uj j→∞ −→ u in Hs(Rn), it follows that uj(ξ)(1 + |ξ|2 )s/2 j→∞ −→ u(ξ)(1 + |ξ|2 )s/2 =: (Asu) (ξ) in L2(Rn ). The fact that vj j→∞ −→ v in H−s(Rn) means that vj(ξ)(1 + |ξ|2 )−s/2 j→∞ −→ v(ξ)(1 + |ξ|2 )−s/2 =: (A−sv) (ξ) in L2(Rn ). Then, by (1), we have  n uj(x)vj(x)dx j→∞ −→  n (Asu) (ξ)(A−sv) (ξ)dξ =:  n u(x)v(x)dx. It is clear that the limit limj→∞  n ujvjdx does not depend on the choice of the sequences {uj} and {vj}. We have:  n uvdx =  n (Asu) (ξ)(A−sv) (ξ)dξ ≤ Asu L2(  n) A−sv L2(  n) = u Hs v H−s . 69 Theorem 3 If v ∈ H−s(Rn), then v H−s = sup 0=u∈Hs(  n)  n uvdx u Hs = sup 0=u∈C∞ 0 (  n)  n uvdx u Hs . (2) Proof 1) The mapping As : Hs(Rn) → L2(Rn), (Asu)(ξ) = u(ξ) 1 + |ξ|2 s/2 is a one–to–one isometric mapping. Indeed, Asu L2 = u Hs . The inverse mapping A−1 s : L2(Rn) → Hs(Rn) is defined as follows: for u∗ ∈ L2 consider w(ξ) = u∗(ξ) (1+|ξ|2)s/2 , and put u = F−1w. Then u(ξ) = w(ξ) and u∗(ξ) = u(ξ)(1 + |ξ|2)s/2 = (Asu)(ξ). Thus, A−1 s u∗ = F−1w = u. The mapping A−s : H−s → L2 is defined similarly. 2) Let v ∈ H−s and v∗(ξ) = (A−sv)(ξ). Then v∗ ∈ L2. It is known that in L2 we have v∗ L2 = sup 0=g∈L2  n g(ξ)v∗(ξ)dξ g L2 . We put u = A−1 s g. Then g(ξ) = (Asu)(ξ), g L2 = u Hs . If g runs over L2, then u runs over Hs. Thus, for v∗ = A−sv we have v H−s = v∗ L2 = sup 0=u∈Hs  n (Asu) (ξ)(A−sv) (ξ)dξ u Hs = sup 0=u∈Hs  n u(x)v(x)dx u Hs From Theorems 2 and 3 it follows that l(u) =  n uvdx is a linear continuous functional on u ∈ Hs(Rn) (if v ∈ H−s(Rn)) and the norm of this functional is equal to v H−s : l = sup 0=u∈Hs |l(u)| u Hs = sup 0=u∈Hs  n uvdx u Hs = v H−s . 70 Riesz Theorem Let H be a Hilbert space and l(u), u ∈ H, be a continuous linear functional on H. Then there exists such element v ∈ H that l(u) = (u, v)H. This element v is unique and l = v H . Proof 1) Let N = Ker l = {z ∈ H : l(z) = 0}. Then N is a closed subspace in H. Indeed, if zj ∈ N and zj j→∞ −→ z in H, then l(zj) j→∞ −→ l(z). Since l(zj) = 0, it follows that l(z) = 0, i. e. , z ∈ N. 2) If N = H, then l(u) = 0, ∀u ∈ H. In this case v = 0. If N = H, then N⊥ = {0} ( where N⊥ is the orthogonal complement of N). So, there exists v0 ∈ N⊥, v0 = 0. Then, l(v0) = 0. 3) For ∀u ∈ H consider u − l(u) l(v0)v0 ∈ N. Indeed, l u − l(u) l(v0) v0 = l(u) − l(u) l(v0)l(v0) = 0. Since v0 ∈ N⊥, we have u − l(u) l(v0) v0, v0 = 0 ⇒ (u, v0) = l(u) v0 2 l(v0) . Denote v = l(v0) v0 2 v0. Then l(u) = (u, v). 4) Uniqueness If (u, v) = (u, ˜v), ∀u ∈ H, then v − ˜v⊥H ⇒ v − ˜v = 0. 5) The norm of l. l = sup 0=u∈H |l(u)| u H = sup 0=u∈H |(u, v)| u H = v H . Indeed, |(u,v)| u H ≤ v H for ∀ 0 = u ∈ H, and for u = v we have |(u,v)| u H = v H. Let l(u) be a continuous linear functional on Hs(Rn). It means that l : Hs → C, a) l(c1u1 + c2u2) = c1l(u1) + c2l(u2), ∀ u1, u2 ∈ Hs, ∀ c1, c2 ∈ C, b) |l(u)| ≤ c u Hs , ∀u ∈ Hs(Rn). The norm l of a functional l is defined by the formula l = sup 0=u∈Hs |l(u)| u Hs . 