Sobolev Spaces

Lecture notes Sobolev spaces

SS 2015

Johanna Penteker Institute of Analysis Johannes Kepler University Linz

These lecture notes are a revised and extended version of the lecture notes written by Roman Strabler and Veronika Pillwein according to a lecture given by Paul F. X. M¨uller

Contents

Chapter 1. Introduction 3 Chapter 2. Weak derivatives and Sobolev spaces 7 2.1. Preliminaries 7 2.2. Weak derivative 8 2.3. The Sobolev spaces W k,p(U) 10 2.4. Examples 14 Chapter 3. Approximation in Sobolev spaces 19 3.1. Smoothing by convolution 19 3.2. Partition of unity 24 3.3. Local approximation by smooth functions 26 3.4. Global approximation by smooth functions 27 3.5. Global approximation by functions smooth up to the boundary 28 Chapter 4. Extensions 33 Chapter 5. Traces 37 Chapter 6. Sobolev inequalities 43 6.1. Gagliardo-Nirenberg-Sobolev inequality 43 1,p 1,p 6.2. Estimates for W and W0 , 1 ≤ p < n 46 6.3. Alternative proof of the Gagliardo-Nirenberg-Sobolev inequality 47 6.4. Hölder spaces 52 6.5. Morrey’s inequality 54 1,p 1,p 6.6. Estimates for W and W0 , n < p ≤ ∞ 57 6.7. General Sobolev inequalities 59 6.8. The borderline case 60 6.9. Trudinger inequality 62 Chapter 7. Compact embeddings 67 Chapter 8. Poincaré’sinequality 71 8.1. General formulation and proof by contradiction 71 8.2. Poincaré’sinequality for a ball 73 8.3. Poincaré’sinequality - an alternative proof 74 Chapter 9. Fourier transform 79 Chapter 10. Exercises 85 1 Chapter 11. Appendix 91 11.1. Notation 91 11.2. Inequalities 93 11.3. Calculus Facts 96 11.4. Convergence theorems for integrals 97 11.5. Absolutely continuous functions 97 Bibliography 99

2 CHAPTER 1

Introduction

The theory of Sobolev spaces give the basis for studying the existence of solutions (in the weak sense) of partial diﬀerential equations (PDEs). As motivation for this theory we give a short introduction on second order elliptic partial diﬀerential equations, but without going deeper into the PDE-theory. For more information about the analytic and numerical theory see [1], [4], [6] and [9].

Boundary value problems for second-order ordinary differential equations. Classical formulation: Find a function u : [0, 1] → R such that −u00(x) + b(x)u0(x) + c(x)u(x) = f(x), x ∈ Ω := (0, 1), (1.1) u(0) = u(1) = 0, with given continuous coefficient functions b, c and given continuous right-hand side f.A function u ∈ C2(Ω) ∩ C(Ω) that satisfies (1.1) is called classical solution. Variational formulation Let v : [0, 1] → R a so-called test function. We can multiply (1.1) with a test function v and integrate over the interval Z Z − u00(x) + b(x)u0(x) + c(x)u(x) v(x)dx = f(x)v(x)dx. (1.2) Ω Ω Every solution of (1.1) is a solution of (1.2) (for every test function v). On the other hand if a function u ∈ C2(Ω) ∩ C(Ω) satisfies equation (1.2) for every test function v then u satisfies the differential equation (1.1). With integration by parts we can rewrite (1.2) as 1 Z Z Z 0 0 0 0 −u (x)v(x) + u (x)v (x)dx+ b(x)u (x) + c(x)u(x) v(x)dx = f(x)v(x)dx . (1.3) 0 Ω Ω Ω Using the boundary conditions v(0) = v(1) = 0, equation (1.3) yields Z Z Z u0(x)v0(x)dx + b(x)u0(x) + c(x)u(x) v(x)dx = f(x)v(x)dx . (1.4) Ω Ω Ω Derivatives occur in equation (1.4) only in terms of the form Z 1 w0(x)ϕ(x)dx, (1.5) 0 1 where ϕ : Ω → R is sufficiently smooth and ϕ|∂Ω = 0. Equation (1.5) is for w ∈ C equal to Z 1 − w(x)ϕ0(x)dx. (1.6) 0 3 1 The existence of the integral in (1.6) is given for w ∈ Lloc(Ω). The problem is, that if w∈ / C1, then w0 in equation (1.5) has no meaning. This leads us to the definition of the weak derivative of w.

1 Definition 1.1. Suppose w, w˜ ∈ Lloc(Ω). We sayw ˜ is the weak derivative of w, written w0 =w ˜ provided Z Z 0 ∞ w(x)ϕ (x) dx = − w˜(x)ϕ(x) dx for all test functions ϕ ∈ Cc (Ω). (1.7) Ω Ω Therefore, the existence of the integral expressions in equation (1.4) (for b, c ∈ L∞(Ω) and f ∈ L2(Ω)) is guaranteed for u, u0, v, v0 ∈ L2(Ω). This suggests the Sobolev space 1 1 0 2 H (Ω) = {w ∈ Lloc(Ω) : w, w ∈ L (Ω)}. 1 1 To incorporate the boundary values of u, v ∈ H we need the Sobolev space H0 . Note that as in L2 pointwise evaluation in H1 does not make sense. Hence, we need the trace theorem (Theorem 5.1) in order to be able to assign ”boundary values” along ∂Ω to a function in the Sobolev space.

1 Definition 1.2. We say u ∈ H0 (Ω) is a weak solution of (1.2) if 1 B(u, v) = hf, vi ∀v ∈ H0 (Ω), (1.8) where h·, ·i denotes the inner product in L2(Ω) and Z Z 0 0 0 1 B(u, v) := u (x)v (x)dx + b(x)u (x) + c(x)u(x) v(x)dx, u, v ∈ H0 (U). Ω Ω The identity (1.8) is called variational formulation of (1.2).

Second order elliptic partial differential equations. The 1-dimensional example brings us to the theory of weak solutions for a greater class of differential equations - second order elliptic partial differential equations. Let U ⊆ Rn be open and bounded. We consider the following boundary value problem on U. Lu = f in U, (1.9) u = 0 on ∂U,

where f : U → R is given and u : U → R is the unknown function. L denotes a second-order elliptic partial differential operator having either the form n n X i,j X i Lu = − a (x)uxixj + b (x)uxi + c(x)u (1.10) i,j=1 i=1 or else n n X i,j X i Lu = − (a (x)uxi )xj + b (x)uxi + c(x)u, (1.11) i,j=1 i=1 i,j n i n for given coefficient functions (a )i,j=1, (b )i=1, c. We assume the symmetry condition ai,j = aj,i for all i, j = 1, . . . , n. (1.12) The differential operator L is in divergence form if it is given by equation (1.11). 4 i,j i Example 1.3. Let a = δi,j, b = 0 and c = 0. Then Lu = −∆u. We consider the problem (1.9) for L given by Lu = − div(a(x)∇u(x)), (1.13) where a ∈ L∞(U) and a > 0. Note that if a ∈ C1(U), then Lu = −a∆u − ∇u∇a. Since a is positive, we have that L is elliptic partial differential operator. We assume for the moment that u is a classical solution of (1.9). We multiply (1.9) by ∞ a smooth test function v ∈ Cc (U), integrate over U and apply integration by parts to the left-hand side to get Z Z a(x)∇u(x)∇v(x) dx = f(x)v(x) dx. (1.14) U U Note that there are no boundary terms since v = 0 on ∂U. By approximation we obtain that ∞ 1 the same identity holds when we replace v ∈ Cc (U) by any v ∈ H0 (U). The left hand-side of (1.14) makes sense if u ∈ H1(U). We choose the Sobolev space to incorporate the boundary 1 condition from (1.9), hence we consider u ∈ H0 (U). This leads to the definition of a weak solution u of (1.9). 1 Definition 1.4. We say u ∈ H0 (U) is a weak solution of problem (1.9) with L given by (1.13) if 1 B(u, v) = hf, vi ∀v ∈ H0 (U), (1.15) where h·, ·i denotes the inner product in L2(U) and Z 1 B(u, v) := a(x)∇u(x)∇v(x) dx, u, v ∈ H0 (U). U More generally, 1 Definition 1.5. We say u ∈ H0 (U) is a weak solution of problem (1.9) with L given by (1.11) if 1 B(u, v) = hf, vi ∀v ∈ H0 (U), (1.16) where h·, ·i denotes the inner product in L2(U) and Z n n X i,j X i 1 B(u, v) := a uxi vxi + b uxi v + cuv dx, u, v ∈ H0 (U). U i,j=1 i=1 Remark 1.6. The identity (1.16) is called variational formulation of (1.9). Theorem 1.7. (1) Let u be a classical solution of (1.9). Let B and hf, ·i in Defini- 1 tion 1.5 be bounded on H0 (U). Then u is a weak solution of (1.9). (2) Let f be continuous, u a weak solution of (1.9) and u ∈ C2(U) ∩ C(U). Then u is a classical solution of (1.9). Existence of weak solutions The central theorem in the theory of existence and uniqueness of weak solutions is the following. Theorem 1.8 (Lax-Milgram). Let H be a Hilbert space and B : H × H → R a bilinear form satisfying the conditions

(1) there exists a constant c1 > 0 such that for all u, v ∈ H

|B(u, v)| ≤ c1kukH kvkH

5 (2) there exists a constant c2 > 0 such that for all u ∈ H 2 kukH ≤ c2B(u, u). Let f : H → R be linear and bounded. Then there exists a unique u ∈ H such that B(u, v) = f(v), for all v ∈ H. (1.17) The following existence theorem for weak solutions of our boundary value problem is 1 based on the Lax-Milgram Theorem applied to the Hilbert space H0 (U). Theorem 1.9 (Existence Theorem for a weak solution). There is a number γ ≥ 0 such that for each µ ≥ γ and each function f ∈ L2(U), there exists a unique weak solution 1 u ∈ H0 (U) of the boundary value problem Lu + µu = f in U u = 0 on ∂U. R Example 1.10. In the case Lu = −∆u we have B(u, v) = U ∇u(x)∇v(x) dx. One can check that γ = 0 is possible. In the case Lu = − div(a(x)∇u(x)), with a ∈ L∞(U), a ≥ c > 0 the same holds true.

6 CHAPTER 2

Weak derivatives and Sobolev spaces

2.1. Preliminaries Notation for derivatives. Let U ⊆ Rn be open. Assume u: U → R, x ∈ U. Let ∂u u(x + he ) − u(x) (x) = lim i , ∂xi h→0 h provided the limit exists. We write ∂u ∂2u uxi = , = uxixj , etc. ∂xi ∂xi∂xj

n A vector of the form α = (α1, . . . , αn) ∈ N0 is called a multiindex of order

|α| = α1 + ··· + αn. Each multiindex α deﬁnes a partial diﬀerential operator of order |α|, given by

∂|α|u Dαu = = ∂α1 ··· ∂αn u. α α x1 xn ∂x1 1 ··· ∂xn n

If k ∈ N0, then Dku = {Dαu : |α| = k} is the set of all partial derivatives of order k. If k = 1, then

Du = ∇u = (ux1 , . . . , uxn ).

Locally integrable functions. The space of locally integrable functions on U, denoted 1 by Lloc(U), is deﬁned as the set of measurable functions f : U → R such that for all compact subsets K ⊆ U the following holds Z |f(x)|dx < ∞. (2.1) K Remark 2.1. Constant functions, piecewise continuous functions and continuous functions are locally integrable. Every function f ∈ Lp(U), 1 ≤ p ≤ ∞ is locally integrable.

∞ Test functions. The space of test functions, denoted by Cc (U), is the space of infinitely differentiable functions φ: U → R with compact support. Note that the support of φ is defined by supp φ = {x ∈ U : φ(x) 6= 0}. 7 ∞ Example 2.2. The following functions are elements in Cc (R). ( − 1 e 1−|x| , falls |x| < 1 φ(x) = 0, falls |x| ≥ 1.

( − 1 e 1−x2 , falls |x| < 1 h(x) = 0, falls |x| ≥ 1.

-2 -1 0 1 2

Figure 2.1. h(x)

2.2. Weak derivative 1 Every locally integrable function u ∈ Lloc(U) determines a regular distribution, i.e. a linear and continuous function Z ∞ Tu : Cc (U) → R,Tu(φ) = u(x)φ(x)dx. (2.2) U

We also use the notation hTu, φi resp. hu, φi instead of Tu(φ). 1 If we assume that u ∈ C (U) then its partial derivatives uxi , 1 ≤ i ≤ n, are continuous 1 and hence, uxi ∈ Lloc(U). Therefore, uxi determines a regular distribution given by Z T : C∞(U) → ,T (φ) = u (x)φ(x)dx. (2.3) uxi c R uxi xi U Obviously, we have by integration by parts (cf. Theorem 11.15) Z Z T (φ) = u (x)φ(x)dx = − u(x)φ (x)dx, φ ∈ C∞(U). (2.4) uxi xi xi c U U 1 The left-hand side integral in (2.4) is deﬁned only when uxi exists a.e. and is in Lloc(U), 1 whereas the right-hand side integral is well deﬁned for every u ∈ Lloc(U).

8 Definition 2.3 (Distributional derivative). th (1) The distributional derivative w.r.t. the i variable, 1 ≤ i ≤ n, of Tu is the distribu- ∞ tion (linear and continuous functional on Cc (U)) ∂xi Tu given by Z ∞ h∂xi Tu, φi = −hu, ∂xi φi = − u(x)∂xi φ(x)dx, φ ∈ Cc (U). (2.5) U (2) Let α be a multiindex of order |α|. Then the αth distributional derivative is the α distribution D Tu given by Z α |α| α ∞ hD Tu, φi = (−1) u(x)D φ(x)dx, φ ∈ Cc (U). (2.6) U If u ∈ Ck(U) and α a multiindex of order k, then Dαu exists in the classical sense and we have by the integration by parts formula Z Z Dαu(x)φ(x)dx = (−1)|α| u(x)Dαφ(x)dx. U U α 1 D u ∈ C(U) ⊆ Lloc(U) deﬁnes again a regular distribution given by Z α TDαu(φ) = D u(x)φ(x)dx U and obviously, α ∞ hD T u, φi = TDαu(φ), for all φ ∈ Cc (U). 1 We say that the classical and the distributional derivative of u ∈ Lloc(U) coincide. What 1 k happens if u ∈ Lloc(U), but u∈ / C (U)? This leads us to the deﬁnition of the weak derivative of u.

1 n Definition 2.4 (Weak derivative). Let u ∈ Lloc(U) and α ∈ N0 a multiindex. If there 1 exists a v ∈ Lloc(U) such that Z Z |α| α ∞ n v(x)φ(x)dx = (−1) u(x)D φ(x)dx ∀φ ∈ Cc (R ) (2.7) U U then v is called the weak αth- partial derivative of u, denoted by Dαu = v.

1 In other words, if we are given u ∈ Lloc(U) and if there happens to exist a function 1 ∞ α v ∈ Lloc(U) satisfying (2.7) for all φ ∈ Cc (U) we say that D u = v in the weak sense. If there does not exist such a function v, then u does not possess a weak αth- partial derivative. Remark 2.5. Classical derivatives are defined pointwise as limit of difference quotients. Weak derivatives, on the other hand, are defined in an integral sense. By changing a function on a set of measure zero we do not affect its weak derivatives. Lemma 2.6 (Uniqueness). A weak αth-partial derivative of u, if it exists, is uniquely defined up to a set of measure zero.

