Part I, Chapter 2 Weak derivatives and Sobolev spaces
We investigate in this chapter the notion of differentiation for Lebesgue inte- grable functions. We introduce an extension of the classical concept of deriva- tive and partial derivative which is called weak derivative. This notion will be used throughout the book. It is particularly useful when one tries to differen- tiate finite element functions that are continuous and piecewise polynomial. In that case one does not need to bother about the points where the classical derivative is multivalued to define the weak derivative. We also introduce the concept of Sobolev spaces. These spaces are useful to study the well-posedness of partial differential equations and their approximation using finite elements.
2.1 Differentiation
We study here the concept of differentiation for Lebesgue integrable functions.
2.1.1 Lebesgue Points Theorem 2.1 (Lebesgue points). Let f L1(D). Let B(x,h) be the ball of radius h> 0 centered at x D. The following∈ holds true for a.e. x D: ∈ ∈ 1 lim f(y) f(x) dy =0. (2.1) h 0 B(x,h) | − | ↓ | | ZB(x,h) Points x D where (2.1) holds true are called Lebesgue points of f. ∈ Proof. See, e.g., Rudin [162, Thm. 7.6]. This result says that for a.e. x D, the averages of f( ) f(x) are small over small balls centered at x∈, i.e., f does not oscillate| · too− much| in the neighborhood of x. Notice that if the function f is continuous at x, then x is a Lebesgue point of f (recall that a continuous function is uniformly continuous over compact sets). 14 Chapter 2. Weak derivatives and Sobolev spaces
Let R be is a countable set with 0 as unique accumulation point (the sign ofH⊂ the members of is unspecified). Let F : R R. We say that F is H F (x+h) F→(x) strongly differentiable at x if the sequence ( − )h converges. h ∈H Theorem 2.2 (Lebesgue’s differentiation). Let f L1(R). Let F (x) := x f(t) dt. Then F is strongly differentiable at every Lebesgue∈ point x of f, −∞ and at these points we have F ′(x)= f(x). R Proof. See Exercise 2.2. ⊓⊔
In the above theorem we have F ′(x) = f(x) for a.e. x in R. Thus it is tempting to move away from the classical sense of differentiation and view 1 1 F ′ as a function in L (R). If we could make sense of F ′ in L (R), then x F (x) = F ′(t) dt would be an extension of the fundamental theorem of −∞ 1 calculus in Lebesgue spaces. As an example of this possibility, let f := [0, ) be the HeavisideR function (i.e., f(x) := 1 if x 0 and f(x) := 0 otherwise).∞ 1 1 ≥ Notice that f L (R) but f Lloc(R) (see Definition 1.25), and F (x) := x f(t) dt is well6∈ defined. Then∈ F (x) = 0 if x < 0 and F (x) = x if x > 0 (notice−∞ that 0 is not a Lebesgue point of f; see Exercise 2.1). We would like R 1 to say that F ′ = f in Lloc(R). The objective of the rest of this section is to make sense of the above argument.
2.1.2 Weak derivatives Definition 2.3 (Weak derivative). Let D be an open set in Rd. Let u, v 1 ∈ Lloc(D). Let i 1:d . We say that v is the weak partial derivative of u in the direction i ∈{if }
u∂ ϕ dx = vϕ dx, ϕ C∞(D), (2.2) i − ∀ ∈ 0 ZD ZD d and we write ∂iu := v. Let α N be a multi-index. We say that v is the ∈ α1 αd weak α-th partial derivative of u and we write ∂1 . . . ∂d u := v if
α1 αd α u∂ . . . ∂ ϕ dx = ( 1)| | vϕ dx, ϕ C∞(D), (2.3) 1 d − ∀ ∈ 0 ZD ZD α α1 αd where α := α1 + ... + αd. Finally we write ∂ u := ∂1 . . . ∂d u, and we set ∂(0,...,0)| u|:= u.
1 Lemma 2.4 (Uniqueness). Let u Lloc(D). If u has a weak α-th partial derivative, then it is uniquely defined.∈
Proof. Let v , v L1 (D) be two weak α-th derivatives of u. We have 1 2 ∈ loc
α α v ϕ dx = ( 1)| | u∂ ϕ dx = v ϕ dx, ϕ C∞(D). 1 − 2 ∀ ∈ 0 ZD ZD ZD Part I. Elements of functional analysis 15