71 Theorem 4 Let l(u) be a linear continuous functional on Hs(Rn). Then there exists unique element v ∈ H−s(Rn), such that l(u) =  n uvdx, ∀ u ∈ Hs (Rn ), (3) and l = v H−s . (4) Proof Consider the mapping As : Hs (Rn ) → L2(Rn ), (Asu) (ξ) = u∗(ξ) = u(ξ) 1 + |ξ|2 s/2 . Then u = A−1 s u∗. We define the functional ˜l(u∗) on L2(Rn) by the formula ˜l(u∗) = l(A−1 s u∗) = l(u). Then ˜l is a linear continuous functional on L2(Rn). By the Riesz theorem for the functional ˜l there exists unique function w ∈ L2(Rn) such that ˜l(u∗) =  n u∗(ξ)w(ξ)dξ, and ˜l = w L2 . Then l(u) = ˜l(u∗) =  n u(ξ) 1 + |ξ|2 s/2 w(ξ)dξ. We denote v(x) = F−1 w(ξ)(1 + |ξ|2)s/2 . Then v(ξ) = w(ξ)(1 + |ξ|2 )s/2 ;  n |v(ξ)|2 (1 + |ξ|2 )−s dξ =  n |w(ξ)|2 dξ. So, v ∈ H−s, and v H−s = w L2 . We have w(ξ) = (1 + |ξ|2)−s/2v(ξ) = (A−sv) (ξ), l(u) = ˜l(u∗) =  n (Asu) (ξ)(A−sv) (ξ)dξ =  n uvdx. For the norm of the functional l we have: l = sup 0=u∈Hs |l(u)| u Hs = sup 0=u∗∈L2 ˜l(u∗) u∗ L2 = ˜l = w L2 = v H−s . Remark Theorem 4 means that H−s(Rn) is dual to Hs(Rn) with respect to L2–duality. 72 3. Mollifications in Hs (Rn ) Let ωρ(x) = ρ−nω x ρ be a mollifier. Recall that ω ∈ C∞ 0 (Rn), ω(x) ≥ 0,  n ω(x)dx = 1. For u ∈ Hs(Rn) consider mollifications: uρ(x) = (ωρ ∗ u)(x), ρ > 0. Theorem 5 If u ∈ Hs(Rn), then uρ − u Hs → 0 as ρ → 0. Proof For the Fourier transform of the convolution uρ = ωρ ∗ u we have uρ(ξ) = (2π)n/2 ωρ(ξ)u(ξ). Next, ωρ(ξ) = (2π)−n/2  n ρ−n ω x ρ e−ixξ dx x ρ =y = (2π)−n/2  n ω(y)e−iyξρ dy = ω(ρξ). Since ω ∈ C∞ 0 (Rn), then ω(ξ) belongs to the Schwartz class S(Rn). Hence, a) |ω(ρξ)| ≤ c, ∀ξ ∈ Rn. b) limρ→0 ω(ρξ) = ω(0) = (2π)−n/2  n ω(y)dy =1 = (2π)−n/2. Let us estimate the norm uρ − u Hs . We have uρ(ξ) − u(ξ) = (2π)n/2 ω(ρξ) − 1 u(ξ); uρ − u 2 Hs =  n (1 + |ξ|2 )s |u(ξ)|2 (2π)n/2 ω(ρξ) − 1 2 →0 as ρ→0∀ ξ dξ. The function under the integral is estimated by C(1 + |ξ|2)s |u(ξ)|2 , which is summable since u ∈ Hs. By the Lebesgue Theorem, uρ − u Hs → 0 as ρ → 0. 73 4. Embedding Hs → Cr Theorem 6 Let s > r + n 2 . Then Hs(Rn) → Cr(Rn). Proof 1) Let u ∈ C∞ 0 (Rn). We have u(x) = (2π)−n/2  n u(ξ)eixξ dξ, ∂α u(x) = (2π)−n/2  n (iξ)α u(ξ)eixξ dξ, ∀α. Then, by the H¨older inequality, |∂α u(x)| ≤  n |ξα | |u(ξ)| dξ ≤  n |ξα |2 (1 + |ξ|2 )−s dξ 1/2  n |u(ξ)|2 (1 + |ξ|2 )s dξ 1/2 . If |α| ≤ r, and s − r > n 2 , then  n |ξα|2dξ (1+|ξ|2)s < ∞. Thus, max |α|≤r max x∈  n |∂α u(x)| ≤ C u Hs , u ∈ C∞ 0 (Rn ), i. e. , u Cr ≤ C u Hs , u ∈ C∞ 0 (Rn ). (5) 2) Let u ∈ Hs(Rn). Then there exists a sequence uj ∈ C∞ 0 (Rn), such that uj j→∞ −→ u in Hs. By (5), uj − ul Cr ≤ C uj − ul Hs j,l→∞ −→ 0. So, {uj} is the Cauchy sequence in Cr(Rn). There exists a limit ˜u ∈ Cr(Rn): uj − ˜u Cr j→∞ −→ 0. In fact, ˜u(x) = u(x), for a.e. x ∈ Rn (check this!). We identify ˜u = u. We have proved that Hs → Cr and u Cr ≤ C u Hs , ∀ u ∈ Hs . Remark Theorem 6 is generalization of the embedding theorem: Wl 2 → Cr if 2(l − r) > n. 74 5. Equivalent norm in Hs with fractional s > 0 Theorem 7 If 0 < s < 1, the norm u Hs is equivalent to the norm u Hs =  n |u|2 dx +  n  n |u(x) − u(y)|2dxdy |x − y|n+2s 1/2 . Proof Note that F : u(x) − u(x + z) → u(ξ)(1 − eizξ ). We have  n  n |u(x) − u(y)|2dxdy |x − y|n+2s y=x+z =  n  n |u(x) − u(x + z)|2dxdz |z|n+2s P arseval =  n  n |u(ξ)|2|1 − eizξ|2dξdz |z|n+2s =  n g(ξ)|u(ξ)|2 dξ, where g(ξ) =  n |1−eizξ|2dz |z|n+2s . The function g(ξ) is homogeneous in ξ of order 2s: g(tξ) =  n |1 − eitzξ|2dz |z|n+2s = t2s  n |1 − eitzξ|2d(tz) |tz|n+2s = t2s g(ξ), ∀ t > 0. The function g(ξ) depends only on |ξ|: g(ξ) =  n |1 − eiz1|ξ||2dz |z|n+2s , where the axis 0z1 has direction of vector ξ. It follows that g(ξ) = A|ξ|2s, A > 0. Then u Hs =  n |u(ξ)|2 (1 + A|ξ|2s )dξ 1/2 . Obviously, c1(1 + |ξ|2)s ≤ 1 + A|ξ|2s ≤ c2(1 + |ξ|2)s, ξ ∈ Rn. Then u Hs u Hs . Corollary If s > 0 , [s] = k , {s} > 0, then the norm u Hs =   |α|≤k  n |∂α u|2 dx + |α|=k  n  n |∂αu(x) − ∂αu(y)|2dxdy |x − y|n+2{s}   1/2 is equivalent to u Hs . 