9 1 Proof. Assume that v, v˜ ∈ Lloc(U) satisfy Z Z Z u(x)Dαφ(x)dx = (−1)|α| v(x)φ(x)dx = (−1)|α| v˜(x)φ(x)dx U U U ∞ for all φ ∈ Cc (U). This implies Z ∞ (v − v˜)φdx = 0 ∀φ ∈ Cc (U). U Hence, v − v˜ = 0 almost everywhere. 1 α Lemma 2.7. Assume that u ∈ Lloc(U) has weak derivatives D u for |α| ≤ k. Then for multiindices α, β with |α| + |β| ≤ k one has Dα(Dβu) = Dβ(Dαu) = Dα+βu. ∞ α ∞ Proof. Let φ ∈ Cc (U), then also D φ ∈ Cc (U). Using the deﬁnition of weak derivatives twice we obtain: Z Z Dα(Dβu)(x)φ(x) = (−1)|α| Dβu(x)Dαφ(x) dx U U Z = (−1)|α|+|β| u(x)Dα(Dβφ(x)) dx U Z = Dα+βu(x)φ(x) dx. U If we change the roles of α and β we obtain Z Z Dβ(Dαu)(x)φ(x) = Dα+βu(x)φ(x) dx. U U 2.3. The Sobolev spaces W k,p(U) Let U ⊆ Rn open. Let 1 ≤ p ≤ ∞ and k be a non-negative integer. Definition 2.8. The Sobolev space W k,p(U) is the space of all locally integrable functions u : U → R such that for every multiindex α with |α| ≤ k the weak derivative Dαu exists and Dαu ∈ Lp(U). Definition 2.9. We deﬁne the norm of u ∈ W k,p(U) to be Z 1 X α p p kukW k,p(U) = |D u(x)| dx , if 1 ≤ p < ∞, |α|≤k U X α kukW k,∞(U) = ess supx∈U |D u(x)|. |α|≤k

Theorem 2.10 (Completeness). For each k ∈ N0 and 1 ≤ p ≤ ∞ the Sobolev space W k,p(U) is a Banach space. Proof. We have to show the following. 10 (1) W k,p is a normed vector space (exercise). (2) W k,p is complete. ∞ k,p We show (2). Assume (um)m=1 is a Cauchy sequence in W (U). It follows from the k,p α ∞ p deﬁnition of the norm on W (U) that (D um)m=1 is a Cauchy sequence in L (U) for each p p |α| ≤ k, cf. Remark 2.11. Since L (U) is complete, there exist functions u, uα ∈ L (U) such that α kum − ukLp → 0, kD um − uαkLp → 0, for all 0 < |α| ≤ k. We show that k,p α u ∈ W (U) and D u = uα, for all 0 < |α| ≤ k. (2.8) ∞ We ﬁx φ ∈ Cc (U). Then Z Z Z Z α α |α| α |α| uD φdx = lim umD φdx = (−1) lim D umφdx = (−1) uαφdx. U m→∞ U m→∞ U U Thus, (2.8) holds and for all |α| ≤ k

α m→∞ α p D um −→ D u in L (U). Hence, m→∞ k,p um −→ u in W (U) . ∞ k,p Remark 2.11. (um)m=1 is a Cauchy sequence in W (U) DEF ⇐⇒ ∀ε > 0 ∃N ∀m, n > N : kum − unkW k,p(U) < ε 1 X α p p ⇐⇒ ∀ε > 0 ∃N ∀m, n > N : kD (um − un)kLp(U) < ε |α|≤k X α α p p ⇐⇒ ∀ε > 0 ∃N ∀m, n > N : kD um − D unkLp(U) < ε |α|≤k α α =⇒ ∀ε > 0 ∃N ∀m, n > N ∀α, |α| ≤ k : kD um − D unkLp(U) < ε α α =⇒ ∀α, |α| ≤ k ∀ε > 0 ∃N ∀m, n > N : kD um − D unkLp(U) < ε α ∞ p ⇐⇒ ∀α, |α| ≤ k is (D um)m=1 a Cauchy sequence in L (U). Remark 2.12. Note that W 0,p(U) = Lp(U). For p = 2, the norm in Deﬁnition 2.9 is induced by the inner product X Z hu, vi = Dαu(x)Dαv(x)dx. (2.9) |α|≤k U Hence, W k,2(U) is a Hilbert space and we write Hk(U) = W k,2(U).

k,p k,p Definition 2.13. The subspace W0 (U) ⊆ W (U) is deﬁned by W k,p(U) k,p ∞ W0 (U) = Cc (U) . 11 k,p ∞ More precisely, u ∈ W0 (U) if and only if there exists a sequence (un)n∈N in Cc (U) such that

kun − ukW k,p(U) → 0. Definition 2.14. Let U, V be open subsets of Rn. We say V is compactly contained in U, written V ⊂⊂ U if V ⊂ K ⊂ U and K is compact.

k,p Definition 2.15. We deﬁne the space Wloc (U) to be the space of functions u : U → R k,p satisfying the following property. Let V ⊂⊂ U, then u V ∈ W (V ). ∞ k,p Notation. Let (um)m=1, u ∈ W (U) k,p • We say (um) converges to u in W (U), written k,p um −→ u in W (U), (2.10) provided

lim kum − uk k,p = 0. (2.11) m→∞ W (U) • We write k,p um −→ u in Wloc (U) (2.12) to mean k,p um −→ u in W (V ) (2.13) for each V ⊂⊂ U.

k,p ∞ Lemma 2.16. Let u ∈ W (U), ζ ∈ Cc (U) and α a multiindex with |α| ≤ k. Then (1) Dαu ∈ W k−|α|,p(U), (2) ζu ∈ W k,p(U) and X α Dα(ζu) = DβζDα−βu (Leibniz formula). (2.14) β β≤α Proof. (1) follows from the deﬁnition of the Sobolev space. We show (2). We know that u ∈ Lp and Dαu ∈ Lp for all |α| ≤ k. Hence, ζu ∈ Lp and Dα(ζu) given by (2.14) is in p k,p ∞ L . Therefore, ζu ∈ W . We prove the Leibniz formula by induction. Let ζ, φ ∈ Cc (U) and |α| = 1. let 1 ≤ i ≤ n. Then, by the deﬁnition of the weak derivative Z Z

(ζu)xi φdx = − ζuφxi dx. (2.15) U U By the chain rule (classical Leibniz formula for the ﬁrst derivative) we have

(ζφ)xi = ζxi φ + ζφxi . (2.16) Combining (2.15) and (2.16) yields Z Z Z

(ζu)xi φ dx = ζxi u φ dx − (ζφ)xi u dx. U U U

12 ∞ Note that ζφ ∈ Cc . Again by the deﬁnition of the weak derivative we have Z Z Z

(ζu)xi φ dx = ζxi u φ dx + ζφ uxi dx U U U Z

= (ζxi u + uxi ζ)φ dx. U

Therefore, (ζu)xi = ζxi u + uxi ζ. Now let |α| = l + 1, then α = β + γ with |β| = l and |γ| = 1. By the definition of the weak derivative we have Z Z Z ζu Dαφ dx = ζu Dβ (Dγφ)dx = (−1)|β| Dβ(ζu)Dγφ dx. (2.17) U U U Using the induction hypothesis and again the definition of the weak derivative we obtain " # Z Z X β (−1)|β| Dβ(ζu)Dγφ dx. = (−1)|β| DσζDβ−σu Dγφ dx σ U U σ≤β " # Z X β = (−1)|β|+|γ| Dγ DσζDβ−σu φ dx σ U σ≤β " # Z X β = (−1)|α| Dγ DσζDβ−σu φ dx σ U σ≤β " # Z X β = (−1)|α| (Dσ+γζDβ−σu + DσζDβ−σ+γu) φ dx. σ U σ≤β (2.18) Note that the last equality in (2.18) is due to the induction basis (|γ| = 1). Let ρ = σ + γ. Then we can rewrite the sum in the right hand side of (2.18) as X β X β X β (Dσ+γζDβ−σu + DσζDβ−σ+γu) = DρζDα−ρu + DσζDα−σu σ ρ − γ σ σ≤β ρ≤β+γ σ≤β X β X β = DσζDα−σu + DσζDα−σu σ − γ σ σ≤α σ≤β X β β = + DσζDα−σu + Dαζ u σ − γ σ σ≤β (2.19) Note that by the definition of the binomial coefficients for multiindices we have α β β = + . (2.20) σ σ − γ σ Combining the equations (2.17), (2.18), (2.19) and (2.20) yields " # Z Z X α ζu Dαφ dx = (−1)|α| DσζDα−σu φ dx. σ U U σ≤α 13

Lemma 2.17. Let u ∈ W k,p(U) and V ⊆ U open, then u ∈ W k,p(V ).

Proof. Obvious consequence of the deﬁnition.

2.4. Examples Weak derivatives.

Example 2.18. Let n = 1, U = (0, 2) and

( x, if 0 < x ≤ 1, u(x) = 1, if 1 < x < 2.

We deﬁne ( 1, if 0 < x ≤ 1, v(x) = 0, if 1 < x < 2.

∞ and show that for all φ ∈ Cc (U)

Z 2 Z 2 u(x)φ0(x)dx = − v(x)φ(x)dx 0 0

holds.

u(x) v(x) 1 1

1 2 1 2

Figure 2.2. u(x) and v(x)

14 Z 2 Z 1 Z 2 u(x)φ0(x)dx = u(x)φ0(x)dx + u(x)φ0(x)dx 0 0 1 Z 1 Z 2 = xφ0(x)dx + 1φ0(x)dx 0 1 1 Z 1 Z 2 0 = xφ(x) − φ(x)dx + φ (x)dx 0 0 1 Z 1 = φ(1) − φ(x)dx + φ(2) −φ(1) 0 |{z} 0 Z 1 Z 2 = − φ(x)dx = − v(x)φ(x) dx. 0 0 Example 2.19. Let n = 1 , U = (0, 2) and ( x, if 0 < x ≤ 1, u(x) = 2, if 1 < x < 2. In order to check, that u does not have a weak derivative we have to show that there does 1 not exist any function v ∈ Lloc(U) satisfying Z 2 Z 2 u(x)φ0(x)dx = − v(x)φ(x)dx, (2.21) 0 0 ∞ 1 for all φ ∈ Cc (U). Assume there exists a v ∈ Lloc(U) satisfying (2.21). Then Z 2 Z 2 Z 1 Z 2 − v(x)φ(x)dx = u(x)φ0(x)dx = xφ0(x) + 2 φ0(x)dx 0 0 0 1 (2.22) 1 Z 1 Z 1 = xφ(x) − φ(x)dx + 2 φ(2) − φ(1) = −φ(1) − φ(x)dx 0 0 0 ∞ ∞ is valid for all φ ∈ Cc (U). We choose a sequence (φm)m=1 of smooth functions satisfying

u(x) 2 1 (φm)m

1 2 1 2

Figure 2.3. u(x) and some elements of the sequence φm(x)

15 m→∞ 0 ≤ φm ≤ 1, φm(1) = 1 and φm(x) −→ 0 ∀x 6= 1.

Replacing φ by φm in (2.22) yields Z 2 Z 1 1 = φm(1) = v(x)φm(x)dx − φm(x)dx. 0 0 We take the limit for m → ∞: " Z 2 Z 1 # 1 = lim φm(1) = lim v(x)φm(x)dx − φm(x)dx = 0. m→∞ m→∞ 0 0

Elements in Sobolev spaces. Note that if n = 1 and U is an open interval in R, then u ∈ W 1,p(U) if and only if u equals a.e. an absolutely continuous function whose derivative (which exists a.e.) and the function itself belong to Lp(U) (Exercise). Such a simple characterization is however only available for n = 1. In general a discontinuous and/or unbounded function can belong to a Sobolev space.

Example 2.20. Let U = {x ∈ Rn : |x| < 1} =: B(0, 1). We ﬁx γ > 0 and consider the function −γ n ! 2 −γ X 2 u(x) = |x| = xi , x ∈ U, x 6= 0. i=1

-1 1

Figure 2.4. |x|−γ

1,p n−p Statement: u ∈ W (U) ⇔ γ < p Proof. Note that u ∈ C1(U \{0}). For x ∈ U \{0} we have

−γxi ux (x) = , 1 ≤ i ≤ n. (2.23) i |x|γ+2 16 Therefore, 1 n ! 2 X 2 γ |Du(x)| = |∇u(x)| = |uxi | = γ+1 i=1 |x| and by Corollary (11.17), Z Z Z 1 Z 1 |Du(x)|dx = γ |x|−γ−1dx = C(n)γ r−γ−1rn−1dr = C(n)γ r−γ−2+ndr U U 0 0 ( (2.24) C(n)γ , if − γ − 1 + n > 0 = −γ−1+n ∞, otherwise. Hence, if γ + 1 < n, then |Du| ∈ L1(U). Analogously we have u ∈ L1(U), if γ + 1 < n.

On the open set U \{0} the function u has weak derivatives Du = (ux1 , . . . , uxn ) and

they coincide with the classical derivatives. We show that under certain circumstances uxi deﬁne weak derivatives on the entire domain U. ∞ n Let φ ∈ Cc (U) and Bε = {x ∈ R : |x| < ε} for a ﬁxed ε > 0 then (by the integration- by-parts formula (Theorem 11.15)) Z Z Z i uφxi dx = − uxi φ dx + uφν dS U\Bε U\Bε ∂Bε

i −xi 1 n with ν (x) = |x| so that ν = (ν . . . ν ) is the inward pointing normal on ∂Bε and dS is the spherical measure on the surface of the ball Bε. The following holds.

Z Z uφνidS ≤ kφk ε−γdS ≤ Cε−γ−1+n −→ε→0 0, if γ + 1 < n (2.25) ∞ ∂Bε ∂Bε Z Z Z lim uφx dx = lim χU\B uφx dx = uφx dx (2.26) ε→0 i ε→0 ε i i U\Bε U U Z Z lim ux φdx = ux φdx. (2.27) ε→0 i i U\Bε U With (2.25),(2.26) and (2.27) it follows that Z Z ∞ uφxi dx = − uxi φdx ∀φ ∈ Cc (U), U U

and the locally integrable function uxi deﬁned in (2.23) is in fact the weak derivative of u on the entire domain U. By an analogous calculation as in (2.24) we have that u, |Du| ∈ Lp(U) ⇔ (γ + 1)p < n. 1,p n−p 1,p Consequently, u ∈ W (U) if and only if γ < p . In particular u∈ / W (U) for each p ≥ n. n n Example 2.21. Let U = {x ∈ R : |x| < 1} and {rk : k ∈ N} = U ∩ Q . This forms ∞ a dense subset in U.({rk}k=1 is dense in U ⇔ for each u ∈ U there exist a subsequence ∞ {rkl }l=1 such that lim rkl → u in U.) 17 For (γ + 1)p < n we deﬁne −k −γ 1,p uk(x) = 2 |x − rk| ∈ W (U) and set ∞ X −k −γ u(x) = 2 |x − rk| . k=1 Then u ∈ W 1,p(U) and is unbounded on each open subset of U.

18 CHAPTER 3

Approximation in Sobolev spaces

In order to study the deeper properties of Sobolev spaces, without returning continually to the deﬁnition of weak derivatives, we need procedures for approximating a function in a Sobolev space by smooth functions. These approximation procedures allow us to consider smooth functions and then extend the statements to functions in the Sobolev space by density arguments. We have to prove that smooth functions are in fact dense in W k,p(U). The method of molliﬁers provides the tool.

3.1. Smoothing by convolution Definition 3.1. (1) Let η ∈ C∞(Rn) be given by

( 2 C e1/(|x| −1), if |x| < 1, η(x) = 0, if |x| ≥ 1, with constant C > 0 chosen such that R η(x)dx = 1. Rn (2) For each ε > 0 we define 1 x η (x) = η . ε εn ε We call η the standard mollifier. Remark 3.2. ∞ n (1) η ≥ 0 and η ∈ Cc (R ). ∞ (2) The functions ηε are C and satisfy Z ηε(x)dx = 1 and supp ηε ⊆ B(0, ε). Rn Remark 3.3. There are other examples of mollifiers, e.g. ( cos(π|t|2) + 1, if |t| < 1, ν(t) = 0, if |t| ≥ 1.

Definition 3.4. Let U ⊆ Rn be open and ε > 0. Let

Uε = {x ∈ U : d(x, ∂U) > ε} = {x ∈ U : B(x, ε) ⊆ U}, where B(x, ε) = {y ∈ Rn : |x − y| < ε}. 19 η(x) ν(t)

-1 -0.5 0.5 1 -1 -0.5 0.5 1

Figure 3.1. Standard molliﬁer and cos-molliﬁer

1 Let f ∈ Lloc(U). Then we define for all x ∈ Uε Z Z ε f (x) := f ∗ ηε(x) = f(y)ηε(x − y)dy = f(x − y)ηε(y)dy (3.1) U B(0,ε) ε 1 f is the mollification of f in Uε. The mollification of a function f ∈ Lloc(U) results from the concept of convolution. Convolution. Let f, g be measurable functions on Rn. The convolution f ∗ g is defined by Z f ∗ g (x) = f(x − y)g(y)dy Rn for all x ∈ Rn such that the integral exists. Proposition 3.5. Assume that all considered integrals exist. Then (1) f ∗ g = g ∗ f, (2) f ∗ g ∗ h = f ∗ (g ∗ h), (3) supp (f ∗ g) ⊆ supp f + supp g. Theorem 3.6. (1) If f ∈ Lp(Rn), 1 ≤ p ≤ ∞, g ∈ L1(Rn) then f ∗ g exists a.e. and f ∗ g ∈ Lp(Rn). In fact kf ∗ gkp ≤ kfkp · kgk1. 1 1 p n q n (2) Let 1 ≤ p ≤ ∞ and 1 = p + q . If f ∈ L (R ) and g ∈ L (R ) then f ∗ g exists on n n R and |f ∗ g (x)| ≤ kfkpkgkq for every x ∈ R . 1 1 1 p n q n (3) Let 1 ≤ p, q, r ≤ ∞ such that 1+ r = p + q . If f ∈ L (R ) and g ∈ L (R ) then f ∗g r n exists a.e. and lies in L (R ). In fact kf ∗ gkr ≤ kfkp · kgkq. (Young’s inequality) n n (4) Let f, g ∈ Cc(R ). Then f ∗ g exists for every x ∈ R and f ∗ g ∈ Cc(R ). (5) If f ∈ L1, g ∈ Ck and Dαg is bounded for all multiindices α with |α| ≤ k. Then f ∗ g ∈ Ck and Dα(f ∗ g) = f ∗ Dαg for |α| ≤ k. Mollification Properties. ε ε ∞ Lemma 3.7. Let Uε ⊆ U and f be as in Definition 3.4. Then f ∈ C (Uε) and for every multiindex α and x ∈ Uε we have Z α ε α ε D f (x) = Dx η (x − y)f(y)dy, U 20 α where Dx denotes the partial derivatives with respect to the variable x = (x1, . . . , xn). Proof. Exercise.