Hence D(v1 v2)ϕ dx = 0. The vanishing integral theorem (Theorem 1.28) implies that v− = v a.e. in D. R 1 2 α If u C| |(D), then the usual and the weak α-th partial derivatives are identical.∈ Moreover it can be shown that if α, β Nd are multi-indices such ∈ that αi βi for all i 1:d , then if the α-th weak derivative of u exists in 1 ≥ ∈{ } Lloc(D), so does the β-th weak derivative. For instance, with d = 1 (writing 1 ∂x instead of ∂1), if ∂xxu exists in Lloc(D), so does ∂xu; see Exercise 2.4. Example 2.5 (1D). Let us revisit the heuristic argument at the end of 2.1.1. Let D := ( 1, 1). (i) Let us first consider a continuous function u§ C0(D; R), e.g.,−u(x) := 0 if x < 0 and u(x) := x otherwise. Then u has∈ a weak derivative. Indeed let v L1(D) be s.t. v(x) := 0 if x < 0 and ∈ v(x) := 1 otherwise. Let ϕ C∞(D). We have ∈ 0 1 1 1 1 u∂xϕ dx = x∂xϕ dx = ϕ dx = vϕ dx. 1 0 − 0 − 1 Z− Z Z Z−
Hence v is the weak derivative of u. (Notice that v2 defined by v2(x) := 0 if 1 x< 0, v2(0) := 2 and v2(x) := 1 if x> 0 is also a weak derivative of u, but v = v2 a.e. in D, i.e., v and v2 coincide in the Lebesgue sense.) (ii) Let us now consider a function u L1(D; R) that is piecewise smooth but exhibits a jump at x = 0, e.g., u(x)∈:= 1 if x< 0 and u(x) := x otherwise. Then u does not have a weak derivative.− Let us prove this statement by contradiction. Assume that there is v L1 (D) s.t. ∂ u = v. We have ∈ loc x 1 1 0 1 1 vϕ dx = u∂xϕ dx = ∂xϕ dx x∂xϕ dx = ϕ(0) + ϕ dx. 1 − 1 1 − 0 0 Z− Z− Z− Z Z
Let ϕn n N be a sequence of functions in C0∞(D) s.t. 0 ϕn(x) 1 for all { } ∈ ≤ ≤ x D, ϕn(0) = 1, and ϕn 0 a.e. in D. Lebesgue’s dominated convergence ∈ → 1 1 theorem implies that 1 = limn ( 1 vϕn dx 0 ϕn dx) = 0, which is a contradiction. →∞ − − R R ⊓⊔ p Lemma 2.6 (Passing to the limit). Let vn n N be a sequence in L (D), { } α∈ p p [1, ], with weak α-th partial derivatives ∂ vn n N in L (D). Assume ∈ ∞ p α p{ } ∈ that vn v in L (D) and ∂ vn gα in L (D). Then v has a weak α-th → α → partial derivative and ∂ v = gα. α Proof. The assumptions imply that limn ∂ vnϕ dx = gαϕ dx and →∞ D D R R α α α α α lim ∂ vnϕ dx = ( 1)| | lim vn∂ ϕ dx = ( 1)| | v∂ ϕ dx, n − n − →∞ ZD →∞ ZD ZD for all ϕ C∞(D). The conclusion follows readily. ∈ 0 ⊓⊔ A function v L1 (D) is said to be locally Lipschitz in D if for all x ∈ loc ∈ D, there is a neighborhood x of x in D and a constant Lx such that N v(z) v(y) Lx z y 2 d for all y, z x. | − |≤ k − kℓ (R ) ∈ N 16 Chapter 2. Weak derivatives and Sobolev spaces
Theorem 2.7 (Rademacher). Let D be an open set in Rd. Let f be a locally Lipschitz function in D. Then f is differentiable in the classical sense a.e. in D. The function f is also weakly differentiable, and the classical and weak derivatives of f coincide a.e. in D.
Proof. See [95, p. 280], [132, p. 44]. ⊓⊔
2.2 Sobolev spaces
In this section we introduce integer-order and fractional-order Sobolev spaces. The scale of Sobolev spaces plays a central role in the finite element error analysis to quantify the decay rate of the approximation error.
2.2.1 Integer-order spaces Definition 2.8 (W m,p(D)). Let m N and p [1, ]. Let D be an open set in Rd. We define the Sobolev space ∈ ∈ ∞
W m,p(D) := v L1 (D) ∂αv Lp(D), α Nd s.t. α m , (2.4) { ∈ loc | ∈ ∀ ∈ | |≤ } where the derivatives are weak partial derivatives. We write W m,p(D; Rq), q 1, for the space composed of Rq-valued functions whose components are all≥ in W m,p(D), and we write W m,p(D) whenever q = d.
Whenever it is possible to identify a length scale ℓD associated with D, m,p e.g., its diameter ℓD := diam(D) if D is bounded, we equip W (D) with the following norm and seminorm: If p [1, ), we set ∈ ∞ p α p α p p α p v := ℓ| | ∂ v , v := ∂ v , k kW m,p(D) D k kLp(D) | |W m,p(D) k kLp(D) α m α =m | X|≤ | X| and if p = , we set ∞ α α α v W m,∞(D) := max ℓD| | ∂ v L∞(D), v W m,∞(D) := max ∂ v L∞(D), k k α m k k | | α =m k k | |≤ | | where the sums and the maxima run over multi-indices α Nd. The advan- ∈ tage of using the factor ℓD is that all the terms in the sums or maxima have the same dimension (note that m,p and m,p have a different k·kW (D) |·|W (D) scaling w.r.t. ℓD). If there is no length scale available or if one works with dimensionless space variables, one sets ℓD := 1 in the above definitions.