75 Proof u ∈ Hs ⇔ u ∈ Hk and ∂αu ∈ H{s} with |α| = k. It is easy to check that u 2 Hs check this! u 2 Hk + |α|=k ∂α u 2 H{s} by Theorem 7 |α|≤k  n |∂α u|2 dx + |α|=k  n  n |∂αu(x) − ∂αu(y)|2dxdy |x − y|n+2{s} . 6. ” ε–inequalities“ Obviously, Hs1 (Rn) → Hs2 (Rn) for s1 > s2. Proposition Let s1 < s < s2. Then for ∀ ε > 0 ∃ C(ε) > 0 such that u 2 Hs ≤ ε u 2 Hs2 + C(ε) u 2 Hs1 (6) Proof (6) is equivalent to the inequality 1 + |ξ|2 s ≤ ε 1 + |ξ|2 s2 + C(ε) 1 + |ξ|2 s1 ⇔ ρs ≤ ερs2 + C(ε)ρs1 , ρ ≥ 1. ⇔ 1 ≤ ερs2−s + C(ε)ρ−(s−s1) , ρ ≥ 1. We denote λ = ε 1 s2−s > 0, and put C(ε) = λ−(s−s1) = ε − s−s1 s2−s . Then ερs2−s + C(ε)ρ−(s−s1) = (λρ)s2−s + (λρ)−(s−s1) obviously ≥ 1 76 §3: Trace embedding theorems We write x ∈ Rn as x = (x , xn), x = (x1, . . . , xn−1). Consider the traces of functions on the hyper–plane xn = 0. We define the trace operator γ0 : C∞ 0 (Rn ) → C∞ 0 (Rn−1 ), (γ0u) (x ) = u(x , 0). Theorem 8 Let s > 1 2. Then the trace operator γ0 : C∞ 0 (Rn) → C∞ 0 (Rn−1) can be extended by continuity to the linear continuous operator γ0 : Hs(Rn) → Hs− 1 2 (Rn−1). We have γ0u Hs− 1 2 (  n−1) ≤ C u Hs(  n) . (7) Proof 1) Let u ∈ C∞ 0 (Rn). Then u(x) = u(x , xn) = (2π)−n/2  n u(ξ , ξn)eixnξn eix ξ dξ dξn; (γ0u) (x ) = u(x , 0) = (2π)− n−1 2  n−1 dξ eix ξ 1 √ 2π ∞ −∞ u(ξ , ξn)dξn = γ0u(ξ ) ⇒ γ0u(ξ ) = 1 √ 2π ∞ −∞ u(ξ , ξn)dξn. Then γ0u(ξ ) 2 ≤ ∞ −∞ |u(ξ)|2 (1 + |ξ|2 )s dξn   ∞ −∞ (1 + |ξ |2 =a2 +ξ2 n)−s dξn   . (8) Here the second integral is finite, since s > 1 2. We have ∞ −∞ dξn (a2 + ξ2 n)s = a a2s ∞ −∞ d ξn a 1 + ξn a 2 s = a1−2s ∞ −∞ dt (1 + t2)s =cs = csa1−2s ⇒ ∞ −∞ dξn (1 + |ξ |2 + ξ2 n)s = cs(1 + |ξ |2 ) 1 2 −s . (9) 77 Thus, from (8) and (9) it follows that 1 + |ξ |2 s− 1 2 γ0u(ξ ) 2 ≤ cs ∞ −∞ |u(ξ)|2 (1 + |ξ|2 )s dξn. Integrate over Rn−1:  n−1 1 + |ξ |2 s− 1 2 γ0u(ξ ) 2 dξ ≤ cs  n |u(ξ)|2 (1 + |ξ|2 )s dξ, i. e. γ0u 2 Hs− 1 2 (  n−1) ≤ cs u 2 Hs(  n) , u ∈ C∞ 0 . (10) 2) C∞ 0 (Rn) is dense in Hs(Rn). Let u ∈ Hs(Rn). Then ∃ {uj}, uj ∈ C∞ 0 (Rn), uj − u Hs(  n) j→∞ −→ 0. By (10), γ0uj − γ0ul 2 Hs− 1 2 (  n−1) ≤ cs uj − ul 2 Hs(  n) j,l→∞ −→ 0. So, {γ0uj} is a Cauchy sequence in Hs− 1 2 (Rn−1). Then there exists a limit: γ0uj j→∞ −→ v ∈ Hs− 1 2 (Rn−1 ) in Hs− 1 2 (Rn−1 ). By definition , v = γ0u. By the limit procedure, the estimate (10) is extended to all u ∈ Hs(Rn). Corollary Let k ∈ N and s > k + 1 2. Then the trace operators γj = γ0 ◦ ∂j xn : Hs(Rn) → Hs−j− 1 2 (Rn−1) are continuous for j = 0, 1, . . . , k. We have γju Hs−j− 1 2 (  n−1) ≤ c u Hs(  n) . Theorem 9 (extension theorem) Let k ∈ Z+, s > k + 1 2. Denote H s− 1 2 (Rn−1) = Hs− 1 2 (Rn−1)×Hs− 3 2 (Rn−1)×. . .×Hs−k− 1 2 (Rn−1). There exists a linear continuous operator P : H s− 1 2 (Rn−1 ) → Hs (Rn ), such that, if ϕ = (ϕ0, ϕ1, . . . , ϕk) ∈ H s− 1 2 (Rn−1), u = Pϕ (∈ Hs(Rn)), then ϕj = γju, j = 0, 1, . . . , k. We have u 2 Hs(  n) ≤ c ϕ 2 H s− 1 2 (  n−1) = c k j=0 ϕj 2 Hs−j− 1 2 (  n) . 