Proof. Fix x ∈ Uε and h so small, that x + eih ∈ Uε. Then ! f ε(x + e h) − f ε(x) Z ηε(x + e h − y) − ηε(x − y) i = i f(y)dy. h U h

Support(x+he-y) U Support(x-y) I

X X+heI

U 2

Figure 3.2. Supports of ηε The support of ! ηε(x + e h − y) − ηε(x − y) y −→ i h is compact in U (cf. ﬁgure 3.2). Therefore, ! ! Z ηε(x + e h − y) − ηε(x − y) Z ηε(x + e h − y) − ηε(x − y) i f(y)dy = i f(y)dy U h V h for some V ⊂⊂ U. By the Heine-Cantor theorem and the mean value theorem we have that ε ε ε η (x + eih − y) − η (x − y) ∂η lim sup − (x − y) = 0. h→0 y∈V h ∂xi Therefore, Z Z ε 1 ε ε ∂η (η (x + eih − y) − η (x − y)) f(y) dy − (x − y)f(y)dy h V U ∂xi ε ε ε Z η (x + eih − y) − η (x − y) ∂η ≤ sup − (x − y) |f(y)|dy y∈V h ∂xi U −→h→0 0.

21 Hence, ε ε Z ∂ ε f (x + eih) − f (x) ε f (x) = lim = ∂x η (x − y)f(y)dy . h→0 i ∂xi h U By the same argument one obtains that for every multiindex α, Dαf ε exists and Z α ε α ε D f (x) = D η (x − y)f(y)dy (x ∈ Uε) . U 1 Corollary 3.8. Let Uε ⊆ U be as in Deﬁnition 3.4. Assume that f ∈ Lloc(U) admits a weak derivative Dαf for some multiindex α. Then α α D (f ∗ ηε)(x) = ηε ∗ D f (x), for all x ∈ Uε.

α Note that the derivative of the molliﬁcation D (f ∗ ηε) exists in the classical sense. Proof. We have by Lemma 3.7 and by the deﬁnition of the weak derivative Z α α ε D (f ∗ ηε)(x) = Dx η (x − y)f(y)dy U Z |α| α ε = (−1) Dy η (x − y)f(y)dy U Z = (−1)|α|+|α| ηε(x − y)Dαf(y)dy U Z = ηε(x − y)Dαf(y)dy U = ηε ∗ Dαf (x).

Theorem 3.9 (Convergence of the molliﬁcation). 1 ε ε→0 (1) f ∈ Lloc(U) =⇒ f (x) −→ f(x) for almost every x ∈ U, i.e.

{x ∈ U : lim f ε(x) 6= f(x)} = 0. ε→0 (2) f ∈ C(U) and K ⊂ U compact. Then

sup |f(z) − f ε(z)| −→ε→0 0. z∈K p (3) f ∈ Lloc(U), 1 ≤ p < ∞, and K ⊂ U compact. Then

(∗) supp ηε(x − ·) ⊆ B(x, ε). (∗∗) R ηε(x − y)dy = 1. Rn (∗ ∗ ∗) 0 ≤ η ≤ 1 (∗ ∗ ∗∗) |B(x, ε)| = εn|B(0, 1)|. Hence,

{x ∈ n : lim f ε(x) 6= f(x)} = 0. R ε→0

ε ∀η > 0 ∃ ε0 > 0 ∀ε < ε0 ∀x ∈ K : |f(x) − f (x)| ≤ C η. (3) Let K ⊂ U compact. Then there exists an open subset W of U with W ⊂⊂ U and an ε0 > 0 such that for all ε < ε0 and for all x ∈ K we have that B(x, ε) ⊆ Uε ⊆ W . Let 0 < ε < ε0. Then ε kf kLp(K) ≤ kfkLp(W ). 1 1 Indeed by H¨older’sinequality with q + p = 1 we have Z ε ε | f (x) | = η (x − y)f(y)dy B(x,ε) Z ε 1 ε 1 ≤ |f(y)| η (x − y) p η (x − y) q dy B(x,ε) 1 1 Z p Z q ≤ ηε(x − y)|f(y)|pdy ηε(x − y)dy . B(x,ε) B(x,ε) | {z } =1

23 Therefore, Z Z Z ε p ε p ε p kf kLp(K) = | f (x) | dx ≤ η (x − y) | f(y) | dy dx K K B(x,ε) Z Z = ηε(x − y) | f(y) |p dy dx K W Z Z = | f(y) |p ηε(x − y)dxdy W K p ≤ kfkLp(W ). W is compactly contained in U. Hence, C(W ) is dense in Lp, i.e. p ∀ f ∈ L (W ) ∀ δ > 0 ∃ g ∈ C(W ): kf − gkLp(W ) ≤ δ.

Fix δ > 0 and choose g ∈ C(W ) such that kf − gkLp < δ. Then ε ε ε ε kf − fkLp(K) ≤ kf − g kLp(K) + kg − gkLp(K) + kg − fkLp(K) ε ≤ kf − gkLp(W ) + kg − gkLp(K) + kf − gkLp(W ) ε ≤ 2δ + kg − gkLp(K). By (2) we have that

ε 1 ε ε→0 kg − gkLp(K) ≤| K | p sup |g (x) − g(x)| −→ 0. x∈K Summarizing we have ε ∀ K ⊂ U compact ∀η > 0 ∃ ε0 > 0 ∀ ε < ε0 : kf − fkLp(K) ≤ η.

3.2. Partition of unity In the following section we use the method of molliﬁcation to construct partitions of unity. We will use these results in the following proofs to obtain global properties from local ones.

Lemma 3.10. Let K be a compact subset of Rn and U ⊆ Rn open such that K ⊂ U. Then ∞ n there exists a function ψ ∈ Cc (R ) such that 0 ≤ ψ ≤ 1, ψ ≡ 1 on K and supp ψ ⊂ U. Proof. Let n Kε = {x ∈ R : d(x, K) ≤ ε}, ε > 0.

Let ε > 0 small enough that K3ε ⊂ U. We set Z ε ε n ψ(x) = η ∗ 1K2ε (x) = η (x − y)1K2ε (y)dy, x ∈ R . Rn By the above properties of molliﬁcation we know that ψ ∈ C∞, ψ ≡ 1 on K and

ε supp ψ ⊆ supp η + K2ε ⊆ K3ε ⊂ U. 24 he

e x Ñn K2e U1

Figure 3.3. Partition of unity

n Sk k Lemma 3.11. Let K ⊂ R be compact with K ⊂ j=1 Vj, where (Vi)i=1 is a sequence n k n of open sets in R . Then there exists a sequence (Ki)i=1 of compact sets in R such that Sk Kj ⊂ Vj and K ⊆ j=1 Kj.

Proof. Since K is bounded we can assume that Vj is bounded. Let 1 V = {x ∈ V : d(x, ∂V ) > }. j,n j j n Then V ⊂ V , for all n ∈ and each sequence (V ) is increasing. The sets (V )k j,n j N j,n n∈N j,n j=1,n∈N form an open cover of K. K is compact. Hence, there exists an N ∈ N such that k k k [ [ [ K ⊆ Vj,N ⊆ Vj,N =: Kj. j=1 j=1 j=1 n Sk k Theorem 3.12. Let U ⊆ R be bounded and U ⊂⊂ i=1 Vi, where (Vi)i=1 is a sequence n of open sets in R . There exists a sequence of smooth functions ξi, 1 ≤ i ≤ k, such that P 0 ≤ ξi ≤ 1, supp ξi ⊂ Vi and ξi ≡ 1 on U. k We call the sequence (ξi)i=1 a smooth partition of unity subordinate to the open k sets (Vi)i=1. n Sk Proof. Let K ⊆ R be compact such that U ⊂ K ⊂ j=1 Vj. According to Lemma 3.11 Sk there exist compact subsets Kj ⊂ Vj such that U ⊂ K ⊂ j=1 Kj. According to Lemma ∞ n 3.10 for each j ∈ {1, . . . , k} there exists a function ψj ∈ Cc (R ) satisfying 0 ≤ ψj ≤ 1, ψj ≡ 1 on Kj and supp ξj ⊂ Vj. Let

ξ1 = ψ1, ξ2 = ψ2(1 − ψ1), . . . , ξk = ψk(1 − ψ1) ... (1 − ψk−1). ∞ Then we have that 0 ≤ ξj ≤ 1 and ξj ∈ Cc (Vj) for all 1 ≤ j ≤ k. Furthermore, k X 1 − ξj = 1 − [ψ1 + ψ2(1 − ψ1) + ... + (1 − ψk−1)] (3.2) j=1

= (1 − ψ1)(1 − ψ2) ... (1 − ψk). (3.3)

25 For each point x ∈ U there is at least one factor (1 − ψj) that vanishes. Hence, the product equals zero on U and thus, k X ξj ≡ 1 on U. j=1 n ∞ Theorem 3.13. Let U ⊆ R be open with locally ﬁnite open cover (Vi)i=1, i.e. U ⊆ S∞ Vi and for every x ∈ U there exist only ﬁnitely many Vi such that x ∈ Vi. Then there i=1 P exists a sequence of smooth functions (ξi)i∈N such that 0 ≤ ξi ≤ 1, supp ξi ⊂ Vi and ξi = 1 on U.

∞ We call the sequence (ξi)i=1 a smooth partition of unity subordinate to the locally ∞ finite cover (Vi)i=1. ∞ Proof. According to Lemma 3.14 we can choose an open cover (Wi)i=1 of U such that Wi ⊆ Vi. Then, analogously to Theorem 3.12 the statement holds. Note that by the local finiteness of the cover we have that for every x ∈ U there are only finitely many ξi such that P x ∈ supp ξi. Hence, ξi(x) is finite for every x ∈ U. n ∞ Lemma 3.14. Let U ⊆ R be open with open cover (Vi)i=1. Then there exists an open ∞ cover (Wi)i=1 of U such that Wi ⊆ Vi for all i ∈ N. S∞ Proof. Let A = U \ i=2 Vi. Then A ⊂ V1 and A is closed in U. There exists an open set W1 such that A ⊂ W1 ⊂ W1 ⊂ V1. The collection (W1,V2,... ) forms a cover of U. Let W1,...,Wk1 be open sets such that {W1, ..., Wk1,Vk,Vk+1, ...} covers U. Let A = Sk−1 S∞ U \ i=1 Wi ∪ i=1 Vk+i . A is closed in U. There exists an open set Wk such that A ⊂ Wk ⊂ Wk ⊂ Vk. Then {W1,...,Wk,Vk+1,... } is an open cover of U.

Using the method of molliﬁcation and the partition of unity we will show in the following that functions in a Sobolev space can be approximated by smooth functions. We start k,p with local approximation (convergence on Wloc (U)), then we extend this idea to global approximation (convergence on W k,p(U)) and ﬁnally, requiring restrictions on the boundary ∂U, we will obtain approximation by functions belonging to C∞(U), and not just C∞(U).

3.3. Local approximation by smooth functions Let U ⊆ Rn open. Remember that for ε > 0

Uε = {x ∈ U : d(x, ∂U) > ε}. Theorem 3.15. Let u ∈ W k,p(U), 1 ≤ p < ∞. Let ε > 0 and set ε ε u (x) = (η ∗ u)(x), x ∈ Uε, where ηε is the mollifier defined in Definition 3.1. Then ε ∞ (1) u ∈ C (Uε) for all ε > 0, ε ε→0 k,p (2) u −→ u in Wloc (U).

26 k,p 1 Proof. u ∈ W (U), therefore u ∈ Lloc(U). Hence, (1) has already be shown in Lemma 3.7. Corollary 3.8 yields that for all |α| ≤ k α ε α D u (x) = ηε ∗ D u (x), for all x ∈ Uε. (3.4) we have to show that for all V ⊂⊂ U ε α ε α lim ku − uk k,p = 0 ⇐⇒ ∀ |α| ≤ k : lim kD u − D uk p = 0. ε→0 W (V ) ε→0 L (V ) Using (3.4) and Theorem 3.9 we obtain α ε α ε α α lim kD u − D ukLp(V ) = lim kη ∗ D u − D ukLp(V ) ε→0 ε→0 α ε α p = lim k(D u) − D uk p = 0. ε→0 L (V ) 3.4. Global approximation by smooth functions Theorem 3.16. Let U ⊆ Rn be open and bounded. Let u ∈ W k,p(U), 1 ≤ p < ∞. Then ∞ k,p there exists a sequence (um)m∈N in C (U) ∩ W (U) such that

lim kum − uk k,p = 0. (3.5) m→∞ W (U) Proof. Let 1 U = {x ∈ U : d(x, ∂U) > }, i ∈ . i i N Then Ui ⊆ Ui+1 and ∞ [ 1 U = {x ∈ U : d(x, ∂U) > }. i i=1

Let Vi = Ui+3 \ U i. Then #{j ∈ N : Vi ∩ Vj 6= ∅} ≤ 3. Therefore, each x ∈ U is element of at least one and at most three sets of the family (V ) . We choose V ⊂⊂ U such that i i∈N 0 [ U = Vi. i∈N0

V k Uk Vk U Vk+1 k+1 Vk+1

Vk+2 Uk+2 Vk+2

Uk+3

V k ' Vk+1 'Vk+2

Figure 3.4. The families Uk,Vk

27 ∞ ∞ Let (ξi)i=0 be a smooth partition of unity subordinate to the family of open sets (Vi)i=0, i.e. ∞ ∞ X 0 ≤ ξi ≤ 1, ξi ∈ Cc (Vi), for all i ∈ N0, ξi = 1 on U . (3.6) i=0 k,p(U) k,p Let u ∈ W . Then by Lemma 2.16 we have that ξiu ∈ W (U) and supp ξiu ⊂⊂ Vi. i ε Let δ > 0 be ﬁxed. By Theorem 3.15 we can choose εi > 0 such that u = η i ∗ (ξi u) satisﬁes

i δ ku − ξ uk k,p ≤ (3.7) i W (U) 2i+1 i supp u ⊂ Wi = Ui+4 \ Ui ⊃ Vi. (3.8) We deﬁne ∞ X v(x) := ui(x), x ∈ U. i=0 ∞ i v ∈ C (U), since for every x ∈ U we have that #{i ∈ N0 : u (x) 6= 0} ≤ 3. We have ∞ X u = u · 1 = ξiu. i=0 Therefore, ∞ ∞ X X i ku − vkW k,p(U) = ξiu − u i=0 i=0 W k,p(U) ∞ ∞ X i X −i−1 ≤ kξi u − u kW k,p(U) ≤ δ 2 = δ. i=0 i=0

Note that kvkW k,p(U) ≤ kv − ukW k,p(U) + kukW k,p(u) < ∞. Summarizing we have that ∀δ > 0 ∃v ∈ W k,p(U) ∩ C∞(U): ku − vk < δ. Wk,p(U) Remark. (1) The assumption of U to be bounded is not absolutely necessary. The same proof n holds for example if U = {x ∈ R : xn > 0}. U is unbounded but has boundary ∂U = {xn = 0}. (2) Note that Theorem 3.16 is also true for U = Rn, see [1, Theorem 3.16].