m,p Proposition 2.9 (Banach space). W (D) equipped with the W m,p(D)- norm is a Banach space. For p =2, the space k·k
Hm(D) := W m,2(D) (2.5) Part I. Elements of functional analysis 17 is a real Hilbert space when equipped with the inner product (v, w)Hm (D) := α α α m D ∂ v ∂ w dx. The same result holds true for the complex space | |≤ m α α H (D; C) with (v, w)Hm (D;C) := α m D ∂ v ∂ w dx. P R | |≤ Proof. We are going to do the proofP for mR = 1. See e.g., [3, Thm. 3.3], [95, p. 249], or [181, Lem. 5.2] for the general case. Let vn n N be a Cauchy 1,p { } ∈ p sequence in W (D). Then vn n N is a Cauchy sequence in L (D) and the { } ∈ sequences of weak partial derivatives ∂ivn n N are also Cauchy sequences p p { } ∈ p in L (D). Hence there is v L (D) and there are g1,...,gd L (D) such p ∈ p ∈ that vn v in L (D) and ∂ivn gi in L (D). We conclude by invoking Lemma 2.6.→ → ⊓⊔ Example 2.10 (H1(D)). Taking m := 1 and p := 2 we have
H1(D) := v L2(D) ∂ v L2(D), i 1:d , { ∈ | i ∈ ∀ ∈{ }} (notice that L2(D) L1 (D)) and ⊂ loc 1 2 2 2 2 2 2 v 1 := v 2 + ℓ v 1 , v 1 := ∂ v 2 . k kH (D) k kL (D) D| |H (D) | |H (D) k i kL (D) i 1: d ∈{X } Let v be the column vector in Rd whose components are the directional ∇ 1 weak derivatives ∂iv of v. Then a more compact notation is H (D) := v 2 2 { ∈ L (D) v L (D) and v 1 := v 2 . | ∇ ∈ } | |H (D) k∇ kL (D) ⊓⊔ Lemma 2.11 (Kernel of ). Let D be open and connected set in Rd. Let v W 1,p(D), p [1, ]. Then∇ v = 0 a.e. on D iff v is constant. ∈ ∈ ∞ ∇ Proof. We prove the result for D := ( 1, 1) and we refer the reader to [132, p. 24], [181, Lem. 6.4] for the general case.− Let u L1 (D) be such that ∂ u = ∈ loc x 0. Fix a function ρ C∞(D) such that ρ dx = 1 and set c := uρ dx. ∈ 0 D ρ D Let now ϕ C0∞(D) and set cϕ := D ϕ dx. Then the function ψ(x) := x ∈ R R (ϕ(y) c ρ(y)) dy is by construction in C∞(D), and we have ∂ ψ(x) = 1 − ϕ R 0 x ϕ−(x) c ρ(x). Since u∂ ψ dx = (∂ u)ψ dx = 0 by assumption on R ϕ D x D x ∂ u, we− infer that − x R R
uϕ dx = u(∂xψ + cϕρ) dx = cϕ uρ dx = cρ ϕ dx, ZD ZD ZD ZD for all ϕ C∞(D). Theorem 1.28 shows that u = c . ∈ 0 ρ ⊓⊔ Remark 2.12 (Lipschitz functions). Let D be an open set in Rd. The space of Lipschitz functions C0,1(D) is closely related to the Sobolev space 1, 0,1 1, W ∞(D). Indeed C (D) L∞(D) is continuously embedded into W ∞(D). 1, ∩ Conversely, if v W ∞(D), then v(y) v(z) d (y, z) v ∞ for all ∈ | − |≤ D k∇ kL (D) y, z D, where dD(y, z) denotes the geodesic distance of y to z in D, i.e., the shortest∈ length of a smooth path connecting y to z in D (if D is convex, 18 Chapter 2. Weak derivatives and Sobolev spaces
d dD(y, z) = y z ℓ2 ); see [181, Lem. 7.8]. A set D R is said to be quasiconvexk if there− k exists C 1 s.t. every pair of points⊂ x, y D can be ≥ ∈ joined by a curve γ in D with length(γ) C x y ℓ2 . If D is a quasiconvex 1, 0,1 ≤ k − k open set, then W ∞(D)= C (D) L∞(D), and if D is also bounded, then 1, 0,1 ∩ W ∞(D)= C (D); see Heinonen [109, Thm. 4.1]. ⊓⊔ Remark 2.13 (Broken seminorms). Let D Rd be an open set and let ⊂ Di i 1: I be a partition of D, i.e., all the subsets Di are open, mutually { } ∈{ } 1,p disjoint, and D i 1: I Di has zero Lebesgue measure. Let v W (D) \ ∈{ } p ∈ p and p [1, ). Then one can write v 1,p = ( v) p . W (D) i 1: I Di L (Di) ∈ ∞ S | | ∈{ } k ∇ |p k In this book we are going to abuse the notation by writing v W 1,p(D) = p P | | i 1: I v Lp(D ). This abuse is justified by observing that ( v) Di = ∈{ } k∇ k i ∇ | (v ) for all v W 1,1(D). We stress that it is important that the P Di loc weak∇ | derivative of ∈v exists to make sense of the above identities. For in- p stance, letting H be the Heaviside function, we have (H ( 1,0)) Lp + | − ( 1,0) p 1,p k∇ k − (H (0,1)) Lp = 0, but H W (D); see Exercise 2.8. k∇ | k (0,1) 6∈ ⊓⊔ 2.2.2 Fractional-order spaces Definition 2.14 (W s,p(D)). Let s (0, 1) and p [1, ]. Let D be an open d s,p ∈ p ∈ ∞ set in R . We define W (D) := v L (D) v s,p < , where ∈ | | |W (D) ∞ 1 v(x) v(y) p p s,p v W (D) := | − sp+d| dx