78 Proof Let h ∈ C∞ 0 (R), h(t) = 1 for |t| ≤ 1, 0 ≤ h(t) ≤ 1. We put V (ξ , xn) = k j=0 1 j! xj nϕj(ξ )h xn 1 + |ξ |2 , ξ ∈ Rn−1 , xn ∈ R. Here ϕj(ξ ) is the Fourier image of ϕj(x ). Clearly, V (ξ , 0) = ϕ0(ξ ), ∂j xn V (ξ , 0) = ϕj(ξ ), j = 1, . . . , k. Let us show that V (ξ , xn) is the Fourier image of the function u(x , xn) such that u ∈ Hs(Rn). We put u(ξ , ξn) = V (ξ , ξn), where V (ξ , ξn) is the Fourier image (in one variable xn → ξn) of V (ξ , xn). Note that xj ng(xn) F → ij g(j) (ξn) = ij dj dξj n g(ξn), g(ρxn) F → 1 ρ g ξn ρ , xj nh(ρxn) F → ij 1 ρj+1 h(j) ξn ρ . Then u(ξ) = k j=0 ij j! ϕj(ξ )(1 + |ξ |2 )− j+1 2 h(j) ξn 1 + |ξ |2 . We have: u 2 Hs(  n) =  n |u(ξ)|2 (1 + |ξ|2 )s dξ ≤ C k j=0  n ϕj(ξ ) 2 (1 + |ξ |2 )−j−1 h(j) ξn 1 + |ξ |2 2 (1 + |ξ|2 )s dξ. We write the integral as  n−1 dξ   dξn . . ., and in the internal integral change variable: τ = ξn√ 1+|ξ |2 . Then 1 + |ξ|2 = 1 + |ξ |2 + ξ2 n = (1 + |ξ |2 )(1 + τ2 ); u 2 Hs(  n) ≤ C k j=0  n−1 dξ ϕj(ξ ) 2 (1+|ξ |2 )s−j− 1 2   h(j) (τ) 2 (1+τ2 )s dτ. Since h ∈ C∞ 0 (R), then h ∈ S(R) and, so,   h(j) (τ) 2 (1 + τ2 )s dτ = C(j, s) < ∞. 79 ⇒ u 2 Hs(  n) ≤ C k j=0  n−1 ϕj(ξ ) 2 (1 + |ξ |2 )s−j− 1 2 dξ . = c k j=0 ϕj 2 Hs−j− 1 2 (  n−1) . So, the operator P : ϕ = (ϕ0, ϕ1, . . . , ϕk) → u is a linear continuous operator from H s− 1 2 (Rn−1) to Hs(Rn) , and γju = ϕj, j = 0, . . . , k. 80 §4: Spaces Hs (Ω) (survey) 1. Definition of Hs (Ω) Let Ω ⊂ Rn be a domain. There are different ways of definition of the Sobolev spaces Hs(Ω). Approach I. Definition 1 Hs(Ω) is the class of restrictions to Ω of functions in Hs(Rn): u ∈ Hs (Ω) ⇔ ∃ v ∈ Hs (Rn ), v|Ω = u. Approach II. Case s ≥ 0. Definition 2 Hs(Ω) is the set of functions in L2(Ω), such that their weak derivatives up to order k = [s] also belong to L2(Ω), and the following norm is finite: u Hs < ∞, u 2 Hs def =    |α|≤s Ω |∂αu|2 dx, if s = [s] |α|≤k Ω |∂αu|2 dx + |α|=k Ω Ω |∂αu(x)−∂αu(y)|2dxdy |x−y|n+2{s} , if s = [s] = k, {s} = s − k. (11) Comments 1) If Ω ⊂ Rn is a bounded domain of Lipschitz class, then both definitions give the same space: Def 1 ⇔ Def 2. If Hs(Ω), s ≥ 0, is the Sobolev space in the sense of Def 2, there exists a linear continuous extension operator Π : Hs(Ω) → Hs(Rn). 2) The spaces Ws p (Ω) with fractional s ≥ 0 and p = 2 can be defined by analogy with Def 2 (with “2“ replaced by “p“). 3) The embedding theorems can be generalized for spaces of fractional order. Next, the space ◦ Hs(Ω) is defined. Definition 3 ◦ Hs(Ω) is the closure of C∞ 0 (Ω) with respect to the norm (11). 81 Definition 4 Let s > 0. Then, by definition, H−s(Ω) = ◦ Hs(Ω) ∗ , i. e. , H−s(Ω) is the space of linear continuous functionals on ◦ Hs(Ω) with the norm u H−s = sup 0=ϕ∈ ◦ Hs(Ω) | u, ϕ | ϕ Hs(Ω) . By analogy with Hs(Rn) and H−s(Rn), for u ∈ H−s(Ω), ϕ ∈ ◦ Hs(Ω), we denote u, ϕ = Ω u(x)ϕ(x)dx. Comments 1) ◦ Hs(Ω) = Hs(Ω) for s < 1 2. 2) Let u ∈ ◦ Hs(Ω), P0u(x) = u(x), x ∈ Ω 0, x ∈ Rn\Ω . Then P0 : ◦ Hs(Ω) → Hs(Rn) is continuous, if s = m + 1 2, m ∈ Z+. 3) Let Ω ⊂ Rn be a bounded domain with Lipschitz boundary. Then H−s(Ω) coincides with the space of restrictions to Ω of distributions ∈ H−s(Rn), if s = m + 1 2, m ∈ Z+. 4) Hs(Ω) is invariant with respect to diffeomorphisms of class Cl, l ≥ |s|, l ∈ N. 2. Trace embedding theorems Theorems 8 and 9 can be extended to the case of bounded domain Ω with smooth boundary. Let Ω ⊂ Rn be a bounded domain of class Cl. Then there exists a covering {Uj}j=1,...,N such that Ω ⊂ N j=1 Uj, either Uj ⊂ Ω, or Uj ∩ ∂Ω = ∅, then ∃ diffeomorphism fj : Uj → K, fj, f−1 j ∈ Cl, fj(Uj ∩ Ω) = K+, fj(Uj ∩ ∂Ω) = ∂K+\Σ+ = Γ. Suppose that domains U1, U2, . . . , UM are of second kind, and UM+1, . . . , UN are strictly interior. There exists a partition of unity {ζj}, such that ζj ∈ C∞ 0 (Rn), supp ζj ⊂ Uj, N j=1 ζj(x) = 1, x ∈ Ω. For x ∈ ∂Ω we have M j=1 ζj(x) = 1. (This is true in some neighbourhood of ∂Ω.) Let u ∈ Cl(∂Ω), uj(x) = u(x)ζj(x), vj = uj ◦ f−1 j . Then vj ∈ Cl(Γ), supp vj ⊂⊂ Γ. We extend vj by zero to Rn−1\Γ: ˜vj(y) = vj(y), y ∈ Γ 0, y ∈ Rn−1\Γ . 