3.5. Global approximation by functions smooth up to the boundary Theorem 3.17. Let U ⊆ Rn be open and bounded and ∂U is C1. Let u ∈ W k,p(U), ∞ ∞ 1 ≤ p < ∞. Then there exists a sequence (um)m=1 in C (U) such that

lim kum − ukW k,p(U) = 0. (3.9) m→∞

28 1 Proof. Step 1: Let x0 ∈ ∂U. ∂U is C , i.e. (cf. Deﬁnition 11.12) there exists r > 0 and γ : Rn−1 −→ R ∈ C1 such that - upon relabeling and reorienting the coordinate axes if necessary - we have

B(x0, r) ∩ U = {x ∈ B(x0, r): xn > γ(x1, . . . , xn−1)}.

B(x,r) x V e n x0 'Ñn-1

r Step 2: Let V = B(x0, 2 ) ∩ U. Let 0 < ε << 1 and λ >> 1 such that for the shifted point ε x = x + λ ε en, x ∈ V, th where en is the n standard unit vector, the following holds ε B(x , ε) ⊆ U ∩ B(x0, r).

Now we define uε(x) = u(xε), x ∈ V . This is the function u translated a distance ελ in en direction. The idea is that we have ”moved up enough” so that ”there is room to mollify within U”. We can mollify the function uε within the ε - ball (i.e. we can mollify it within U). Let ε ε v (x) := η ∗ uε(x). Clearly, vε ∈ C∞(V ). The mollification vε converges towards u in W k,p(V ). This is true if and only if for all |α| ≤ k we have α ε α lim kD v − D uk p = 0. ε→0 L (V ) Indeed, α ε α α ε α α α kD v − D ukLp(V ) ≤ kD v − D uεkLp(V ) + kD uε − D ukLp(V ) −→ 0, when ε → 0. The first term on the right-hand side vanishes for ε → 0 by the argument of Theorem 3.15. The second term vanishes for ε → 0 by the fact that translation is continuous in Lp (Exercise). 1 Step 3: ∂U is C . Hence, by definition, for every x ∈ ∂U there exists rx > 0 and a n−1 continuous function γx : R −→ R such that

B(x, rx) ∩ U = {ω ∈ B(x, rx): ωn > γx(ω1, . . . , ωn−1)}. 29 Therefore, {B(x, rx): x ∈ ∂U} forms an open covering of ∂U. ∂U is compact (bounded and closed). Hence, there exists N ∈ N and x1, . . . , xN ∈ ∂U such that N [ ri ∂U ⊆ B(x , ). (3.10) i 2 i=1

ri Let Vi = B(xi, 2 ) ∩ U. Choose δ > 0. Steps 1-2 show that for every 1 ≤ i ≤ N there exists ∞ a function vi ∈ C (V i) with

k,p kvi − ukW (Vi) ≤ δ. SN We choose V0 ⊂⊂ U such that U ⊆ i=0 Vi. By Theorem 3.15 we get that there exists a ∞ function v0 ∈ C (V0) such that

kv − uk k,p ≤ δ. (3.11) 0 W (V0)

V0 ¶U

N Let (ξi)i=0 be a smooth partition of unity subordinate to the open cover r r {V ,B(x , 1 ),...,B(x , N )} 0 1 2 N 2

∞ ∞ ri PN of U, i.e. 0 ≤ ξi ≤ 1, ξ0 ∈ Cc (V0), ξi ∈ Cc (B(xi, 2 )), 1 ≤ i ≤ N and i=0 ξi = 1 on U. We deﬁne N X ∞ v := ξivi ∈ C (U). i=0 Then we have for all | α |≤ k: N N X X kv − uk = ξ (v − u) ≤ kξ (v − u)k W k,p(U) i i i i W k,p(U) i=0 W k,p(U) i=0 N X ≤ C(N, k, p) kv − uk k,p ≤ C N δ =: δ . i W (Vi) 0 i=0

Note that kvkW k,p(U) < ∞. Summarizing we have that ∞ k,p ∀δ0 > 0 ∃v ∈ C (U) ∩ W (U): kv − ukW k,p(U) < δ0. 30 Remark 3.18. 1 p X α p kξi(vi − u)k k,p = kD (ξi(vi − u))k p W (Vi) L (Vi) |α|≤k and by Leibniz’s formula

X α |Dα(ξ (v − u))(x)| = Dα−βξ (x) Dβ(v − u)(x) i i β i i β≤α X α α−β β ≤ D ξi(x) D (vi − u)(x) β β≤α α−β X α β ≤ sup D ξi(x) D (vi − u)(x) . β≤α β β≤α Hence, p α p α−β p X α β kD (ξi(vi − u))k p ≤ sup sup D ξi(x) D (u − vi) L (Vi) x∈Vi β≤α β β≤α p L (Vi) p p α−β p X α β p ≤ Ci(p, α) sup sup D ξi(x) D (u − vi) Lp(V ). x∈V β≤α β i i β≤α Summarizing we have   1   1 p p p X α p α X X α β p kD (ξi(vi − u))k p ≤ Ci(p, k) sup sup |D ξi(x)| D (u − vi)  L (Vi)  Lp(V ) |α|≤k x∈V β i |α|≤k i |α|≤k β≤α 1   p α X α p ≤ Ci(p, k) sup sup |D ξi(x)| k D (u − vi)k p  L (Vi) |α|≤k x∈V i |α|≤k 1   p X α p = Ci(p, k, Vi, ξi) k D (u − vi)k p .  L (Vi) |α|≤k

Actually an analogous proof (see [1, Theorem 3.18]) gives the following statement Theorem 3.19. Let U ⊆ Rn be open and let it have the segment property (see [1, p.54]). ∞ n k,p Then the set of restrictions to U in Cc (R ) is dense in W (U), 1 ≤ p < ∞. and the following corollary Corollary 3.20. Let 1 ≤ p < ∞. Then k,p n k,p n W0 (R ) = W (R ). (3.12)

CHAPTER 4

Extensions

In general, many properties of W k,p(U) can be inherited from W k,p(Rn) provided U is ”nice”. The goal of this section is to extend functions in the Sobolev space W k,p(U) to become functions in the Sobolev space W k,p(Rn). Indeed, we need a strong theorem. Observe for instance that extending u ∈ W 1,p(U) by setting it zero in Rn \ U will not in general work, as we thereby create such a strong discontinuity along ∂U that the extended function no longer has a weak partial derivative. We must invent a way to extend u that preserves the weak derivatives across ∂U. n Theorem 4.1 (Extension Theorem). Let k ∈ N0. Let 1 ≤ p ≤ ∞. Let U ⊂ R open and bounded and assume ∂U is Ck. Let V ⊂ Rn be open such that U ⊂⊂ V . Then there exists a linear and bounded operator k,p k,p n E : W (U) −→ W (R ) such that for all u ∈ W k,p(U): (1) Eu = u a.e. on U, (2) supp Eu ⊂ V ,

(3) kE(u)kW k,p(Rn) ≤ C(p, k, U, V ) kukW k,p(U). Definition 4.2. We call Eu an extension of u to Rn. Definition 4.3 (essential support). Let u ∈ W k,p(U). Then the support of u is given by [ supp(u) = U \ {V ⊆ U open : u = 0 a.e. on V }. If necessary we write ess supp(u) (essential support) for the support of u ∈ W k,p(U) to avoid confusion with the classical support of a continuous function. Note that for a continuous function the essential support and the classical support coincide (Exercise). Proof. Let k = 1 and 1 ≤ p < ∞. Step 1: Let x0 ∈ ∂U. Suppose that ∂U is flat near x0 and lies in the plane {xn = 0}, see figure 4.1. Then we may assume there exists δ > 0 such that + B : = U ∩ B(x0, δ) = B ∩ {xn > 0} − n B : = (R \ U) ∩ B(x0, δ) = B ∩ {xn ≤ 0}, where B = B(x0, δ). Assume that u ∈ C1(U). We define ( u(x), if x ∈ B+, u(x) = (4.1) xn − u(x1, . . . , xn−1, −xn) + 4 · u(x1, . . . , xn−1, − 2 ), if x ∈ B . 33 U

X B

X 0 X B

Figure 4.1. half-ball at the boundary

This is called a higher-order reﬂection of u from B+ to B−. Step 2: We show that u ∈ C1(B). − + We use the notation u := u B− and u := u B+ . Then we have − + lim u (x1, . . . , xn−1, xn) = lim u (x1, . . . , xn−1, xn). − + xn→0 xn→0 Obviously, for 1 ≤ i ≤ n − 1, − ∂u (x) xn = −3∂xi u(x1,..., −xn) + 4∂xi u(x1,..., − ) ∂xi 2 Hence, by the above ∂u− ∂u ∂u+ lim (x1, . . . , xn) = (x1, . . . , xn−1, 0) = lim (x1, . . . , xn) − + xn→0 ∂xi ∂xi xn→0 ∂xi For i = n we have

∂ − xn u (x) = 3∂xn u(x1,..., −xn) − 2∂xn u(x1,..., − ) ∂xn 2 and therefore, ∂u− ∂u+ lim (x1, . . . , xn) = ∂x u(x1, . . . , xn−1, 0) = lim (x1, . . . , xn) − n + xn→0 ∂xn xn→0 ∂xn Summarizing we have for all multiindices |α| ≤ 1 α + α − D u |{xn=0} = D u |{xn=0}. Step 3 Using these calculations we have

kukW 1,p(B) ≤ Cp kukW 1,p(B+), (4.2) where Cp is some constant that does not depend on u. (Exercise). Step 4 What happens if ∂U is not ﬂat near x0? We reduce the general case to the case 1 where ∂U is ﬂat near x0. We need the assumption that ∂U is C . Then for every x0 ∈ ∂U there exists an r > 0 and a C1-function γ : Rn−1 → R such that we have

U ∩ B(x0, r) = {x ∈ B(x0, r): xn > γ(x1, ..., xn−1)} 34 and ∂U ∩ B(x0, r) = {x ∈ B(x0, r): xn = γ(x1, ..., xn−1)}. There exists a bijective C1-function φ and its inverse ψ such that φ straightens out ∂U near x0”. Explicitly φ resp. ψ are given by n φ : B = B(x0, r) −→ W ⊆ R , (x1, . . . , xn) 7→ (x1, . . . , xn−1, xn − γ(x1, . . . , xn−1)).

ψ : W −→ B, (y1, . . . , yn) 7→ (y1, . . . , yn−1, yn − γ(y1, . . . , yn−1)). These two maps have the following properties:

(1) φ(∂U ∩ B) −→ {xn = 0} ∩ W , (2) φ ◦ ψ = IdB and ψ ◦ φ = IdW

U B S B W

Q B Q

Q W

Figure 4.2. Straightening out the boundary

Let y = φ(x), x = ψ(y) and u0(y) = u(ψ(y)) for y ∈ W +. Then choose a small ball Q 0 1 0 around y0 = φ(x0) and use the steps 1-3 to obtain a function u ∈ C (Q) as extension of u from Q+ to Q satisfying 0 0 ku kW 1,p(Q) ≤ C · ku kW 1,p(Q+). Let R := ψ(Q). We transform back to x-variables, and obtain an extension u : R → R, x 7→ u0(φ(x)) satisfying kukW 1,p(R) ≤ C · kukW 1,p(U), (4.3) where C is independent of u. 1 Step 5: ∂U is compact and C . Therefore, there exist ﬁnitely many open sets Ui such that N [ ∂U ⊆ Ui. i=1 U ⊂⊂ V . Hence, we can arrange that Ui ⊆ V for all i. We choose U0 ⊂⊂ U such that N [ U ⊂ Ui ⊆ V. i=0 35 N Subordinate to the cover (Ui)i=0 of U there exists a smooth partition of unity, i.e. a sequence N P of smooth functions (ξi)i=0 such that 0 ≤ ξi ≤ 1, supp ξi ⊆ Ui and ξi = 1 on U. 1 1 Note that since u ∈ C (U) we have that ξiu ∈ C (U). According to Steps 1-4 there exists an extension ξiu: Ui −→ R, such that

1,p 1,p kξiukW (Ui) ≤ CikξiukW (U). (4.4)

Note that ξ0u = u, since supp ξ0 ⊂ U0 ⊂⊂ U. We deﬁne N N X [ u(x) = ξiu(x), x ∈ Ui ⊂ V i=0 i=0 n SN and u ≡ 0 for x ∈ R \ ( i=0 Ui). Then supp u ⊂ V and

kukW 1,p(Rn) ≤ CkukW 1,p(U) (4.5) for some constant C, independent of u. Step 6: For u ∈ C1(U) and we can deﬁne our extension operator E as follows Eu = u. (4.6) The operator is linear and satisﬁes: Eu = u on U, supp Eu ⊂ V and

kukW 1,p(Rn) ≤ CkukW 1,p(U). Step 7: Let u ∈ W 1,p(U) and 1 ≤ p < ∞. We deﬁne the extension operator using a ∞ 1 density argument. By Theorem 3.17 we can choose a sequence (um)m=1 ∈ C (U) such that m→∞ kum − ukW 1,p(U) −→ 0. (4.7) By the linearity of E and equation (4.5) we have

kE(um) − E(uk)kW 1,p(Rn) ≤ C · kum − ukkW 1,p(U). ∞ 1,p ∞ 1,p n (um)m=1 is Cauchy-sequence in W (U). Hence, (E(um))m=1 is Cauchy sequence in W (R ). 1,p n 1,p n Since W (R ) is complete there exists a limit limm→∞ T um in W (R ) and we can deﬁne

Eu = lim Eum. (4.8) m→∞ The operator E : W 1,p(U) → W 1,p(Rn) defined in (4.8) is well-defined (i.e. does not depend on the choice of the sequence (um)) and satisfies the properties of the theorem. Step 8: The case p = ∞ is left to the reader. Remark 4.4. n n−1 + (1) Theorem 4.1 is also true for the half-space R+ = R × R . This is obtained by Step 1-3 of the proof. (2) Assume that ∂U is C2. Then the extension operator E constructed above is also a bounded linear operator from W 2,p(U) to W 2,p(Rn). (3) The above construction does not provide an extension for the Sobolev spaces W k,p(U), k > 2. This requires a more complicated higher-order reflection tech- nique, see e.g. [1, Chapter 4].

36 CHAPTER 5

Traces

In the following chapter we discuss the possibility of assigning ”boundary values” along ∂U to a function u ∈ W 1,p(U), assuming that ∂U is C1. If u ∈ C(U), then u clearly has values on ∂U in the usual sense, but a typical function u ∈ W 1,p(U) is in general not continuous and only defined almost everywhere in U. Since ∂U has n-dimensional Lebesgue-measure zero, there is no direct meaning we can give to the expression ”u restricted to the boundary”. To resolve this problems we need the trace operator. Theorem 5.1 (Trace Theorem). Let U ⊆ Rn be open, bounded and ∂U is C1. Let 1 ≤ p < ∞. Then there exists a linear bounded operator T : W 1,p(U) −→ Lp(∂U) such that 1,p (1) T u = u ∂U for all u ∈ W (U) ∩ C(U) (2) kT ukLp(∂U) ≤ CkukW 1,p(U) for each u ∈ W 1,p(U), with the constant depending only on p and U. Definition 5.2. We call T u the trace of u on ∂U. Remark 5.3. k,p n (1) There is a version of Theorem 5.1 for the Sobolev spaces W (U) for 1 < k < p , see e.g. [1, Theorem 5.22]. (2) There does not exist a bounded linear operator T : Lp(U) −→ Lp(∂U) p such that Tu = u ∂U whenever u ∈ C(U) ∩ L (U). Proof. Let u ∈ C1(U). Step 1: Let x0 ∈ ∂U. As in the proof of Theorem 4.1 we assume that ∂U is flat near x0 lying in the plane {xn = 0}. Choose an open ball B = B(x0, r) such that + B := U ∩ B = B ∩ {xn > 0}, − n B := (R \ U) ∩ B = B ∩ {xn ≤ 0}. ˆ r ˆ Let B = B(x0, 2 ) and Γ = B ∩ ∂U, see figure 5.1. We show that + kukLp(Γ) ≤ C(p, B )kukW 1,p(B+). (5.1) ∞ ˆ + We choose ξ ∈ Cc (B) such that ξ ≥ 0 on B and ξ = 1 on B. Then supp ξu ⊆ B 0 n−1 and ξu = u on Γ. Let x = (x1, . . . , xn−1) ∈ R . Then, by the fundamental theorem and Hölder’sinequality, we have 37 U

r B ^B

Figure 5.1.

Z ∞ p 0 p 0 |ξu (x , 0)| ≤ |(ξu)xn (x , t)|dt 0 Z ∞ p−1 0 p ≤ r |(ξu)xn (x , t)| dt 0 Z ∞ p−1 p−1 0 p 0 p ≤ r 2 |ξxn u(x , t)| + |ξuxn (x , t)| dt. 0 We integrate over Γ and obtain Z Z Z ∞ 0 p 0 p−1 0 p 0 p 0 |ξu (x , 0)| dx ≤ (2r) |ξxn u(x , t)| + |ξuxn (x , t)| dtdx Γ Rn−1 0 Z + p p ≤ C(p, B ) |u(x)| + |uxn (x)| dx B+ + ≤ C(p, B )kukW 1,p(B+).