dy , p< , (2.6) | | D D x y ∞ Z Z k − kℓ2 v(x) v(y) and v W s,∞(D) := esssupx,y D | x −y s | . Letting now s> 1, we define | | ∈ k − kℓ2 W s,p(D) := v W m,p(D) ∂αv W σ,p(D), α, α = m , (2.7) { ∈ | ∈ ∀ | | } where m := s and σ := s m. Finally we denote Hs(D) := W s,2(D). We write W s,p⌊ (⌋D; Rq), q 1,− for the space composed of Rq-valued functions whose components are all in≥ W s,p(D), and we write W s,p(D) whenever q = d. Definition 2.15 (Sobolev–Slobodeckij norm). Let s = m + σ with m := s and σ := s m (0, 1). For all v W s,p(D), we set, ⌊ ⌋ p − ∈ p sp p ∈ for p [1, ), v W s,p(D) := v W m,p(D) + ℓD v W s,p(D) with seminorm p ∈ ∞ k k α p k k | | v W s,p(D) := α =m ∂ v W σ,p(D), and we set | | | | | | P s v s,∞ := max( v m,∞ ,ℓ v s,∞ ), k kW (D) k kW (D) D| |W (D) α with seminorm v W s,∞(D) := max α =m ∂ v W σ,∞(D). Equipped with this norm, W s,p(D) is| | a Banach space (and| | a| Hilbert| space if p =2). Example 2.16 (Power functions). Let D := (0, 1) and consider the func- α 2 1 tion v(x) := x with α R. One can verify that v L (D) if α > 2 , v H1(D) if α> 1 , and,∈ more generally v Hs(D) if∈α>s 1 . − ∈ 2 ∈ − 2 ⊓⊔ Part I. Elements of functional analysis 19
Example 2.17 (H¨older functions). If D is bounded and p [1, ), then ∞ ∋.C0,α(D) ֒ W s,p(D) provided 0 s < α 1; see Exercise 2.9 → ≤ ≤ ⊓⊔ Example 2.18 (Step function). Let D := ( 1, 1) and consider v(x) :=0 if x< 0 and v(x) := 1 if x 0. Then v W s,p(−D) iff sp < 1 as shown by the following computation (notice≥ that sp >∈ 0):
0 1 0 p 1 1 1 1 v W s,p =2 sp+1 dx dy =2 sp sp dx. | | 1 0 y x 1 −sp (1 x) − x Z− Z | − | Z− − | | 0 1 The integral 1 x sp dx is convergent if and only if sp < 1. − | | ⊓⊔ Remark 2.19R (Definition by interpolation). Fractional-order Sobolev spaces can also be defined by means of the interpolation theory between Banach spaces (see A.5). Let p [1, ) and s (0, 1). Then we have § ∈ ∞ ∈ s,p p 1,p W (D) = [L (D), W (D)]s,p,
m+s,p m,p m+1,p and more generally W (D) = [W (D), W (D)]s,p for all m N, with equivalent norms in all the cases; see Tartar [181, Lem. 36.1]. ∈ ⊓⊔
2.3 Key properties: density and embedding
This section reviews some key properties of Sobolev spaces: the density of smooth functions and the (compact) embedding into Lebesgue spaces or into spaces composed of H¨older continuous functions.
2.3.1 Density of smooth functions Theorem 2.20 (Meyers–Serrin). Let D be an open set in Rd. Let s 0 s,p s,p ≥ and p [1, ). Then C∞(D) W (D) is dense in W (D). ∈ ∞ ∩ Proof. See Meyers and Serrin [136] and Adams and Fournier [3, Thm. 3.17]; see also Evans [95, p. 251] for bounded D. ⊓⊔ m, Remark 2.21 (p = ). Let m N. The closure of C∞(D) W ∞(D) ∞ ∈ m,∩ with respect to the Sobolev norm W m,∞(D) differs from W ∞(D) since it is composed of functions whose derivativesk·k up to order m are continuous and bounded on D. ⊓⊔ The density of smooth functions in Sobolev spaces allows one to derive many useful results. We list here some of the most important ones.
Corollary 2.22 (Differentiation of a product). Let D be an open subset d 1,p of R . Then we have uv W (D) L∞(D) and (uv) = v (u)+ u (v) 1,p ∈ ∩ ∇ ∇ ∇ for all u, v W (D) L∞(D) and all p [1, ]. ∈ ∩ ∈ ∞ 20 Chapter 2. Weak derivatives and Sobolev spaces
Proof. See, e.g., [46, Prop. 9.4, p. 269]. ⊓⊔ Corollary 2.23 (Differentiation of a composition). Let D Rd be an open set. Let G C1(R). Assume that G(0) = 0 and there is⊂ M < ∈ 1,p ∞ such that G′(t) M for all t R. Then we have G(u) W (D) and | | ≤ ∈1,p ∈ (G(u)) = G′(u) u for all u W (D) and all p [1, ]. ∇ ∇ ∈ ∈ ∞ Proof. See, e.g., [46, Prop. 9.5, p. 270]. ⊓⊔ Corollary 2.24 (Change of variable). Let D,D′ be two open subsets of d 1 R . Assume that there exists a bijection T : D′ D s.t. T C (D′; D), 1 1 d d →1 ∈ d d T − C (D; D′), DT L∞(D′; R × ), and DT − L∞(D; R × ), where ∈ 1 ∈ ∈ 1 DT and DT − are the Jacobian matrices of T and T − , respectively. Then 1,p 1,p we have u T W (D′) for all u W (D) and all p [1, ], and ◦ ∈ ∈ ∈ ∞ ∂ ′ (u T )(x )= ∂ u(T (x))∂ ′ T (x ) for all