82 Then ˜vj ∈ Cl(Rn−1), supp ˜vj ⊂ Γ. Consider   M j=1 ˜vj 2 Hs(  n−1)   1/2 def = u Hs(∂Ω) , s ≤ l. (12) Definition Hs(∂Ω) is the closure of Cl(∂Ω) with respect to the norm (12). This norm depends on the choice of covering {Uj}, diffeomorphisms {fj}, and partition of unity {ζj}. It can be proved that all such norms (for different {Uj} , {fj} , {ζj}) are equivalent to each other. So, the class Hs(∂Ω) is well– defined. Theorem 10 (trace embedding theorem) Let Ω ⊂ Rn be a bounded domain of class Cl, l ∈ N. Let k ∈ Z+, s > k + 1 2, s ≤ l. Let γj : Cl(Ω) → Cl−j(∂Ω) be the trace operator: γju = ∂j u ∂νj ∂Ω , j = 0, . . . , k (where ∂j ∂νj are ” normal“ derivatives of u). Then the operator γj can be extended (uniquely) to linear continuous operator γj : Hs(Ω) → Hs−j− 1 2 (∂Ω), j = 0, 1, . . . , k. The proof is based on Theorem 8, and using covering {Uj}, diffeomorphisms {fj} and partition of unity {ζj}. Theorem 11 (extension theorem) Let Ω ⊂ Rn be a bounded domain of class Cl, s ≤ l, s > k + 1 2, where k ∈ Z+. We denote H s− 1 2 (∂Ω) = Hs− 1 2 (∂Ω) × Hs− 3 2 (∂Ω) × . . . × Hs−k− 1 2 (∂Ω). There exists a linear continuous operator PΩ : H s− 1 2 (∂Ω) → Hs (Ω) such that, if ϕ = (ϕ0, ϕ1, . . . , ϕk), ϕj ∈ Hs−j− 1 2 (∂Ω), and u = PΩϕ, then γju = ϕj, j = 0, 1, . . . , k, and u 2 Hs(Ω) ≤ c k j=0 ϕj 2 Hs−j− 1 2 (∂Ω) = c ϕ 2 H s− 1 2 (∂Ω) . Theorem 12 Let Ω ⊂ Rn be a bounded domain of class Cl, l ∈ N. Then u ∈ ◦ Hl(Ω) = ◦ Wl 2(Ω) if and only if u ∈ Hl(Ω) and γ0u = γ1u = . . . = γl−1u = 0. 83 Proof For simplicity we prove Theorem 12 in the case l = 1. u ∈ ◦ H1 (Ω) ⇔ u ∈ H1 (Ω) and γ0u = 0. “⇒“ Obvious. “⇐“ Using covering, diffeomorphisms and partition of unity, we reduce the question to the following. Let K+ = {x ∈ Rn : |x| < 1, xn > 0} be the half–ball. Suppose that u ∈ H1(K+), u(x) = 0 near Σ+, γ0u = u|Γ = 0. We have to prove that u ∈ ◦ H1(K+). We have the following representation for u(x): u(x , xn) = xn 0 ∂u ∂xn (x , t)dt, for a. e. (x , xn) ∈ K+. (13) We fix a cut–off function h(t) such that h ∈ C∞(R+), h(t) = 0, 0 ≤ t ≤ 1 2, h(t) = 1 for t ≥ 1, and 0 ≤ h(t) ≤ 1. We put hm(t) = h(mt), then hm(t) = 1 for t ≥ 1 m. Consider um(x) = u(x , xn)hm(xn). Then um(x) = 0 near ∂K+, um ∈ H1(K+). Let us check that um − u H1(K+) m→∞ −→ 0. We have: u(x , xn) − um(x , xn) = (1 − hm(xn))u(x , xn), ∂ ∂xj (u(x) − um(x)) = (1 − hm(xn)) ∂u(x) ∂xj , j = 1, . . . , n − 1, ∂ ∂xn (u(x) − um(x)) = (1 − hm(xn)) ∂u(x) ∂xn − ∂hm ∂xn u(x). Then K+ |u(x) − um(x)|2 dx = K+ |1 − hm(xn)|2 ≤1 |u(x)|2 dx ≤ K+∩{0 0. Then (um)ρ ∈ C∞ 0 (K+) for sufficiently small ρ and (um)ρ − um H1(K+) ρ→0 −→ 0. Thus, we can approximate function u(x) by functions (um)ρ ∈ C∞ 0 (K+) in H1(K+)–norm ⇒ u ∈ ◦ H1(K+). 