This yields equation (5.1). Step 2: Analogously to the proof of Theorem 4.1 we can straighten out the boundary near x0 if necessary to obtain the setting in Step 1 (cf. ﬁgure 4.2). After transforming back to the original setting we obtain the estimate Z p p |u| dS ≤ CkukW 1,p(U), (5.2) Γ where Γ is some open subset of ∂U containing x0. 38 Step 3: ∂U is compact. Therefore, there exist ﬁnitely many open subsets Γi of ∂U such that N [ ∂U = Γi i=1 and by Step 1 and Step 2 we obtain

kuk p ≤ C kuk 1,p . (5.3) L (Γi) i W (U) Therefore, Z N Z N p X p X kuk p = |u| dx ≤ |u| dx = kuk p ≤ C(N, p)kuk 1,p . (5.4) L (∂U) L (Γi) W (U) ∂U i=1 Γi i=1 We deﬁne

T u = u ∂U (5.5) then our previous estimate implies

kT ukLp(∂U) ≤ C kukW 1,p(U) . (5.6) Hence, Theorem 5.1 is proven for u ∈ C1(U). 1,p ∞ ∞ 1,p Step 4: Assume now u ∈ W (U). Choose (um)m=1 ∈ C (U) ∩ W (U) such that m→∞ kum − ukW 1,p(U) −→ 0 . By the equations (5.5) and (5.6) we have

kT (um − ul)kLp(∂U) ≤ C kum − ulkW 1,p(U). ∞ p p Hence, (T um)m=1 is a Cauchy sequence in L with limit limm→∞ T um ∈ L . We deﬁne

T u = lim (T um). m→∞ ∞ The operator T u is well deﬁned (does not depend on the choice of the sequence (um)m=1), bounded and linear. ∞ The sequence (um)m=1 is constructed from u by smoothing by convolution (Theorem 1,p 3.17). If u ∈ C(U) ∩ W (U) we have by Theorem 3.9 that (um) converges uniformly to u on compact subsets of U, especially on ∂U. Therefore, (T u )∞ converges uniformly to m m=1 u ∂U on ∂U. Of special interest are functions which have trace zero. In the following theorem we examine more closely what it means for a Sobolev function to have zero trace.

Theorem 5.4 (Trace Zero Theorem). Let U ⊆ Rn be open. Assume U is bounded and ∂U is C1. Let 1 ≤ p < ∞ and u ∈ W 1,p(U). Then 1,p T u ≡ 0 ⇔ u ∈ W0 (U) Recall that 1,p ∞ ∞ m→∞ u ∈ W0 (U) ⇐⇒ ∃ sequence (um)m=1 in Cc such that kum − ukW 1,p(U) −→ 0.

39 1,p Proof. 1. ”⇐”: By the deﬁnition of the trace operator we have for u ∈ W0 (U) with ∞ (um)m=1 as above that

kT ukLp(∂U) = kT u − T um + T umkLp(∂U)

≤ kT u − T umkLp(∂U) + kT umkLp(∂U)

= kT u − T umkLp(∂U) → 0. Therefore, T u ≡ 0 with equality in Lp. We have that 1,p 1,p W0 (U) ⊆ {u ∈ W (U): T u ≡ 0}. 2. ”⇒”: We show T u ≡ 0 ⇒ u ∈ W 1,p(U). By deﬁnition T u ≡ 0 if and only if there 1 exists a sequence (um) in C (U) such that 1,p um → u in W (U) and kT umkLp(∂U) → 0, if m → ∞. (5.7) Using partition of unity and ﬂattening out ∂U as before, we may assume that n n−1 + n−1 U = R+ = R × R = R × {x ∈ R, x > 0}, (5.8) 1,p n u ∈ W (U) with compact support in R+. Recall that [ supp(u) = U \ {V ⊆ U open : u = 0 a.e. on V }. 0 n−1 + 1 Step 1: Let x ∈ R , xn ∈ R . Let (um) in C (U) be as in equation (5.7). Then Z xn 0 0 0 um(x , xn) = um(x , 0) + (um)xn (x , t)dt 0 and by the triangle inequality and inequality (11.2)

0.4

0.2 0 1/m 2/m

-5 -4 -3 -2 -1

0.8

1-zm(x) 0.6

0.4

0.2 0 1/m 2/m 0 2/m 1/m2 4 6 8 10

-0.05

-0.1 -0.15 z' (x) -0.2 m

Figure 5.2. ζm(x)

The function

wm(x) = (1 − ζm(x))u(x)

1,p n is in W (R+) and 1 supp w ⊆ {(x0, t) ∈ supp u : t > }. m m We show that

kw − uk 1,p n → 0, m → ∞. (5.10) m W (R+)

The weak partial derivatives of wm are given by

0 0 0 0 0 (wm)xn (x , xn) = uxn (x , xn)(1 − ζm(x , xn)) − mζ (mxn)u(x , xn) and for 1 ≤ i < n

=: Cp (Am + Bm) . 1,p n p n Since u ∈ W (R+) we have that uxn ∈ L (R+). By deﬁnition ζm ≤ 1 and ζm → 0 for all n x ∈ R+. Therefore, by Lebesgue domination theorem

Am → 0, if m → ∞.

We use Step 1 in order to get an estimate for Bm. By equation (5.9) we have 2 Z Z m p 0 0 p 0 Bm = m |ζ (mt)u(x , t)| dtdx Rn−1 0 2 Z m Z p 0 p 0 ≤ Cpm |u(x , t)| dx dt 0 Rn−1 2 s Z m Z Z p p−1 0 p 0 ≤ Cp m s |uxn (x , t)| dx dt ds 0 0 Rn−1 2 ! 2 Z m Z m Z p p−1 0 p 0 ≤ Cpm s ds |uxn (x , t)| dx dt 0 0 Rn−1 2 Z m Z 0 p 0 m→∞ = Cp |uxn (x , t)| dx dt −→ 0. 0 Rn−1 Hence, Z p |(wm)xn − uxn | dx → 0, ifm → ∞. n R+ Summarizing we have equation (5.10). ∞ n Step 4: We use smoothing by convolution to construct a sequence (˜um) in Cc (R+). 1 0 n−1 1 Let εm = m . We know that wm = 0 for all (x , xn) ∈ R × (0, m ) and wm has compact n n support in R+. Hence, ωm ∗ ηεm is well deﬁned on R+ and has compact support. Therefore, ∞ n ω ∗ ηεm ∈ Cc (R+) and by Theorem 3.9

kwm ∗ ηεm − wmkW 1,p → 0.

Letu ˜m = wm ∗ ηεm . Then

ku˜m − ukW 1,p ≤ ku˜m − wmkW 1,p + kwm − ukW 1,p → 0, m → ∞.

42 CHAPTER 6

Sobolev inequalities

In this chapter we prove a class of inequalities of the form

kukX ≤ CkukW k,p(U), (6.1) where X is a Banach space, i.e. we consider the question: ”If u ∈ W k,p(U), does u belong automatically to a certain other Banach space X?” Inequalities of the form (6.1) are called Sobolev type inequalities. This kind of estimates give us information on the embeddings of Sobolev spaces into other spaces. Recall that we say that a Banach space E is continuously embedded into another Banach space F , written E,→ F if there exists a constant C such that for all x ∈ E

kxkF ≤ CkxkE. (6.2) This means that the natural inclusion map i : E → F, x 7→ x is continuous. We start the investigations with the Sobolev spaces W 1,p(U) and will observe that these Sobolev spaces indeed embed into certain other spaces, but which other spaces depends upon whether 1 ≤ p < n, (6.3) p = n, (6.4) n < p ≤ ∞. (6.5) The case (6.3) is covered by the Gagliardo-Nirenberg-Sobolev inequality, see Section 6.1 and the case (6.5) is covered by the so called Morrey’s inequality, see Section (6.5).

6.1. Gagliardo-Nirenberg-Sobolev inequality Definition 6.1. If 1 ≤ p < n, the Sobolev conjugate of p is np p∗ = . n − p Note that 1 1 1 = − and p∗ > p. (6.6) p∗ p n Theorem 6.2 (Gagliardo-Nirenberg-Sobolev Inequality). Let 1 ≤ p < n. There exists a constant C, depending only on n and p such that

∗ kukLp (Rn) ≤ CkDukLp(Rn) (6.7) 1 n for all u ∈ Cc (R ).

43 Motivation. We ﬁrst demonstrate that if any inequality of the form kuk ≤ CkDuk , (6.8) Lq(Rn) Lp(Rn) ∞ n for certain constants C > 0, 1 ≤ q < ∞ and functions u ∈ Cc (R ) holds, then the number ∞ n q cannot be arbitrary. Let u ∈ Cc (R ), u 6≡ 0 and deﬁne for λ > 0 n uλ(x) := u(λx)(x ∈ R ) .

We assume that (6.8) holds and apply it to uλ, i.e. there exists a constant C such that for all λ > 0

kuλkLq(Rn) ≤ CkDuλkLp(Rn). (6.9) Now Z Z Z q q 1 q |uλ(x)| dx = |u(λx)| dx = n |u(y)| dy Rn Rn λ Rn and Z Z p Z p p p λ p |Duλ(x)| dx = λ |Du(λx)| dx = n |Du(y)| dy. Rn Rn λ Rn Hence, by (6.9) we get 1 p 1 1 q λ p kuk q n ≤ C kDuk p n λn L (R ) λn L (R ) and therefore, 1− n + n p q kukLq(Rn) ≤ Cλ kDukLp(Rn). n n If 1 − p + q 6= 0 we can obtain a contradiction by sending λ to 0 or ∞, depending on n n n n whether 1 − p + q > 0 or 1 − p + q < 0. Thus, if in fact the desired inequality (6.1) holds, n n 1 1 1 np we must necessarily have 1 − p + q = 0. This implies that q = p − n and therefore, q = n−p . Proof. Assume p = 1. Note that u has compact support. Therefore, we have for each i = 1, . . . , n and x ∈ Rn Z xi

u(x) = uxi (x1, . . . , xi−1, yi, xi+1, . . . , xn)dyi −∞ and Z ∞ |u(x)| ≤ |Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)|dyi . −∞ Then 1 n Z ∞ ! n−1 n Y |u(x)| n−1 ≤ |Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)| dyi . i=1 −∞ We integrate the above inequality with respect to x1 and obtain: 1 Z ∞ Z ∞ n Z ∞ ! n−1 n Y |u(x)| n−1 dx1 ≤ |Du(x1, . . . , xi−1, yi, xi+1, . . . , xn)| dyi dx1 −∞ −∞ i=1 −∞ 1 1 ! −1 n ! −1 Z ∞ n Z ∞ Y Z ∞ n = |Du| dy1 |Du| dyi dx1. −∞ −∞ i=2 −∞

44 1 Applying the general H¨olderinequality (Theorem 11.6) with pi = n−1 , i = 1, . . . , n − 1 we obtain

1 1 Z ∞ Z ∞ ! n−1 n Z ∞ Z ∞ ! n−1 n Y |u(x)| n−1 dx1 ≤ |Du| dy1 |Du| dx1 dyi . −∞ −∞ i=2 −∞ −∞

Now we integrate with respect to x2 and obtain

1 1 Z ∞ Z ∞ Z ∞ Z ∞ ! n−1 Z ∞ n Z ∞ Z ∞ ! n−1 n Y |u(x)| n−1 dx1dx2 ≤ |Du| dy1 dx2 |Du| dx1 dyi dx2 −∞ −∞ −∞ −∞ −∞ i=2 −∞ −∞ 1 ! −1 n Z ∞ Z ∞ n Z ∞ 1 Y n−1 = |Du| dx1 dy2 Ii dx2, −∞ −∞ −∞ i=1,i6=2 where Z ∞ Z ∞ Z ∞ I1 = |Du|dy1 and Ii = |Du|dx1dyi for i = 3, . . . , n. −∞ −∞ −∞ Applying the general H¨olderinequality once more we obtain

1 1 ∞ ∞ ∞ ∞ ! n−1 ∞ ∞ ! n−1 Z Z n Z Z Z Z |u(x)| n−1 dx1dx2 ≤ |Du| dx1 dy2 |Du| dy1 dx2 −∞ −∞ −∞ −∞ −∞ −∞

1 n ! −1 Y Z ∞ Z ∞ Z ∞ n |Du| dx1 dx2 dyi . i=3 −∞ −∞ −∞

We continue by integrating with respect to x3, . . . , xn and and using H¨older’sgeneral inequality to obtain ﬁnally

1 n Z n Z ∞ Z ∞ ! n−1 Z ! n−1 n Y |u| n−1 dx ≤ ··· |Du| dx1 . . . dyi . . . dxn = |Du|dx . (6.10) n n R i=1 −∞ −∞ R This is the Gagliardo-Nirenberg-Sobolev inequality for p=1. We consider now the case 1 < p < n. Let v := |u|γ for some γ > 1. We apply (6.10) to v. Then, by H¨older’sinequality (Theorem 11.5)

n−1 Z n Z Z γn γ γ−1 |u| n−1 dx ≤ |D|u| | dx = γ |u| |Du| dx Rn Rn Rn p−1 1 (6.11) Z ! p Z ! p (γ−1) p p ≤ γ |u| p−1 dx |Du| dx . Rn Rn γn p We choose γ so that n−1 = (γ − 1) p−1 . That is, we set p(n − 1) γ = n − p 45 γn p np ∗ in which case n−1 = (γ − 1) p−1 = n−p = p . Therefore, (6.11) becomes p−1 1 −1 n ! p ! p Z n Z Z |u|p∗ dx ≤ γ |u|p∗ dx |Du|pdx Rn Rn Rn what is equal to 1 −1 p−1 n − ! p Z n p Z |u|p∗ dx ≤ γ |Du|pdx . Rn Rn Note that n − 1 p − 1 1 − = and γ = C(n, p). n p p∗

1,p 1,p 6.2. Estimates for W and W0 , 1 ≤ p < n The Gagliardo-Nirenberg-Sobolev inequality (Theorem 6.2) gives the continuous embedding of W 1,p(U), 1 ≤ p < n, into the space Lp∗ , where p∗ is the Sobolev conjugate of p. Theorem 6.3. Let U ⊆ Rn open and bounded and suppose ∂U is C1. Assume 1 ≤ p < n and u ∈ W 1,p(U). Then u ∈ Lp∗ (U) with the estimate

kukLp∗ (U) ≤ CkukW 1,p(U), (6.12) where the constant C depends on n,p and U. In particular, we have for all 1 ≤ q ≤ p∗

kukLq(U) ≤ CkukW 1,p(U). (6.13) Proof. The Extension Theorem (Theorem 4.1) yields that there exists an extension u = Eu ∈ W 1,p(Rn) such that u = u in U, u has compact support and

kukW 1,p(Rn) ≤ CkukW 1,p(U). (6.14) Because u has compact support we know from Theorem 3.15 that there exists a sequence ∞ ∞ n (um)m=1 of functions in Cc (R ) such that 1,p n um → u in W (R ). (6.15) Now according to Theorem 6.2 we have that for all l, m ≥ 1

ku − u k ∗ ≤ CkDu − Du k . (6.16) m l Lp (Rn) m l Lp(Rn) Thus, by equation (6.15) and (6.16), p∗ um → u in L . (6.17) By the Gagliardo-Nirenberg-Sobolev inequality we have

∗ kumkLp (Rn) ≤ CkDumkLp(Rn) and hence,

kuk p∗ n ≤ CkDuk p n ≤ Ckuk . (6.18) L (R ) L (R ) W 1,p(Rn) Therefore, by the properties of the extension u we have

kuk ∗ = kuk ∗ ≤ kuk ∗ ≤ Ckuk ≤ Ckuk . Lp (U) Lp (U) Lp (Rn) W 1,p(Rn) W 1,p(U) 46 Since |U| < ∞ we obtain by H¨older’sinequality equation (6.13). Remark 6.4. The boundedness of U is not essential. Remark 4.4 gives the statement n n−1 + n (6.12) for U = R+ = R × R . Corollary 3.20 gives the statement (6.12) for U = R . n 1,p Theorem 6.5. Let U ⊆ R be open and bonded. Let u ∈ W0 (U), 1 ≤ p < n. Then we have the estimate kukLq(U) ≤ CkDukLp(U) for each q ∈ [1, p∗]. The constant depends only on p, q, n and U. In particular , for all 1 ≤ p < n,

kukLp(U) ≤ CkDukLp(U).