i 1:d and x D . xi ′ j 1: d xj xi ′ ′ ′ ◦ ∈{ } ∈{ } ∈ Proof. See, e.g.,P [46, Prop. 9.6, p. 270]. ⊓⊔ 2.3.2 Embedding We use the notation V ֒ W to mean that the embedding of V into W is → continuous, i.e., there is c such that v W c v V for all v V (see A.2). The main idea of the results in thisk sectionk ≤ isk thatk functions∈ in the Sobo§ lev space W s,p(D) with differentiability index s > 0 do have an integrability index larger than p (i.e., they belong to some Lebesgue space Lq(D) with q>p), and if s is sufficiently large, for all u W s,p(D) (recall that u is actually a class of functions that coincide almost∈ everywhere in D), there is a representative of u that is continuous (or even H¨older continuous). How large s must be for these properties to hold true depends on the space di- mension. The case d = 1 is particularly simple since W 1,1(R) ֒ C0(R) and W 1,1(D) ֒ C0(D) for every bounded interval D; see [181, Lem.→ 8.5] (see also Exercise→ 5.7). In the rest of this section, we assume that d 2. We first consider the case where D = Rd. ≥ Theorem 2.25 (Embedding of W 1,p(Rd)). Let d 2 and let p [1, ]. The following holds true: ≥ ∈ ∞ (i) (Gagliardo–Nirenberg–Sobolev): If p [1, d), then ∈
1,p d q d pd (W (R ) ֒ L (R ), q [p,p∗], p∗ := . (2.8 → ∀ ∈ d p − p∗ ∗ d In particular u Lp (Rd) 1∗ u Lp(Rd) with 1∗ := d 1 for all u k k ≤ k∇∗ k − ∈ (W 1,p(Rd). Hence W 1,p(Rd) ֒ Lp (Rd), and the embedding into Lq(Rd → for all q [p,p∗) follows from Corollary 1.39. (ii) If p = d,∈ then
(W 1,d(Rd) ֒ Lq(Rd), q [d, ). (2.9 → ∀ ∈ ∞ Part I. Elements of functional analysis 21
(iii) (Morrey): If p (d, ], then ∈ ∞
1,p d d 0,α d d (W (R ) ֒ L∞(R ) C (R ), α :=1 . (2.10 → ∩ − p
Proof. See [46, Thm. 9.9, Cor. 9.11, Thm. 9.12], [95, p. 263-266], [172, I.7.4, I.8.2], [181, Chap. 8-9]. § § ⊓⊔ Remark 2.26 (Continuous function). The embedding (2.10) means that there is c, only depending on p and d, such that
α d u(x) u(y) c x y 2 d u p d , for a.e. x, y R , (2.11) | − |≤ k − kℓ (R )k∇ kL (R ) ∈ for all u W 1,p(Rd). In other words there is a continuous function v C0,α(Rd) such∈ that u = v almost everywhere. It is then possible to replace ∈u by its continuous representative v. We will systematically do this replacement in this book when a continuous embedding in a space of continuous functions is invoked. ⊓⊔ The above results extend to Sobolev spaces of arbitrary order. Theorem 2.27 (Embedding of W s,p(Rd)). Let d 2, s > 0, and p [1, ]. The following holds true: ≥ ∈ ∞ q d pd L (R ) if sp
Theorem 2.29 (Extension from Lipschitz sets). Let s > 0 and p [1, ]. Let D be an open, bounded subset of Rd. If D is a Lipschitz set, then∈ it has∞ the (s,p)-extension property. Proof. See Calder´on [52], Stein [173, p. 181] (for s N), [106, Thm. 1.4.3.1] and [83, Prop. 4.43] (for p (1, )), [134, Thm.∈ A.1&A.4] (for s [0, 1], p [1, ] and s> 0, p = 2)∈ [181,∞ Lem. 12.4] (for s = 1). ∈ ∈ ∞ ⊓⊔ Theorem 2.30 (Embedding of W s,p(D)). Let d 2, s > 0, and p [1, ]. Let D be an open, bounded subset of Rd. If D ≥is a Lipschitz set, then∈ we∞ have
q pd L (D) if sp 0, and := ǫ. Notice that pα < p∗ and also pα < pβ 1+α p pβ pα − since ǫ> 0. Since one can choose β s.t. ǫ is arbitrarily close to zero, we pick q β so that pβ < p∗. Hence pβ (pα,p∗). But u L (D) for all q (pβ,p∗), which would contradict Theorem∈ 2.30 if the Lipschitz6∈ property∈ had been omitted. Hence D cannot be a Lipschitz set in R2 (α > 1 means that D has a cusp at the origin). This counterexample shows that some smoothness assumption on D is needed for Theorem 2.30 to hold true. ⊓⊔ We conclude this section with important compactness results. Recall -from A.4 that the embedding V ֒ W between two Banach spaces is com pact iff§ from every bounded sequence→ in V, one can extract a converging subsequence in W. Theorem 2.34 (Rellich–Kondrachov). Let s > 0 and p [1, ]. Let D be an open, bounded subset of Rd. If D is a Lipschitz set, then∈ the∞ following embeddings are compact: Part I. Elements of functional analysis 23 s,p q pd .