85 §5: Application to elliptic boundary value problems 1. Dirichlet problem for the Poisson equation Let Ω ⊂ Rn be a bounded domain. Consider the classical Dirichlet problem: − u = F, x ∈ Ω u|∂Ω = g. (1) If Φ(x) is arbitrary function in Ω such that Φ|∂Ω = g, then the function v(x) = u(x) − Φ(x) is solution of the problem − v = f, x ∈ Ω v|∂Ω = 0 (2) where f(x) = F(x) + Φ(x). First, we’ll study problem (2) with homogeneous boundary condition. In the classical setting of problem (2), the boundary is sufficiently smooth, f ∈ C(Ω) and solution v ∈ C2(Ω). Now we want to define ” weak“ solution of problem (2) under wide conditions on ∂Ω and f. Let us formally multiply equation − v = f by the test function ϕ ∈ C∞ 0 (Ω) and integrate over Ω. Then v(x) satisfies the integral identity Ω v ϕdx = Ω fϕdx, ∀ϕ ∈ C∞ 0 (Ω). (3) The left–hand side is well–defined for any v ∈ ◦ H1(Ω) = ◦ W1 2 (Ω), ϕ ∈ ◦ H1(Ω); and the right–hand side is well–defined for f ∈ H−1(Ω), ϕ ∈ ◦ H1(Ω) (since H−1(Ω) is the dual space to ◦ H1(Ω) with respect to L2–duality). The boundary condition v|∂Ω = 0 we understand in the sense that v ∈ ◦ H1(Ω). Then we can consider arbitrary domains. Definition Let Ω ⊂ Rn be arbitrary bounded domain. A function v ∈ ◦ H1(Ω) is called a weak solution of the Dirichlet problem (2) with f ∈ H−1(Ω), if v satisfies the identity (3) for any ϕ ∈ ◦ H1(Ω). Theorem 1 Let Ω ⊂ Rn be a bounded domain. Then, for any f ∈ H−1(Ω), there exists unique (weak) solution v ∈ ◦ H1(Ω) of the Dirichlet problem (2). We have v H1(Ω) ≤ C f H−1(Ω). 86 Proof 1) The form [v, ϕ] := Ω v ϕdx, v, ϕ ∈ ◦ H1 (Ω), defines an inner product in the space ◦ H1(Ω). The corresponding norm [v, v]1/2 is equivalent to the standard norm v H1(Ω) = Ω |v|2 + | v|2 dx 1/2 . This follows from the Friedrichs inequality: Ω |v|2 dx ≤ CΩ Ω | v|2 dx, ∀v ∈ ◦ H1 (Ω) (here it is important that Ω is bounded). 2) The right–hand side of (3) is lf (ϕ) = Ω fϕdx. lf (ϕ) is antilinear continuous functional on ϕ ∈ ◦ H1(Ω): |lf (ϕ)| ≤ f H−1(Ω) ϕ H1(Ω) . We rewrite (3) in the following form: [v, ϕ] = lf (ϕ). (4) By the Riesz Theorem, for antilinear continuous functional lf on ◦ H1(Ω) there exists unique function v ∈ ◦ H1(Ω) such that lf (ϕ) = [v, ϕ], and the norm of this functional is equal to the norm of v. (Now we consider ◦ H1(Ω) as the Hilbert space with the inner product [·, ·].) Then, by the Riesz Theorem, lf = sup 0=ϕ∈ ◦ H1(Ω) |lf (ϕ)| [ϕ, ϕ]1/2 = [v, v]1/2 . (5) Thus, v is the unique solution of (4) (⇔ (3)). Since, by definition of the class H−1(Ω), f H−1(Ω) = sup 0=ϕ∈ ◦ H1(Ω) |lf (ϕ)| ϕ H1(Ω) , and ϕ H1(Ω) [ϕ, ϕ]1/2 , it follows from (5) that v H1(Ω) ≤ C f H−1(Ω) . 87 2. Now we return to the problem (1) with non–homogeneous boundary condition u|∂Ω = g. Suppose that Ω ⊂ Rn is a bounded domain of class C1. Then, by Theorem 10 (trace embedding theorem) the trace operator γ0 (γ0u = u|∂Ω) is continuous from H1(Ω) onto H1/2(∂Ω): γ0 : H1 (Ω) → H1/2 (∂Ω). Consider the problem − u = F, x ∈ Ω, γ0u = u|∂Ω = g(x), for given F ∈ H−1(Ω) and g ∈ H1/2(∂Ω). We look for solution u ∈ H1(Ω). Equation − u = F in Ω is understood in the sense of distributions: u(x) is a weak solution of (1), if u ∈ H1(Ω), u(x) satisfies the identity Ω u ϕdx = Ω Fϕdx, ∀ϕ ∈ C∞ 0 (Ω), and γ0u = g. Theorem 2 Let Ω ⊂ Rn be a bounded domain of class C1. Let F ∈ H−1(Ω), g ∈ H1/2(∂Ω). Then there exists unique weak solution u ∈ H1(Ω) of problem (1). We have u H1(Ω) ≤ C F H−1(Ω) + g H1/2(∂Ω) . (6) Proof 1) By Theorem 11 (extension theorem), for g ∈ H1/2(∂Ω), there exists extension G = PΩg ∈ H1(Ω) such that γ0G = g and G H1(Ω) ≤ C1 g H1/2(∂Ω) . (7) If u ∈ H1(Ω) and γ0u = g. Then v = u − G ∈ H1(Ω) and γ0v = 0. This is equivalent to the fact that v ∈ ◦ H1(Ω). Function v is solution of the problem − v = f v|∂Ω = 0, (8) where f = F + G. From G ∈ H1(Ω) it follows that G ∈ H−1(Ω) and G H−1(Ω) ≤ C2 G H1(Ω). Then f ∈ H−1(Ω) and f H−1 ≤ F H−1 + C2 G H1(Ω) ≤ F H−1 + C1C2 g H1/2 . 