1,p ∞ ∞ Proof. Let u ∈ W0 . Then there exists a sequence (um)m=1 in Cc (U) such that 1,p n um → u in W (U). Now we extend each function um to be 0 on R \ U. Analogously to the above proof we get from the Gagliardo-Nirenberg-Sobolev inequality (Theorem 6.2) the following estimate

kukLp∗ (U) ≤ CkDukLp(U). Since U is bounded we have |U| < ∞ and therefore, for every 1 ≤ q ≤ p∗ the following estimate holds

kukLq(U) ≤ CkukLp∗ (U). 6.3. Alternative proof of the Gagliardo-Nirenberg-Sobolev inequality 1 n Definition 6.6 (Maximal function). Let f ∈ Lloc(R ). Then 1 Z M(f)(x) := sup |f(y)|dy r>0 |B(x, r)| B(x,r) is the maximal function of f. Theorem 6.7 (Hardy-Littlewood maximal inequality). If f ∈ L1(Rn), then for every λ > 0 kfk |{M(f) > λ}| ≤ C 1 , (6.19) 1 λ where C1 is a constant which depends only on the dimension n. We say that M(f) is of weak type (1, 1). If f ∈ Lp(Rn), 1 < p ≤ ∞, then

kM(f)kp ≤ Cp kfkp , (6.20)

where Cp depends only on n and p. Proof. See [10]. Remark 6.8. The proof of inequality (6.20) is based on a typical interpolation argument: If we have

kM(f)k1 ≤ C1 kfk1 (6.21) and

kM(f)k∞ ≤ Cp kfk∞ (6.22) 47 then it follows that

kM(f)kp ≤ Cp kfkp 1 < p < ∞. Note that M(f) does not satisfy (6.21), but the weaker estimate (6.19). Hence, we need a stronger interpolation argument: Estimate (6.19) and

kM(f)k∞ ≤ C kfk∞ yield

kM(f)kp ≤ Cp kfkp 1 < p < ∞. 1 n Definition 6.9 (Riesz potentials). Let f ∈ Lloc(R ) be a non-negative function. The Riesz potential of f of order 1 is given by Z 1−n f(y) I1(f)(x) = (| · | ∗ f)(x) = n−1 dy. Rn |x − y| α−n (The Riesz potential of order α > 0 would be Iα(f)(x) = (| · | ∗ f)(x).) Proposition 6.10. If f ∈ L1(Rn), then for all λ > 0 n kfk n−1 |{I (|f|) > λ}| ≤ C (n) 1 , (6.23) 1 1 λ where C1(n) is some positive constant that depends only on n. If f ∈ Lp(Rn), 1 < p < n, then

kI1(|f|)k np ≤ Cp(n) kfkp , (6.24) n−p where Cp(n) is some positive constant that depends only on n and p.

Proof. We may assume that f ≥ 0. Given δ > 0 we divide the integral deﬁning I1(f) into a good part and a bad part. Z f(y) Z f(y) I1(f)(x) = n−1 dy + n−1 dy = bδ(x) + gδ(x). (6.25) B(x,δ) |x − y| Rn\B(x,δ) |x − y|

We get an estimate for the ”good part” gδ(x) by H¨older’sinequality. For p > 1 we have 1 Z Z q f(y) q(1−n) gδ(x) = n−1 dy ≤ kfkp |x − y| dy , Rn\B(x,δ) |x − y| Rn\B(x,δ) p where q = p−1 . Substituting z = y − x in the second term on the right-hand side and applying integration in polar coordinates (Theorem 11.16) yields 1 1 Z q Z q |x − y|q(1−n)dy = |z|q(1−n)dz Rn\B(x,δ) Rn\B(0,δ) 1 Z ∞ Z q = rq(1−n)+n−1dS(ω)dr δ Sn−1 p−1 p−n = |∂B(0, 1)| p c(n, p)δ p . Summarizing we obtain for p > 1 p−n p gδ(x) ≤ C(n, p) kfkp δ . (6.26) 48 For p = 1 we obtain: Z f(y) 1−n 1−n gδ(x) = n−1 dy ≤ kfk1 sup |x − y| ≤ kfk1δ . Rn\B(x,δ) |x − y| y∈Rn\B(x,δ)

The ”bad part” bδ(x) can be dealt with by a maximal function argument. Let −j Bj = B(x, 2 δ) and Aj = Bj \ Bj+1, j ≥ 0.

The sets Aj, j ≥ 0 form a partition of the ball B(x, δ). Since n −(j+1) −j Aj = {y ∈ R : 2 δ < |x − y| ≤ 2 δ}, we have Z f(y) bδ(x) = n−1 dy B(x,δ) |x − y| X Z f(y) = dy |x − y|n−1 j≥0 Aj (6.27) X Z ≤ 2(j+1)(n−1)δ1−n f(y)dy j≥0 Aj X Z = 2n−1δ 2−j(2−jδ)−n f(y)dy. j≥0 Aj −j n Since f ≥ 0, |Aj| ≤ |Bj| and |Bj| = (2 δ) |B(0, 1)|, we get the following estimate X Z X 1 Z 2−j(2−jδ)−n f(y)dy ≤ c(n) 2−j f(y)dy, |B | (6.28) j≥0 Aj j≥0 j Bj where c(n) = |B(0, 1)|. Using the deﬁnition of the maximal function (Deﬁnition 6.6) we obtain X 1 Z 2−j f(y)dy ≤ 2M(f)(x). |B | (6.29) j≥0 j Bj Summarizing the estimates (6.27), (6.28) and (6.29) we obtain

bδ(x) ≤ C(n)δM(f)(x). (6.30) Putting the estimates (6.26) and (6.29) into (6.25) yields the upper bound

1− n p I1(f)(x) ≤ C(n, p) δM(f)(x) + δ kfkp . (6.31)

p/n kfkp Observe that the minimum of the right-hand side is attained with δ = C(n, p) M(f)(x) . We get, by putting the minimum into equation (6.31), the following

p 1− p n n I1(f)(x) ≤ C(n, p) kfkp (M(f)(x)) . (6.32) Hence, 2 Z np p Z n−p p |I1(f)(x)| n−p dx ≤ C(n, p) kfkp |M(f)(x)| dx. Rn Rn 49 Applying Theorem 6.7 for p > 1 yields

Z np np n−p |I1(f)(x)| n−p dx ≤ C(n, p) kfkp . Rn This is what we wanted to prove. The case p = 1 is similar. We use the weak conclusion of Theorem 6.7, which asserts that kfk ∀λ > 0 : |{M(f) > λ}| ≤ C 1 . 1 λ Hence, by equation (6.32) we have n −1 n−1 |{I1(f) > λ}| ≤ C(n)(λ kfk1) . Gagliardo-Nirenberg inequality - another proof. Proposition 6.10 gives an alternative proof for Theorem 6.2. 1 n n Proof II Theorem 6.2. Let u ∈ Cc (R ) and 1 < p < n. For every x ∈ R , s ∈ R and ω ∈ ∂B(0, 1) we have Z s d u(x + sω) − u(x) = u(x + rω)dr 0 dr Z s (6.33) = Du(x + rω) · ω dr. 0 Since u has compact support, we have lim u(x + sω) = 0 s→∞ and therefore, Z ∞ u(x) = − Du(x + rω) · ω dr. (6.34) 0 Integration over ∂B(0, 1) yields Z Z Z ∞ u(x)dS(ω) = − Du(x + rω) · ω drdS(ω). (6.35) ∂B(0,1) ∂B(0,1) 0 The left-hand side equals c(n)u(x), (6.36) where c(n) = |∂B(0, 1)|. Using Fubini’s theorem we get for the right-hand side of (6.35) Z Z ∞ Z ∞ Z Du(x + rω) · ω drdS = Du(x + rω) · ω dSdr. (6.37) ∂B(0,1) 0 0 ∂B(0,1) Using the transformation formula in Theorem 11.16 yields Z ∞ Z Z Du(x + z)z Z Du(y)(y − x) Du(x + rω) · ω dS(ω)dr = n dz = n dy. (6.38) 0 ∂B(0,1) Rn |z| Rn |x − y| Summarizing the equations (6.35), (6.36), (6.37) and (6.38) we obtain Z Du(y)(x − y) u(x) = C(n) n dy (6.39) Rn |x − y| 50 and therefore, Z |Du(y)| |u(x)| ≤ C(n) n−1 dy. (6.40) Rn |x − y| Hence, by Deﬁnition 6.9

|u(x)| ≤ C(n)I1(|Du|)(x). (6.41) Proposition 6.10 yields

kI1(|Du|)k np ≤ Cp(n) kDukp . (6.42) n−p Equations (6.41) and (6.42) give the statement

kuk np ≤ C(n, p) kDukp . n−p It remains to show the statement for p = 1. We may assume that u is non-negative. The support of u can be written as union of the sets n j j+1 Aj := x ∈ R : 2 < u(x) ≤ 2 , j ∈ Z. We consider the function  0, if u(x) ≤ 2j  j j j+1 vj(x) = u(x) − 2 , if 2 < u(x) ≤ 2 (6.43) 2j, if 2j+1 < u(x).

j−1 j j−1 Since vj(x) > 2 if and only if u(x) − 2 > 2 , we obtain j+1 j+2 j+1 j−1 |Aj+1| = | 2 < u ≤ 2 | ≤ | u > 2 | = | u > 4 · 2 | (6.44) j−1 j−1 ≤ | u > 3 · 2 | = | vj > 2 |. n The function vj is continuous on R and compactly supported. Hence, by smoothing by con- ∞ n volution we can construct a sequence in Cc (R ), which converges by Theorem 3.9 uniformly to vj. This approximation argument allows us to apply the potential estimate (6.41) to vj:

|vj(x)| ≤ C(n)I1(|Dvj|)(x). (6.45) Equation (6.45) and (6.44) yield j−1 |Aj+1| ≤ | vj > 2 | −1 j−1 ≤ | I1(|Dvj|) > C(n) 2 |. Using the weak estimate (6.23) in Proposition 6.10 for λ = C(n)−12j−1 we get

n Z n−1 −j+1 |Aj+1| ≤ C1(n) C(n)2 |Dvj|dx . Rn

The deﬁnition of vj yields that the support of Dvj is contained in Aj and Dvj = Du on Aj. Hence,

n Z ! n−1 −j |Aj+1| ≤ C(n) 2 |Du|dx . (6.46) Aj 51 By the deﬁnition of Aj we obtain Z Z n X n |u(x)| n−1 dx = |u(x)| n−1 dx n A R j∈Z j n X j+1 n−1 ≤ 2 |Aj| (6.47) j∈Z n n X j+1 n−1 = 2 n−1 2 |Aj+1|. j∈Z Equation (6.46) yields

n ! n−1 n n Z X j+1 n−1 X 2 n−1 2 |Aj+1| ≤ C(n) |Du(x)|dx A j∈Z j∈Z j n ! −1 X Z n ≤ C(n) |Du(x)|dx (6.48) A j∈Z j n Z n−1 = C(n) |Du(x)|dx . Rn Summarizing equation (6.47) and (6.48) yields the statement:

kuk n ≤ C(1, n) kDuk1 . n−1

6.4. Hölderspaces Morrey’s inequality (Section 6.5) gives the continuous embedding of the Sobolev spaces W 1,p(U), p > n into spaces of Höldercontinuous functions, the so called Hölderspaces.

Throughout this chapter let U ⊆ Rn be open, 0 < γ ≤ 1. Definition 6.11 (H¨oldercontinuous). A function u : U → R is said to be H¨older continuous with exponent γ, if there exists a constant C > 0 such that for all x, y ∈ U |u(x) − u(y)| ≤ C|x − y|γ. For γ = 1 the function is said to be Lipschitz continuous and C is called Lipschitz constant.

√ 1 Example 6.12. f(x) = x, x ∈ [0, 1], is H¨oldercontinuous with exponent γ = 2 , but it is not Lipschitz continuous. We show that 1 ∀ x, y ∈ [0, 1] : |f(x) − f(y)| ≤ |x − y| 2 , i.e. √ √ 1 | x − y| ≤ |x − y| 2 . (6.49) 52 √ √ √ √ Since |x − y| = | x + y|| x − y| we have p √ √ √ √ x − y |x| + |y| 1 x + y 1 1 | x − y| = √ √ ≤ √ √ |x − y| 2 ≤ √ √ |x − y| 2 = |x − y| 2 . x + y x + y x + y

Now assume f is Lipschitz continuous, i.e. √ √ ∃L > 0 ∀x, y ∈ [0, 1] : | x − y| ≤ L|x − y|. 1 1 Let x = n2 , y = n4 , then the following holds 1 1 1 1 ∃ L > 0 ∀ n ∈ : | − | ≤ L| − |. N n n2 n2 n4 This is equivalent to 1 ∃ L > 0 ∀ n ∈ : |n − 1| ≤ L|1 − | ≤ L. N n2 Such constant L can not exist. Therefore, f is not Lipschitz continuous. Definition 6.13. (1) If u : U → R is bounded and continuous, we write

kuk0,γ := kuk∞ + [u]0,γ.

k,γ Definition 6.14 (H¨olderspace). Let k ∈ N0 and 0 < γ ≤ 1. The H¨olderspace C (U) consists of all functions Ck(U) for which the norm

X α X α kukk,γ := kD uk∞ + [D u]0,γ |α|≤k |α|=k is finite. So the Hölderspace consists of all the functions that are k-times continuously differentiable and whose k-th partial derivatives are bounded and Hölder continuous.

k,γ Theorem 6.15. (C (U), k · kk,γ) is a Banach space.

Proof. First we need to verify that k · kk,γ indeed is a norm, so one has to check the norm properties: (1) kuk = 0 ⇒ u = 0 (2) kλuk = |λ| · kuk (3) ku + vk ≤ kuk + kvk 53 P α which follow directly from kukk := α≤k kD uk∞ being a norm and the seminorm properties of [ · ]0,γ. k,γ Now let (un)n∈N be a Cauchy sequence in C (U), i.e.

∀ ε > 0 ∃ n(ε) > 0 ∀ m, n ≥ n(ε): kum − unkk,γ ≤ ε.

k Then (un)n∈N also is a Cauchy sequence in (C (U), k · kk), where X α kukk := kD uk∞. α≤k This is a Banach space. Therefore, there exists a limit u ∈ Ck(U).

What is left to show is that for any multiindex α with |α| ≤ k: α α lim [D un − D u]0,γ = 0. n→∞ We know that

6.5. Morrey’s inequality Theorem 6.16 (Morrey’s inequality). Let n < p ≤ ∞. Then there exists a constant C, depending only on n and p such that

kukC0,γ (Rn) ≤ CkukW 1,p(Rn), (6.50) 1 n 1,p n n for all u ∈ C (R ) ∩ W (R ), where γ = 1 − p .