( i) If sp d, W (D) ֒ L (D) for all q [1, d sp) ≤ → ∈ − .(ii) If sp>d, W s,p(D) ֒ 0(D) s,p s′,p →C .′iii) W (D) ֒ W (D) for all s>s) → Proof. See [3, Thm. 6.3], [46, Thm. 9.16], [95, p. 272], [132, p. 35], [106, Thm. 1.4.3.2]. ⊓⊔ Exercises Exercise 2.1 (Lebesgue point). Let a R. Let f : R R be defined by f(x) := 0 if x < 0, f(0) := a, and f(x) :=∈ 1 if x > 0. Show→ that 0 is not a Lebesgue point of f for all a. Exercise 2.2 (Lebesgue differentiation). The goal is to prove Theo- rem 2.2. (i) Let h (the sign of h is unspecified). Show that R(x, h) := F (x+h) F (x) ∈ H1 x+h − f(x)= (f(t) f(x)) dt. (ii) Conclude. h − h x − Exercise 2.3 (LebesgueR measure and weak derivative). Let D := (0, 1). Let C be the Cantor set (see Example 1.5). Let f : D R be defined by f(∞x) := x if x C , and f(x) :=2 5x if x C . (i) Is f→measur- 6∈ ∞ − ∈ ∞ able? (Hint: use Corollary 1.11.) (ii) Compute supx D f(x), ess supx D f(x), ∈ ∈ infx D f(x), ess infx D f(x), and f L∞(D) (iii) Show that f is weakly dif- ∈ ∈ k k x ferentiable and compute ∂ f(x). (iv) Compute f(x) ∂ f(t) dt for all x − 0 t x D. (iv) Identify the function f c C0(D) that satisfies f = f c a.e. on D? R Compute∈ f c(x) x ∂ f(t) dt for all∈x D. − 0 t ∈ Exercise 2.4 (WeakR derivative). Let D := ( 1, 1). Prove that if u 1 − ∈ Lloc(D) has a second-order weak derivative, it also has a first-order weak x derivative. (Hint: consider ψ(x) := 1(ϕ(t) cϕρ(t)) dt for all ϕ C0∞(D), − − ∈ with c := ϕ dx, ρ C∞(D), and ρ dx = 1.) ϕ D ∈ 0 R D R R 1 d Exercise 2.5 (Clairaut’s theorem). Let v Lloc(D). Let α, β N and assume that the weak derivatives ∂αv, ∂βv exist∈ and that the weak derivative∈ ∂α(∂βv) exists. Prove that ∂β(∂αv) exists and ∂α(∂βv)= ∂β(∂αv). Exercise 2.6 (Weak and classical derivatives). Let k N, k 1, and let v Ck(D). Prove that, up to the order k, the weak derivatives∈ ≥ and the classical∈ derivatives of v coincide. Exercise 2.7 (H1(D)). (i) Let D := ( 1, 1) and u : D R s.t. u(x) := 3 − 1 → x 2 1. Determine whether u is a member of H (D; R). (ii) Let u1 |C1|((−1, 0]; R) and u C1([0, 1); R) and assume that u (0) = u (0). Let∈ − 2 ∈ 1 2 u be such that u ( 1,0) := u1 and u (0,1) := u2. Determine whether u is a member of H1(D;| R−). Explain why u| H1(D; R) if u (0) = u (0). 6∈ 1 6 2 24 Chapter 2. Weak derivatives and Sobolev spaces Exercise 2.8 (Broken seminorm). Let D be an open set in Rd. Let D1,...,Dn be a partition of D as in Remark 2.13. (i) Show that ( v) Di = { } 1,1 ∇ | (v Di ) for all i 1:n and all v Wloc (D). (ii) Let p [1, ) and ∇ | 1,p ∈ { } n p ∈ p ∈ ∞ v W (D). Show that i=1 v Di 1,p = v 1,p . (iii) Let s (0, 1), | W (Di) W (D) ∈ s,p | | n | | p p ∈ p [1, ), and v W (D). Prove that i=1 v Di s,p v s,p . ∈ ∞ ∈ P | | |W (Di) ≤ | |W (D) Exercise 2.9 (W s,p). Let D be a boundedP open set in Rd. Let α (0, 1]. ∋ .(Show that C0,α(D; R) ֒ W s,p(D; R) for all p [1, ) if s [0, α → ∈ ∞ ∈ 1 1 2 Exercise 2.10 (Unbounded function in H (D)). Let D := B(0, 2 ) R 1 ⊂ be the ball centered at 0 and of radius 2 . (i) Show that the (unbounded) function u(x) := ln ln( x ℓ2 ) has partial weak derivatives. (Hint: work − k k1 on D B(0,ǫ) with ǫ (0, 2 ), and use Lebesgue’s dominated convergence theorem.)\ (ii) Show that∈ u is in H1(D). Exercise 2.11 (Equivalent norm). Let m N, m 2, and let p [1, ). ∈ ≥1 ∈ ∞ p mp p p Prove that the norm v := ( v p + ℓ v ) is equivalent to the k k k kL D | |W m,p(D) canonical norm in W m,p(D). (Hint: use Theorem 2.34 and the Peetre–Tartar lemma (Lemma A.20).) Part I. Elements of functional analysis 25 Solution to exercises Exercise 2.1 (Lebesgue point). Let r> 0. We have 1 r 1 0 1 r 1 f(t) a dt = a dt + 1 a dt = ( a + 1 a ). 2r r | − | 2r r | | 2r 0 | − | 2 | | | − | Z− Z− Z 1 1 1 r Since 2 ( a + 1 a ) 2 for all a R, this proves that 2r r f(t) a dt cannot converge| | | − to zero| ≥ a r 0. Hence∈ 0 is not a Lebesgue point− | of f.