88 By Theorem 1, there exists unique solution v ∈ ◦ H1(Ω) of the problem (8), and v H1(Ω) ≤ C3 f H−1 . Then u = v + G is unique solution of the problem (1), and u H1 ≤ v H1 + G H1 ≤ C3 f H−1 + C1 g H1/2(∂Ω) ≤ C3 F H−1 + (C1C2C3 + C1) g H1/2(∂Ω) . 3. Dirichlet problem with spectral parameter Now we consider the problem − u = λu + f(x), x ∈ Ω u|∂Ω = 0, (9) with spectral parameter λ. Here Ω is bounded. Definition Let Ω ⊂ Rn be arbitrary bounded domain. Let f ∈ H−1(Ω). A function u ∈ ◦ H1(Ω) satisfying identity Ω u ϕdx = λ Ω uϕdx + Ω f(x)ϕdx, ∀ϕ ∈ ◦ H1 (Ω), (10) is called a weak solution of problem (9). As before, we denote [u, ϕ] = Ω u ϕdx. This is inner product in ◦ H1(Ω). The form Ω uϕdx, u, ϕ ∈ ◦ H1(Ω) is continuous sesquilinear form in ◦ H1(Ω). By the Riesz theorem for such forms it can be represented as [Au, ϕ], where A is a linear continuous operator in ◦ H1(Ω). Obviously, Ω uϕdx = Ω ϕudx , so [Au, ϕ] = [Aϕ, u] = [u, Aϕ], ∀u, ϕ ∈ ◦ H1(Ω). It follows that A = A∗. Next, [Au, u] = Ω |u|2dx > 0 if u = 0. So, A > 0. Lemma The operator A is compact operator in ◦ H1(Ω). Proof This follows from the embedding theorem: ◦ H1(Ω) is compactly embedded in L2(Ω). 89 We’ll use the following property of compact operators: T is a compact operator in the Hilbert space H, if and only if for any sequence {uk} which converges weakly in H, the sequence {Tuk} converges strongly in H. Let {uk} be a weakly convergent sequence in ◦ H1(Ω). Since the embedding operator J : ◦ H1(Ω) → L2(Ω) is compact, {uk} converges strongly in L2(Ω). We want to check that {Auk} converges strongly in L2(Ω). Since {uk} weakly converges in ◦ H1(Ω), it follows that uk H1(Ω) is uniformly bounded. A is a continuous operator; then also Auk H1(Ω) is uniformly bounded. We have [A(uk − ul), A(uk − ul)] = Ω (uk − ul)(Auk − Aul)dx ≤ uk − ul L2(Ω) →0 Auk − Aul L2(Ω) ≤C → 0 as k, l → ∞. {Auk} converges strongly in ◦ H1(Ω). It follows that A is compact operator. As before, the functional lf (ϕ) = Ω fϕdx (where f ∈ H−1(Ω)) is continuous antilinear functional on ϕ ∈ ◦ H1(Ω). By the Riesz Theorem, there exists unique element v ∈ ◦ H1(Ω) such that Ω fϕdx = [v, ϕ], ∀ϕ ∈ ◦ H1(Ω), and f H−1(Ω) v H1(Ω). Now, we can rewrite identity (10) in the form [u, ϕ] = λ [Au, ϕ] + [v, ϕ] , ∀ϕ ∈ ◦ H1 (Ω), (11) which is equivalent to the equation u − λAu = v, (12) where v ∈ ◦ H1(Ω) is given, and we are looking for solution u ∈ ◦ H1(Ω). Thus, we reduced the problem (9) to the abstract equation (12) with compact operator A in the Hilbert space ◦ H1(Ω). We analyse equation (12), using the properties of compact operators. The case v = 0 (which corresponds to f = 0): − u = λu x ∈ Ω u|∂Ω = 0 (13) 90 ⇔ u − λAu = 0 ⇔ Au = µu (where µ = 1 λ ) It is known that the spectrum of a compact operator is discrete: it consists of eigenvalues µj, j ∈ N, that may accumulate only to point µ = 0; each eigenvalue is of finite multiplicity (i. e. , dim ker(A−µjI) < ∞). In our case A = A∗ > 0, then