54 Proof. 1. We show that there exists a constant C(n) such that for any B(x, r) ⊆ Rn 1 Z Z |Du(y)| |u(y) − u(x)|dy ≤ C(n) n−1 dy. (6.51) |B(x, r)| B(x,r) B(x,r) |x − y| Let x ∈ Rn, r > 0 be ﬁxed. Let w ∈ ∂B(0, 1) and s < r. Then Z s Z s Z s d |u(x + sw) − u(x)| ≤ u(x + tw) dt = |Du(x + tw) · w|dt = |Du(x + tw)|dt. 0 dt 0 0 Hence, Z Z Z s |u(x + sw) − u(x)|dS(w) ≤ |Du(x + tw)|dt dS(w). (6.52) ∂B(0,1) ∂B(0,1) 0 We apply Fubini to the right hand side and apply integration in polar coordinates (Theorem 11.16) to obtain Z Z s Z s Z |Du(x + tw)|dt dS(w) = |Du(x + tw)|dS(w)dt ∂B(0,1) 0 0 ∂B(0,1) Z |Du(y)| = n−1 dy. B(x,s) |y − x| Now, multiplying equation (6.52) by sn−1 and integrating from 0 to r with respect to s, yields the inequality: Z r Z Z r Z n−1 n−1 |Du(y)| |u(x + sw) − u(x)|dS(w)s ds ≤ s n−1 dy ds. (6.53) 0 ∂B(0,1) 0 B(x,r) |y − x| On the left-hand side of (6.53) we apply integration in polar coordinates to obtain Z Z r Z n Z n−1 |Du(y)| r |Du(y)| |u(v) − u(x)|dv ≤ s ds n−1 dy = n−1 dy. B(x,r) 0 B(x,r) |y − x| n B(x,r) |y − x| Note that |B(x, r)| = rn|B(0, 1)| = rnC(n). Hence, we have Z Z |Du(y)| |u(v) − u(x)|dv ≤ C(n)|B(x, r)| n−1 dy. B(x,r) B(x,r) |y − x| This is equation (6.51). 2. Now ﬁx x ∈ Rn. We apply equation (6.51) as follows 1 Z 1 Z |u(x)| ≤ |u(x) − u(y)|dy + |u(y)|dy |B(x, 1)| B(x,1) |B(x, 1)| B(x,1) Z |Du(y)| 1 Z ≤ n−1 dy + |u(y)|dy B(x,1) |y − x| |B(x, 1)| B(x,1) Z |Du(y)| Z dy = n−1 dy + |u(y)| B(x,1) |y − x| B(x,1) |B(x, 1)| 1 Z Z p |Du(y)| p dy ≤ n−1 dy + |u(y)| . B(x,1) |y − x| B(x,1) |B(x, 1)|

55 dy The last inequality holds, since (B(x, 1), |B(x,1)| ) is a probability space. We apply H¨older’s inequality to the ﬁrst term on the right-hand side and obtain

p−1 1 ! p Z p Z 1 |u(x)| ≤ |Du(y)|pdy dy + C(n, p)kuk . n−1 p Lp(B(x,1)) B(x,1) B(x,1) |y − x| p−1

Hence, by integration in polar coordinates we have

Z Z 1 n−1 Z 1 Z 1 1 r (n−1)−(n−1) p − n−1 dy = C(n) dr = r p−1 dr = r p−1 dr n−1 p (n−1) p B(x,1) |y − x| p−1 0 r p−1 0 0

n−1 Since p > n, we have p−1 < 1. Therefore,

Z 1 1 − n−1 p−n r p−1 dr = C(n, p)r p−1 = C(n, p). 0 0

Summarizing we have

|u(x)| ≤ C(n, p)kukW 1,p(Rn). Since x was arbitrary, we can conclude

sup |u(x)| ≤ CkukW 1,p(Rn). (6.54) x∈Rn

3. Choose any two points x, y ∈ Rn and write r := |x − y|. Let W = B(x, r) ∩ B(y, r).

W x y z

Then 1 Z 1 Z |u(x) − u(y)| ≤ |u(x) − u(z)|dz + |u(y) − u(z)|dz |W | W |W | W C Z C Z ≤ |u(x) − u(z)|dz + |u(y) − u(z)|dz =: A + B |B(x, r)| B(x,r) |B(y, r)| B(y,r)

1 Z p 1− n p |u(x) − u(y)| ≤ Cr p |Du(z)| dz . B(x,2r) for all u ∈ C1(B(x, 2r)), y ∈ B(x, r) ⊆ Rn, n < p < ∞. The estimate is indeed valid if we integrate on the right hand side over B(x, r) instead of B(x, 2r), but the proof is a bit trickier.

1,p 1,p 6.6. Estimates for W and W0 , n < p ≤ ∞ Definition 6.18. We say u∗ is a version of a given function u provided u = u∗ a.e.

Theorem 6.19. Assume u ∈ W 1,p(Rn), n < p ≤ ∞. Then u has a version u∗ ∈ C0,γ(Rn) n for γ = 1 − p , with the estimate ku∗k ≤ Ckuk . (6.56) C0,γ (Rn) W 1,p(Rn) Proof. Use Corollary 3.20 and follow the proof of Theorem 6.20. Theorem 6.20. Let U ⊆ Rn open, bounded and suppose ∂U is C1. Assume u ∈ W 1,p(U), ∗ 0,γ n n < p ≤ ∞. Then u has a version u ∈ C (U) for γ = 1 − p , with the estimate ∗ ku kC0,γ (U) ≤ CkukW 1,p(U). (6.57)

57 Proof. According to Theorem 4.1 there exists a compactly supported function u = Eu ∈ W 1,p(Rn) such that u = u on U and kuk ≤ Ckuk . (6.58) W 1,p(Rn) W 1,p(U)

Since u has compact support, we obtain from Theorem 3.15 the existence of functions um ∈ ∞ n Cc (R ) such that ku − uk → 0. (6.59) m W 1,p(Rn) Now according to Theorem 6.16 we have for all m, l ∈ N ku − u k ≤ Cku − u k . (6.60) m l C0,γ (Rn) m l W 1,p(Rn) ∞ 1,p n 0,γ n (um)m=1 converges to u in W (R ), therefore it is Cauchy sequence in C (R ). Since this is a complete Banach space, there exists a function u∗ ∈ C0,γ(Rn) such that ku − u∗k → 0. (6.61) m C0,γ (Rn) Owing to the equations (6.60) and (6.61) we see that u = u∗ a.e. on Rn, i.e. u∗ is a version of u. Note that u = u a.e. on U hence, u∗ is a version of u on U.

∞ n Theorem 6.16 can be applied to the functions um ∈ Cc (R ), i.e. ku k ≤ Cku k m C0,γ (Rn) m W 1,p(Rn) and therefore, by the equations (6.60),(6.61) and (6.58) we have ku∗k ≤ Ckuk ≤ Ckuk . C0,γ (Rn) W 1,p(Rn) W 1,p(U)

By the deﬁnition of the norm k·kC0,γ we have ku∗k ≤ ku∗k . C0,γ (U) C0,γ (Rn) Remark 6.21. (1) Note that for p = ∞ we have the following inequality

∗ ku kC0,1(U) ≤ CkukW 1,∞(U). Hence, every u ∈ W 1,∞(U) has a Lipschitz continuous version u∗. n (2) Let n < p < ∞. Inequality (6.57) does not hold for γ ∈ (1 − p , 1]. (3) If U ⊆ Rn is bounded then ∗ ku kC0,β (U) ≤ CkukW 1,p(U) for all 0 < β ≤ γ. n 1,p (4) Let U ⊆ R open and bounded. Assume u ∈ W0 (U), n < p ≤ ∞. Then u has a ∗ 0,γ n version u ∈ C (U) for γ = 1 − p , with the estimate ∗ ku kC0,γ (U) ≤ CkDukLp .

58 6.7. General Sobolev inequalities Theorem 6.22. Let U ⊆ Rn be open and bounded with a C1 boundary. Assume u ∈ W k,p(U), 1 ≤ p < ∞. n q 1 1 k (1) If k < p , then u ∈ L (U), where q = p − n with

kukLq(U) ≤ CkukW k,p(U), (6.62) where the constant C depends only on k, p, n and U. n m,γ n (2) If k > p , then u ∈ C , where m = k − b p c − 1 and ( n n n b p c + 1 − p , if p is not an integer, γ = n any positive number < 1, if p is an integer. We have the estimate

kukCm,γ (U) ≤ CkukW k,p(U). (6.63) n k,p β 1,p Proof. Case 1: Let k < p . u ∈ W (U), then D u ∈ W (U) for all |β| ≤ k − 1 and β D u W 1,p(U) ≤ kukW k,p(U). (6.64) We apply Theorem 6.3 and obtain β β D u Lp∗ (U) ≤ C D u W 1,p(U) ≤ kukW k,p(U), for all |β| ≤ k − 2. (6.65) Using this equation we obtain

kukW k−1,p∗ (U) ≤ CkukW k,p(U). (6.66)

∗ k−1,p1 Set p1 = p and apply the same step again to u ∈ W (U). Then we obtain β D u ∗ ≤ Ckuk , for all |β| ≤ k − 2. (6.67) Lp1 (U) W k,p(U) Applying the step k-times yields the estimate

kukLq(U) ≤ CkukW k,p(U), (6.68) 1 1 k where q = p − n . n n n n Case 2: Let k > p and p is no integer. Choose ` such that ` < p < ` + 1, i.e. ` = b p c. Then ` < k. Since u ∈ W k,p(U) we have Dβu ∈ W `,p(U) for all |β| ≤ k − ` and β D u W `,p(U) ≤ kukW k,p(U). (6.69) n Since ` < p we can apply case 1 and obtain β β D u Lq(U) ≤ C D u W `,p(U), for all |β| ≤ k − `, 1 1 ` where q = p − n . Therefore, β D u Lq(U) ≤ CkukW k,p(U), for all |β| ≤ k − `. This equation yields

kukW k−`,q(U) ≤ CkukW k,p(U). (6.70) Hence, u ∈ W k−`,q(U). 59 1 1 ` np n Note that q = p − n . Hence, q = n−`p . Since p < ` + 1, we have q > n. We can apply Theorem 6.20 to Dβu ∈ W 1,q, |α| ≤ k − ` − 1 and obtain β β D u C0,γ (U) ≤ C D u W 1,q(U) ≤ CkukW k−`,q(U), (6.71) n n n where γ = 1 − q = b p c + 1 − p . Using the equations (6.70) and (6.71) yields

X α X α kuk k−b n c−1,γ = kD uk + [D u] C p ∞ 0,γ |α|≤k−b n c−1 |α|=k−b n c−1 p p

≤ CkukW k−`,q(U) ≤ CkukW k,p(U). . n n n n Case 3: Let k > p and p is an integer. Set ` = p − 1. Then ` < p < k. Analogously k−`,q 1 1 ` to the case 2 we get that u ∈ W , where q = p − n . This implies q = n. Since |U| < ∞ we have u ∈ W k−`,r for all r < n. Hence, Dαu ∈ W 1,r(U) for all |α| ≤ k − ` − 1. We apply Theorem 6.3 and obtain α α kD ukLr∗ ≤ CkD ukW 1,r(U)

≤ CkukW k−`,r(U) ≤ CkukW k−`,n . n Note that the last constant depends on |U|. Therefore, for all |α| ≤ k − ` − 1 = k − p the functions Dαu are in Ls, n ≤ s < ∞. k− n ,s p α 1,s n Hence, u ∈ W (U) and D u ∈ W for all α ≤ k − p − 1. Let n < s < ∞ and apply Theorem 6.20 n kDαuk ≤ CkDαuk , for all |α| ≤ k − − 1, C0,γ (U) W 1,s(U) p n where γ = 1 − s . Analogously to the case 2 we obtain k− n −1,γ u ∈ C p (U), 0 < γ < 1 and

kuk k− n −1,γ ≤ Ckuk ≤ Ckuk . C p W k−`,n(U) W k,p(U) n 1 Corollary 6.23. Let U ⊆ R be open and bounded with a C boundary. Let j, k ∈ N0 k+j,p n j,q 1 1 k and 1 ≤ p < ∞. Assume u ∈ W (U). If k < p , then u ∈ W (U), where q = p − n with

kukW j,q(U) ≤ CkukW k+j,p(U), (6.72) where the constant C depends only on k, p, n and U.

6.8. The borderline case We established Sobolev inequalities for n (1) k < p (Gagliardo-Nirenberg-Sobolev inequality) n (2) k > p (Morrey’s inequality) 60 k What can we say about the case n = p ? ∗ np Considering the fact that the Sobolev conjugate index p = n−p converges to ∞ as p → n, we might expect by the Gagliardo-Nirenberg-Sobolev inequality that W 1,n(U) is continuously embedded in L∞(U). This is however false for n > 1. Let U = B(0, 1) ⊆ Rn. The function 1 1,n ∞ u(x) = log log 1 + |x| belongs to W (U) but not to L (U). Lemma 6.24. Let U ⊆ Rn be open and bounded. Let u ∈ W 1,n(U). Then there exists a constant C such that

kukLp(U) ≤ CkukW 1,n(U), (6.73) where ( p = ∞, if n = 1, 1 ≤ p < ∞, if n > 1. Remark 6.25. (1) We will see in the proof that the boundedness of U is not necessary for the case n = 1. The same argument holds for U = Rn with constant equal to 1. (2) If U ⊆ Rn bounded, then the constant C depends on U and n and in the case of n > 1 additionally on some arbitrarily chosen parameter q ∈ [max(n, p), ∞). 1 Proof. 1. Let n = 1 and u ∈ Cc (U). Let x ∈ U. Then Z ∞ |u(x)| ≤ |Du(y)|dy. −∞ Therefore,

kuk∞ ≤ kDukL1 . By the same argument as in the proof of Theorem 6.3 we obtain that for u ∈ W 1,n(U) the following equation holds kukL∞(U) ≤ CkukW 1,n(U), (6.74) where C depends on n and U. 1 1 1 ns 2. Let n ≥ 2 and choose n ≤ q < ∞. Set s = p + q . Then 1 ≤ s < n and q = n−s . Note that since |U| < ∞ the following estimate holds 1 − 1 1− s s n n kukW 1,s(U) ≤ n |U| kukW 1,n(U), Theorem 6.3 yields

kukLs∗ ≤ C(n, s, U)kukW 1,s(U). Note that s∗ = q. Therefore,

kukLq(U) ≤ C(n, q, U)kukW 1,n(U). Again, since |U| < ∞ the following estimate holds

kukLp(U) ≤ C(n, q, U)kukW 1,n(U). for all 1 ≤ p ≤ q. Since q < ∞ was arbitrarily chosen, we have that for all 1 ≤ p < ∞ there exists a q ∈ [n, ∞) with p ≤ q such that

kukLp(U) ≤ C(n, q, U)kukW 1,n(U). Note that the constant does depend on the choice of q. 61 Remark 6.26. (1) The space W 1,n(Rn) embeds into the space BMO (the space of bounded mean oscillation), cf. Remark 8.4. 1,n (2) Trudinger’s inequality (Theorem 6.28) gives the embedding of W0 (U) in another Sobolev space.

n n 1 Theorem 6.27 (General case k = p ). Let U ⊆ R be open and bounded with a C k,p n q boundary. Assume u ∈ W (U), 1 ≤ p < ∞. If k = p , then u ∈ L (U) for all p ≤ q < ∞ and

kukLq(U) ≤ CkukW k,p (6.75) where the constant C depends on k, n, p, q.

6.9. Trudinger inequality Theorem 6.28 (Trudinger Inequality). Let U ⊆ Rn be open and bounded. Let u ∈ 1,n W0 (U). Then ( n ) Z |u| n−1 exp ≤ C|U|, (6.76) U c kDukn where c > 0 and C ≥ 1 are constants which depend only on n. 1,n Remark 6.29. This theorem yields that the Sobolev space W0 (U), U bounded, is n embedded in the Orlicz space Lϕ(U) with ϕ(t) = exp |t| n−1 − 1.Lϕ(U) is the space of all measurable functions u : U → R such that Z |u(x)| kukLϕ(U) = inf{c > 0 : ϕ dx ≤ 1} < ∞. U c See [1, Theorem 8.25]

n 1 Proof of Trudinger’s inequality. Let U ⊆ R be open and bounded. Let f ∈ Lloc(U). Recall (Deﬁnition 6.9) that the Riesz potential of f of order 1 is given by Z 1−n f(y) I1(f)(x) = (| · | ∗ f)(x) = n−1 dy. (6.77) U |x − y| Proposition 6.30. Let U ⊆ Rn be open and bounded, n ≤ q < ∞ and f ∈ Ln(U). Then 1− 1 + 1 1− 1 1 n q n q kI1(f)kq ≤ q |B(0, 1)| |U| kfkn. (6.78) Remark 6.31. Proposition 6.30 is the special case p = n of the following statement: Let U ⊆ Rn be open and bounded, 1 ≤ p ≤ n and np p ≤ q < . n − p p 1 1 Then, for f ∈ L (U) and δ = p − q ≥ 0, 1−δ 1 − δ 1− 1 1 −δ n n kI1(f)kq ≤ 1 |B(0, 1)| |U| kfkp. (6.79) n − δ 62 Note that Proposition 6.10 states that for 1 ≤ p < n

kI1(|f|)k np ≤ Cp(n) kfkp . n−p

and for p > n

kI1(f)k∞ ≤ C(n, p, |U|)kfkp.