− | ↓ R Exercise 2.2 (Lebesgue differentiation). (i) Let x R. We have ∈ F (x + h) F (x) 1 x+h − = f(t) dt, h h Zx for all h . We infer that ∈ H 1 x+h 1 x+h R(x, h)= f(t) dt f(x)= (f(t) f(x)) dt. h − h − Zx Zx F (x+h) F (x) (ii) We want to prove that we have h− f(x) 0 as h 0, for every Lebesgue point x of f. That is| to say, R(x, h−) 0.| →Recalling→ that the sign of h is unspecified and using the above expression→ for R(x, h), we have 1 x+h 1 x+h R(x, h) f(t) f(x) dt 2 f(t) f(x) dt, | |≤ h x | − | ≤ 2h x h | − | | | Z | | Z − which shows that limh 0 R(x, h) = 0 since x is a Lebesgue point of f. Hence F is strongly differentiable→ | at x. | Exercise 2.3 (Lebesgue measure and weak derivative). (i) Yes, f is measurable according to Corollary 1.11, since f(x)= x for a.e. x D. (ii) We have ∈ sup f(x) = max sup x, sup (2 5x) =2, x D x D C x C∞ − ∈ ∈ \ ∞ ∈ ! ess sup f(x) = sup x =1, x D x D C ∈ ∈ \ ∞ inf f(x) = min inf x, inf (2 5x) = 3, x D x D C∞ x C∞ − − ∈ ∈ \ ∈ ess inf f(x) = inf x =0, x D x D C ∈ ∈ \ ∞ f L∞(D) = esssup f(x) = sup x =1. k k x D | | x D C | | ∈ ∈ \ ∞ (iii) For all φ C∞(D) we have ∈ 26 Chapter 2. Weak derivatives and Sobolev spaces f(x)∂ φ(x) dx = x∂ φ(x) dx = φ(x) dx. x x − ZD ZD ZD Hence f is weakly differentiable and ∂xf(x) = 1 for a.e. x D. (iv) For all x D C , we have ∈ ∈ \ ∞ x f(x) ∂ f(t) dt = x x =0. − t − Z0 For all x C , we have ∈ ∞ x f(x) ∂ f(t) dt =2 5x x =2 6x. − t − − − Z0 Hence the fundamental theorem of calculus for f holds true only a.e. on D. (v) We have f(x) = x for a.e. x D, hence f c(x) = x. (Observe that, in accordance with Theorem 2.25 (with∈ d = 1 and all p 1), we indeed c 0 c ≥ have f C (D).) Since ∂tf = ∂tf a.e. in D, the fundamental theorem of calculus∈ implies that x x f c(x) ∂ f(t) dt = f c(x) ∂ f c(t) dt = f c(0) = 0. − t − t Z0 Z0 Exercise 2.4 (Weak derivative). Let ρ C∞(D) with ρ dx = 1. For ∈ 0 D all ϕ C∞(D), the function ψ in the hint is in C∞(D) and ∂ ψ = ϕ c ρ ∈ 0 0 R x − ϕ with cϕ = D ϕ dx. Letting v := ∂xxu and Cρ := D u∂xρ dx, we have R R u∂xϕ dx = u∂xxψ dx + cϕCρ = vψ dx + cϕCρ D D D Z Z x Z = v(x) (ϕ(y) cϕρ(y)) dy dx + cϕCρ D 1 − Z Z− 1 = v(x) dx (ϕ(y) c ρ(y)) dy + c C . − ϕ ϕ ρ ZD Zy 1 Setting C′ := C v(x) dx ρ(y) dy, we thus have ρ ρ − D y R R 1 u∂xϕ dx = v(x) dx ϕ(y) dy + Cρ′ ϕ(y) dy, ZD ZD Zy ZD which shows that u has a first-order weak derivative. Exercise 2.5 (Clairaut’s theorem). Let ϕ C0∞(D). Using Clairaut’s theorem for ϕ, we infer that ∈ Part I. Elements of functional analysis 27 α α β α β ( 1)| | ∂ v∂ ϕ dx = v∂ (∂ ϕ) dx − ZD ZD β α β β α = v∂ (∂ ϕ) dx = ( 1)| | ∂ v∂ ϕ dx, − ZD ZD α β β where we used the definition of ∂ v (and ∂ ϕ C0∞(D)) and ∂ v (and α α β ∈ ∂ ϕ C0∞(D)). Using the definition of ∂ (∂ v), the above identity shows that ∈ α β β α β ∂ v∂ ϕ dx = ( 1)| | ∂ (∂ v)ϕ dx, − ZD ZD β α for all ϕ C0∞(D), which in turn implies that the weak derivative ∂ (∂ v) exists and∈ this weak derivative is indeed equal to ∂α(∂βv). Exercise 2.6 (Weak and classical derivatives). Let α Nd be a multi- index of length α k. Let (∂αv) , (∂αv) denote the classical∈ and weak | | ≤ cl wk derivatives, respectively. For all ϕ C0∞(D), integrating by parts the classical derivative (there are no boundary∈ terms since ϕ has compact support), we infer that α α α α (∂ v) ϕ dx = ( 1)| | v∂ ϕ dx = (∂ v) ϕ dx, cl − wk ZD ZD ZD and we conclude invoking the vanishing integral theorem 1.28. Exercise 2.7 (H1(D)). (i) Let us set D := ( 1, 1). Clearly u L2(D). Let us determine whether u has a weak derivative− and whether∈ the weak 2 derivative is in L (D). Let φ C∞(D). We have ∈ 0 1 0 1 3 3 u(x)∂xφ(x) dx = (( x) 2 1)∂xφ(x) dx + (x 2 1)∂xφ(x) dx 1 1 − − 0 − Z− Z− Z 0 1 1 1 3 2 3 2 = 2 ( x) φ(x) dx φ(0) 2 x φ(x) dx + φ(0) − 1 − − − − 0 Z− Z 0 1 1 1 3 2 3 2 = 2 x φ(x) dx 2 x φ(x) dx − 1 − | | − 0 | | Z− Z 1 = w(x)φ(x) dx − 1 Z− 3 1 where w(x)= x 2 sgn(x) with 2 | | 1 if x< 0 sgn(x)= − (1 otherwise. Since w L2(D), we infer that u H1(D). ∈ 2 ∈2 2 (ii) Since u1 L (( 1, 0)), u2 L ((0, 1)) and u L2(D) = ( u1 L2(( 1,0)) + ∈ − ∈ k k k k − 2 1 2 u 2 ) 2 , we infer that u L (D). Let φ C∞(D). Using that u k 2kL ((0,1)) ∈ ∈ 0 1 ∈ 28 Chapter 2. Weak derivatives and Sobolev