all eigenvalues µj are positive: µj > 0. We enumerate eigenvalues in non–increasing order counting multiplicities µ1 ≥ µ2 ≥ . . . Then each eigenvalue corresponds to one eigenfunction uj : Auj = µjuj, j ∈ N. Eigenfunctions {uj} are linearly independent. We have : µj → 0 as j → ∞. Then for the eigenvalues λj = 1 µj of the Dirichlet problem (13) we have the following properties: 0 < λ1 ≤ λ2 ≤ . . ., λj → ∞ as j → ∞. Thus, we have the following theorem. Theorem 3 The spectrum of the Dirichlet problem (13) is discrete. There exists non–trivial solution only if λ = λj, j ∈ N. All eigenvalues are positive and have finite multiplicities. The only accumulation point is infinity: λj → ∞ as j → ∞. The case v = 0 (f = 0) − u = λu + f x ∈ Ω u|∂Ω = 0 (14) ⇔ u − λAu = v For compact operator A it is known that, if λ = λj(= 1 µj ), ∀j ∈ N, then the operator (I − λA)−1 is bounded. We can find unique solution u = (I − λA)−1 v, and u H1(Ω) ≤ (I − λA)−1 =Cλ v H1(Ω) . Since v H1(Ω) f H−1(Ω), we arrive at the following theorem. Theorem 4 If λ /∈ {λj}j∈  (λ is not eigenvalue), then for any f ∈ H−1(Ω) there exists unique (weak) solution u ∈ ◦ H1(Ω) of the problem (14), and u H1(Ω) ≤ Cλ f H−1(Ω) . 91 Now, suppose that λ = λj, and v = 0 (f = 0). Then, solution of the equation u − λjAu = v exists, if v satisfies the solvability condition: v⊥ker(I − λjA). It means that v is orthogonal (with respect to the inner product [·, ·]) in ◦ H1(Ω) to all eigenfunctions ϕ (k) j , k = 1, . . . , p, corresponding to the eigenvalue λj (here p is the multiplicity of λj). Since [v, ϕ] = Ω f(x)ϕ(x)dx, this solvability condition is equivalent to: Ω f(x)ϕ (k) j (x)dx = 0, k = 1, . . . , p. (15) The solution u(x) is not unique, but is defined up to a summand p j=1 cjϕ (k) j with arbitrary constants cj. Theorem 5 If λ = λj is eigenvalue of the Dirichlet problem, and ϕ (k) j , k = 1, . . . , p, are corresponding (linearly independent) eigenfunctions, then problem (14) has solution for any f ∈ H−1(Ω), which satisfies the solvability conditions (15). Solution is not unique and is represented as u = u0 + p j=1 cjϕ (k) j , where u0 is a fixed solution, and cj are arbitrary constants. 4. Hilbert–Schmidt Theorem Finally, we can apply the Hilbert–Schmidt Theorem for compact operators and obtain the following result. Let λ1 ≤ λ2 ≤ . . . be eigenvalues of the Dirichlet problem. Here we repeat each λj according to its multiplicity. There exists an orthogonal system of eigenfunctions {ϕj}j∈  : ϕj − λjAϕj = 0, j ∈ N, [ϕj, ϕl] = 0, j = l. By the Hilbert–Schmidt Theorem, {ϕj}j∈  is orthogonal basis in ◦ H1(Ω), i. e. , for any F ∈ ◦ H1(Ω), F = ∞ j=1 [F, ϕj] [ϕj, ϕj] ϕj It is important that ϕj⊥ϕl also in L2(Ω). Indeed, [Au, ϕ] = Ω uϕdx (by Definition of operator A). Next, Aϕj = µjϕj (where µj = 1 λj ). Thus, [Aϕj, ϕl] = Ω ϕjϕldx = µj [ϕj, ϕl] = 0, j = l. 92 We have [F, ϕj] = λj [F, Aϕj] = λj Ω Fϕjdx, [ϕj, ϕj] = λj [Aϕj, ϕj] = λj Ω |ϕj|2 dx. Then [F, ϕj] [ϕj, ϕj] = Ω Fϕjdx Ω |ϕj|2dx = (F, ϕj)L2(Ω) ϕj 2 L2(Ω) , and F = ∞ j=1 (F, ϕj)L2(Ω) ϕj 2 L2(Ω) ϕj. The last fomula can be extended to all F ∈ L2(Ω). Theorem 6 Let Ω ⊂ Rn be a bounded domain. Then there exists an ortogonal system of eigenfunctions {ϕj}j∈  of the Dirichlet problem. This system forms an orthogonal basis in L2(Ω) and in ◦ H1(Ω) (with respect to the inner product [·, ·]). 93