spaces C1(( 1, 0]), u C1([0, 1)) and setting v(x) := u (x) if x < 0 and v(x) := − 2 ∈ 1 u2(x) otherwise, we infer that 0 0 u(x)∂xφ(x) dx = u1(x)∂xφ(x) dx + u2(x)∂xφ(x) dx D 1 1 Z Z− Z− = v(x)φ(x) dx + φ(0)(u (0) u (0)) = v(x)φ(x) dx, − 1 − 2 − ZD ZD 2 since u1(0) = u2(0). We infer that ∂xu = v which is in L (D). Hence u 1 ∈ H (D). If u1(0) = u2(0), we infer from Example 2.5 that there is no function w L1 (D) s.t.6 vφ dx = φ(0). Hence u H1(D). ∈ loc D 6∈ Exercise 2.8 (BrokenR seminorm). (i) Let v W 1,1(D). Let k 1:d , ∈ loc ∈ { } let i 1:n , and let ϕ C0∞(Di). Lettingϕ ˜ denote the zero-extension of ϕ to D∈{, we have} ∈ (∂kv) Di ϕ dx = ∂kvϕ˜ dx = v∂kϕ˜ dx = v Di ∂kϕ dx, | − − | ZDi ZD ZD ZDi which shows that (∂kv) Di = ∂k(v Di ). | | (ii) The identity is a direct consequence of ( v) Di = (v Di ) since ∇ | ∇ | n n n p p p v Di 1,p = (v Di ) Lp = ( v) Di Lp | | |W (Di) k∇ | k (Di) k ∇ | k (Di) i=1 i=1 i=1 X X X = v p = v p . k∇ kLp(D) | |W 1,p(D) (iii) The definition of W s,p(D) implies that p p v(x) v(y) v W s,p(D) = | − sp+d| dx dy | | D D x y Z Z k − kℓ2 n n v(x) v(y) p = | − sp+d| dx dy i=1 j=1 Di Dj x y 2 X X Z Z k − kℓ n p n v(x) v(y) p | − | dx dy = v s,p . ≥ sp+d | |W (Di) i=1 Di Di x y 2 i=1 X Z Z k − kℓ X s,p 0,α Exercise 2.9 (W ). Note first that C (D) ֒ L∞(D) since α> 0. Then 0,α p → 0,α C (D) ֒ L (D) since D is bounded. Let v C (D) and let cα be the → α∈ constant such that v(x) v(y) cα x y ℓ2 . Let ℓD = diam(D). Since D B(x,ℓ ) for all| x D−, we infer| ≤ thatk − k ⊂ D ∈ Part I. Elements of functional analysis 29 p p v(x) v(y) p 1 v W s,p(D) = | − sp+d| dx dy cα (s α)p+d dx dy | | D D x y 2 ≤ D D x y − Z Z k − kℓ Z Z k − kℓ2 p 1 cα (s α)p+d dx dy ≤ D B(x,ℓD) x y − Z Z k − kℓ2 1 p (α s)p (α s)p 1 = c S(0, 1) ℓ − r − − dr dx, α| | D ZD Z0 where S(0, 1) is the unit sphere in Rd and S(0, 1) is the (d 1)-dimensional | | − measure of S(0, 1). The integral is finite, i.e., v W s,p(D) is finite, if and only if s < α. | | Exercise 2.10 (Unbounded function in H1(D)). (i) First we observe that 2 1 u L (D). Let ǫ (0, 2 ). Then u is of class C∞ in D B(0,ǫ). Using radial ∈ ∈ cos(θ) \ sin(θ) coordinates, we set w1(x) := ∂1u(x)= r ln(r) and w2(x) := ∂2u(x)= r ln(r) 1 for x = 0. One can verify that wi L (D) for i 1, 2 . Let ϕ C0∞(D). We obtain6 ∈ ∈ { } ∈ wiϕ dx = u∂iϕ dx + T (ǫ), D B(0,ǫ) − D B(0,ǫ) Z \ Z \ with T (ǫ) := (e n)uϕ ds, where n is the unit normal at ∂B(0,ǫ) ∂B(0,ǫ) i· pointing outward and e is the unit canonical vector defining the ith direction. R i Since T (ǫ) 2πǫ ln( ln(ǫ)) ϕ L∞(D), we infer that T (ǫ) 0 as ǫ 0. | | ≤ −1 k k → → Since wi and u are in L (D), letting ǫ 0 in the above equality, we infer using Lebesgue’s dominated convergence theorem→ that w ϕ dx = u∂ ϕ dx. D i − D i Since ϕ is arbitrary in C0∞(D), we conclude that wi is the weak derivative of u in the ith direction. R R (ii) Let us now show that w L2(D), i 1, 2 . We have i ∈ ∈{ } 1 2 ln(2) 2 2 r − 1 2π w1 L2(D)+ w2 L2(D) =2π 2 2 dr =2π 2 dt = < . k k k k 0 r ln(r) t ln(2) ∞ Z Z−∞ Exercise 2.11 (Equivalent norm). For every integer n 0, let n,d := d p ≥ bn,d B α N α = n and bn,d := card( n,d). Let Yn := [L (D)] equipped with{ ∈ some| | product| } norm. Let us considerB the integer m 2 and let Y := Lp(D) Y be equipped with the product norm ≥ × m 1 p p mp p (z, (yα)α ) Y := z p + ℓ yα p . k ∈Bm,d k k kL (D) D k kL (D) α ∈BXm,d m,p α We define the operator A : W (D) Y by A(v) := (v, (∂ v)α ). → ∈Bm,d Let Z := Y1 ... Ym 1 be equipped with some product norm. We define T : W m,p(D)× Z×by − → 30 Chapter 2. Weak derivatives and Sobolev spaces α1 αm−1 T (v) := ((∂ v)α ,..., (∂ v)αm ). 1∈B1,d −1∈Bm−1,d The equivalence of norms in finite-dimensional spaces implies that there exists c such that m,p c v m,p A(v) + T (v) , v W (D). k kW (D) ≤k kY k kZ ∀ ∈ Both A and T are linear and bounded. The operator A is injective. The op- ′ erator T is compact since the embedding W m,p(D) ֒ W m ,p(D) is compact → for all m′ 1:m 1 . Then the assertion follows from Lemma A.20. ∈{ − }