Proof. Let 1 1 1 = 1 + − . s q n

Since n ≤ q < ∞ we have n 1 ≤ s < . n − 1

We show that the function h(y) = |y|1−n is in Ls(U). Let R > 0, so that |U| = |B(0,R)| = |B(0, 1)|Rn. Then Z 1 Z 1 s(n−1) dy ≤ s(n−1) dy. (6.80) U |y| B(0,R) |y|

In order to prove equation (6.80) we have to consider two cases. Case 1: B(0,R)∩U 6= ∅. Then Z Z Z |y|s(1−n)dy = |y|s(1−n)dy + |y|s(1−n)dy U U\B U∩B Z Z ≤ Rs(1−n)dy + |y|s(1−n)dy U\B U∩B Z = |U \ B|Rs(1−n) + |y|s(1−n)dy U∩B Z = |B \ U|Rs(1−n) + |y|s(1−n)dy U∩B Z Z ≤ |y|s(1−n)dy + |y|s(1−n)dy B\U U∩B Z = |y|s(1−n)dy B

Case 2: B(0,R) ∩ U = ∅. Then Z Z Z |y|s(1−n)dy ≤ Rs(1−n)dy = |U|Rs(1−n) = |B(0,R)|Rs(1−n) ≤ |y|s(1−n)dy. U U B

63 Applying integration in polar coordinates (Theorem 11.16) to equation (6.80) yields Z Z R Z |y|s(1−n)dy = rs(1−n)+n−1dS(ω)dr B(0,R) 0 Sn−1 Rs(1−n)+n = |∂B(0, 1)| s(1 − n) + n n = |B(0, 1)|Rs(1−n)+n s(1 − n) + n 1 1 = |B(0, 1)|Rsn(1/n−1+1/s) s 1/n − 1 + 1/s q sn = |B(0, 1)|R q , s 1 1 1 n 1 1 1 where we used s = 1 + q − n . Since |U| = |B(0, 1)|R , s ≥ 1 and s = 1 + q − n the following estimates hold 1 q s 1 n khk ≤ |B(0, 1)| s R q s s 1 1 − 1 1 ≤ q s |B(0, 1)| s q |U| q 1+ 1 − 1 1− 1 1 = q q n |B(0, 1)| n |U| q .

1 1 1 We apply Young’s theorem (Theorem 11.11) with s + n = 1 + q and obtain by the above estimate

kI1(f)kq = kh ∗ fkq ≤ khkskfkn 1+ 1 − 1 1− 1 1 q n n q ≤ q |B(0, 1)| |U| kfkn. n n qn Proof of Theorem 6.28. Let U ⊆ R and f ∈ L (U). Let q ≥ n, then n−1 ≥ n and therefore, Proposition 6.30 asserts

1− 1 + 1 1− 1 1 0 n qn0 n qn0 kI1(f)kqn0 ≤ (qn ) |B(0, 1)| |U| kfkn, (6.81) 0 n 0 1 where n = n−1 . Since qn (1 − n ) = q, we get

Z 0 0 qn0 0 qn0− qn +1 qn0− qn qn0 n n |I1(f)(x)| dx ≤ (qn ) |B(0, 1)| |U|kfkn U (6.82) 0 q+1 q qn0 = (qn ) |B(0, 1)| |U|kfkn . Hence, Z qn0 0 q |I1(f)(x)| 0 qn |B(0, 1)| dx ≤ qn n0 |U|. (6.83) U c1kfkn c1 We show the following estimate by applying the equation (6.83) Z N kn0 X 1 |I1(f)(x)| dx ≤ C2|U|. (6.84) k! c1kfk U k=0 n 64 We divide the sum on the left-hand side into two parts

Z N kn0 N Z kn0 X 1 |I1(f)(x)| X 1 |I1(f)(x)| dx = dx k! c1kfk k! c1kfk U k=0 n k=0 U n n−1 Z kn0 N Z kn0 X 1 |I1(f)(x)| X 1 |I1(f)(x)| = dx + dx k! c1kfk k! c1kfk k=0 U n k=n U n =: I + II. (6.85)

Applying (6.83) yields

N Z kn0 ∞ k+1 0 k X 1 |I1(f)(x)| 0 X k n |B(0, 1)| II = dx ≤ |U|n n0 k! U c1kfkn k! c1 k=n k=1 (6.86) ∞ k X kk n0|B(0, 1)| = |U|n0 . (k − 1)! cn0 k=1 1 Stirling’s formula yields

kk k ≤ √ ek ≤ e2k, k ≥ 1. (k − 1)! 2πk Summarizing we obtain

∞ k X e2n0|B(0, 1)| II ≤ |U|n0 . (6.87) cn0 k=1 1

2 0 n0 We choose c1(n) so that e n |B(0, 1)| < c1 . Then the sum on the right-hand side converges and there exists a constant C(n) such that

II ≤ C(n)|U|. (6.88)

n0 In order to get an estimate for I we set g(x) = |I1(f)(x)| . Then, by H¨older’sinequality, c1kfkn we obtain for 1 ≤ k < n:

0 Z |I (f)(x)|kn Z 1 dx = |g(x)|kdx U c1kfkn U k Z n n 1 − 1 ≤ |g(x)| dx |U| k n U k Z nn0 ! n 1 − 1 |I1(f)(x)| = |U| k n dx . U c1kfkn

65 We apply equation (6.83) to the integral on the right-hand side and obtain kn0 n k Z 0 n |I1(f)(x)| 1− k nn |B(0, 1)| n 0 dx ≤ |U| nn n0 |U| U c1kfkn c1 k 0 k+ k |B(0, 1)| n = |U|(nn ) n0 c1 k 0 k+ k |B(0, 1)| ≤ |U|(nn ) n e2n0|B(0, 1)| = c(n)ke−2k|U|. Summarizing we obtain n−1 Z kn0 X 1 |I1(f)(x)| I = |U| + dx k! c1kfk k=1 U n n−1 X c(n)ke−2k (6.89) ≤ |U| + |U| k! k=1 = |U|C(n), where C(n) ≥ 1. Collectin the equations (6.85), (6.88) and (6.89) we obtain that there exist constants c1(n) and C2(n) such that Z N kn0 X 1 |I1(f)(x)| dx ≤ C2|U|. (6.90) k! c1kfk U k=0 n The monotone convergence theorem yields ( n ) Z n−1 |I1(f)(x)| exp dx ≤ C2|U|. (6.91) U c1kfkn ∞ Let u ∈ Cc (U). Then equation (6.41) yields

|u(x)| ≤ C(n)|I1(Du)(x)| and by (6.91) we obtain ( n ) ( n ) Z n−1 Z n−1 |u| I1(|Du|) exp ≤ exp dx ≤ C2(n)|U|, (6.92) U c1 kDukn U c2kDukn where C2(n) ≥ 1.

66 CHAPTER 7

Compact embeddings

The Gagliardo-Nirenberg-Sobolev inequality shows that W 1,p(U) is continuously embedded into Lp∗ (U), if 1 ≤ p < n. Now we show that W 1,p(U) is in fact compactly embedded into some Lq(U) space. Definition 7.1 (compactly embedded). Let X and Y be Banach spaces, X ⊂ Y . We say X is compactly embedded in Y if and only if the operator Id: X → Y, x 7→ x is continuous and compact, i.e.

(1) ∃ C ∀x ∈ X : kxkY ≤ CkxkX , ∞ (2) for all sequences (xn)n=1 in X with supn kxnkX ≤ ∞ there exists a subsequence ∞ i→∞ (xni )i=1 and y ∈ Y such that kI(xni ) − ykY −→ 0. Theorem 7.2 (Rellich-Kondrachov Compactness Theorem). Let U ⊆ Rn open and bounded and let ∂U be C1. Let 1 ≤ p < n. Then W 1,p(U) ⊂⊂ Lq(U), for all 1 ≤ q < p∗. Proof. We ﬁx q ∈ [1, p∗). Let u ∈ W 1,p(U). Theorem 6.3 yields

kukLq ≤ CkukW 1,p(U). Hence, the operator Id: W 1,p → Lq is continuous. ∞ 1,p We have to show compactness. Let (ûm)m=1 ∈ W (U) and supm kuˆmkW 1,p(U) ≤ A. ∞ ∞ We show that there exists a subsequence (ûmk )k=1 of the bounded sequence (ûm)m=1 and a q k→∞ q u ∈ L (U) so that kuˆmk − ukL (U) −→ 0. By the extension theorem we may assume that ∞ 1,p n (1)( um)m=1 is in W (R ) with um U =u ˆm, (2) for all m ∈ N there exists V with U ⊂⊂ V such that supp um ⊂ V , (3) supm kumkW 1,p(Rn) < ∞. We first consider the smooth functions

ε ∞ n um = ηε ∗ um ∈ Cc (R )(ε > 0, m ∈ N). ε We may assume that for all m ∈ N the support of um is in V . Statement 1: ε ε→0 lim sup kum − umkLq(V ) −→ 0 . ε→0 m∈N 67 Veriﬁcation: If um is smooth, then Z ε um(x) − um(x) = η(y)(um(x − εy) − um(x))dy B(0,1) Z Z 1 d = η(y) um(x − εty)dt dy B(0,1) 0 dt Z Z 1 = −ε η(y) Dum(x − εt) · y dt dy. B(0,1) 0 Thus, Z Z Z 1 Z ε |um(x) − um(x)|dx ≤ ε η(y) |Dum(x − εty)| dx dt dy V B(0,1) 0 V Z ≤ ε |Dum(z)|dz. V ∞ n ε Summarizing we have for um ∈ Cc (R ) with supp um ∈ V the estimate ε kum − umkL1(V ) ≤ εkDumkL1(V ). (7.1) 1,p By approximation (Theorem 3.17) this estimate holds for um ∈ W (V ). Since V is open and bounded, we obtain ε kum − umkL1(V ) ≤ εkDumkL1(V ) ≤ εCkDumkLp(V ).

By assumption we have that supm kumkW 1,p(V ) < ∞. Therefore, ε lim sup kum − umk 1 = 0. (7.2) ε→0 L (V ) m∈N Note that 1 ≤ q < p∗. Let 0 ≤ θ ≤ 1 such that 1 1 − θ θ = + . q 1 p∗ We apply the interpolation theorem for Lp-norms (Theorem 11.10) to obtain

ε ε 1−θ ε θ kum − umkLq(V ) ≤ kum − umkL1(V )kum − umkLp∗ (V ). Theorem 6.3 yields

ε ε 1−θ ε θ kum − umkLq(V ) ≤ kum − umkL1(V )kum − umkW 1,p(V ) and by equation (7.2) ε lim sup kum − umk q = 0. (7.3) ε→0 L (V ) m∈N

ε ∞ ∞ n Statement 2: Let ε > 0 be ﬁxed. The sequence (um)m=1 in Cc (R ) is uniformly bounded and uniformly equicontinuous, i.e. ε (1) supm kumk∞ < ∞ n ε ε (2) ∀η > 0 ∃δ > 0 ∀m ∈ N ∀x, y ∈ R : |x − y| < δ ⇒ |um(x) − um(y)| < η. 68 Veriﬁcation: Let x ∈ Rn. Z Z ε ε ε |um(x)| ≤ η (x − y)|um(y)|dy ≤ sup |η (x)| |um(y)|dy B(x,ε) x∈Rn V 1 1 1 C ≤ ku k 1 ≤ |V | p ku k p = . εn m L (V ) εn m L (V ) εn Hence, ε C sup kumk∞ ≤ n . (7.4) m∈N ε By Lemma 3.7 and the Minkowski inequality for integrals (Theorem 11.4) we have Z ε ε 1 1 C p p |Dum(x)| ≤ |Dη (x − y)| |um(y)|dy ≤ n+1 |V | kumkL (V ) = n+1 . (7.5) B(x,ε) ε ε Hence, ε C sup kDumk∞ ≤ n+1 . (7.6) m∈N ε Equation (7.6) yields n ε ε ∀η > 0 ∃δ > 0 ∀m ∈ N ∀x, y ∈ R : |x − y| < δ ⇒ |um(x) − um(y)| < η.

ε ∞ The sequence (um)m=1 satisﬁes the requirements of the Arzela-Ascoli compactness criterion (Theorem 11.18), which asserts that for the uniformly bounded and uniformly equicon- ε tinuous family of functions (um) there exists a subsequence that converges uniformly to a continuous function on compact subsets of Rn, i.e. ε ∀ε > 0 ∃Nε ⊆ N, #Nε = ∞ :(uj)j∈Nε converges uniformly on V. (7.7) ε q This implies that (uj)j∈N ε converges in L (V ) (1 ≤ q ≤ ∞). Summarizing we have 1 ∀` ∈ ∃ε > 0 ∀ε < ε ∀m ∈ : kuε − u k ≤ (7.8) N ` ` N m m Lq(V ) ` ε ε 1 ∀ε > 0 ∃Nε ⊆ ∀` ∈ ∃N` ⊆ Nε ∀i, j ∈ N` : u − u ≤ . (7.9) N N j i Lq(V ) ` We combine equation (7.8) and (7.9) to obtain 3 ∀` ∈ ∃N ⊆ ∀i, j ∈ N : ku − u k ≤ . (7.10) N ` N ` i j Lq(V ) ` We apply Cantor’s diagonal argument. Note that by deﬁnition

N1 ⊃ N2 ⊃ N3 ⊃ · · · ⊃ N` ⊃ ....

Let di = min Ni. Then dj ∈ Ni for all j ≥ i and by equation (7.10) we have for all i, j ∈ N: 1 1 ud − ud q ≤ max{ , }. i j L (V ) i j Then we have ∀ε > 0 ∃N ∈ ∀i, j ≥ N : u − u < ε. N di dj Lq(V ) 69 q q q Hence, (udi )i∈N is Cauchy sequence in L (V ) with limit u ∈ L (V ), i.e. there exists u ∈ L (V ) such that

kudi − ukLq(V ) → 0, if i → ∞. Since U ⊆ V we have

kudi − ukLq(U) → 0, if i → ∞. (7.11) The Arzela-Ascoli theorem (Theorem 11.18) gives the compact embedding of W 1,p(U), n < p ≤ ∞ in Lq, 1 ≤ q ≤ ∞. Theorem 7.3. Let U ⊆ Rn open and bounded and let ∂U be C1. Let n < p ≤ ∞. Then W 1,p(U) ⊂⊂ Lq(U), for all 1 ≤ q ≤ ∞. Sketch of the proof. By the Arzela-Ascoli theorem (Theorem 11.18) we obtain that C0,γ(U) ⊂⊂ C(U) for all 0 < γ ≤ 1. Then use Morrey’s inequality (Theorem 6.20) to obtain the statement. Lemma 6.24 and Theorem 7.2 give the statement for the borderline case p = n: Theorem 7.4. Let U ⊆ Rn open and bounded and let ∂U be C1. Then W 1,n(U) ⊂⊂ Lq(U), for all 1 ≤ q ≤ n. Proof. Lemma 6.24 yields the continuous embedding. Choose a bounded sequence ∞ 1,n (um)m=1 in W (U). Then, since U is bounded, we have that for every p < n the sequence ∞ 1,p p∗ (um)m=1 is bounded in W (U). Theorem 7.2 asserts that there exists a limit u ∈ L (U) such that

kum − ukLp∗ (U) → 0. n ∗ If we choose 2 < p < n, then n < p and we have

kum − ukLn(U) → 0. Remark 7.5. Summarizing we have by Theorem 7.2, Theorem 7.3 and Theorem 7.4 the following statement: W 1,p(U) ⊂⊂ Lp, (7.12) for all 1 ≤ p ≤ ∞. Note also that 1,p p W0 (U) ⊂⊂ L , (7.13) for all 1 ≤ p ≤ ∞, even if we do not assume ∂U to be C1.

70 CHAPTER 8

Poincar´e’sinequality

The inequality

kukLp∗ ≤ CkDukLp (8.1) does not hold for every u ∈ W 1,p(U), where U ⊆ Rn is open and bounded. The Gagliardo- 1 n Nirenberg inequality (Theorem 6.2) states that it holds for u ∈ Cc (R ) and hence for u ∈ 1,p n 1,p n W (R ). Theorem 6.13 states that it holds for u ∈ W0 (U), U ⊆ R open. If we consider an arbitrary smooth function that is constant and nonzero on the unit ball B(0, 1) and zero outside, we have 1 p∗ kDukLp = 0 and kukLp∗ = |B(0, 1)| .

However, if we replace the integrand on the left-hand side of (8.1) by ku − (u)U kLp , where Z dx (u)U = u(x) (mean value of u in U) U |U| we obtain an inequality that holds for all u ∈ W 1,p(U).

8.1. General formulation and proof by contradiction Theorem 8.1 (Poincar´e’sInequality). Let U ⊆ Rn be open, bounded and connected with a C1-boundary ∂U. Let 1 ≤ p ≤ ∞. Then there exists a constant C, depending only on n, p and U, such that

ku − (u)U kLp(U) ≤ CkDukLp(U) (8.2) for all u ∈ W 1,p(U). Proof. By contradiction. We assume that the statement is not true, i.e. 1,p ∀k ∈ N ∃uk ∈ W (U): kuk − (uk)U kLp(U) > kkDukkLp(U). (8.3) We deﬁne uk − (uk)U vk := . kuk − (uk)U kLp(U)

Then kvkkLp(U) = 1 and (vk)U = 0. The gradient of vk

Duk Dvk = , kuk − (uk)U kLp(U) satisﬁes by assumption (8.3)

kDukkLp(U) 1 kDvkkLp(U) = < . (8.4) kuk − (uk)U kLp(U) k 71 Hence, 1 kv k 1,p ≤ C(n, p) kDv k p + kv k p ≤ C(n, p) 1 + k W (U) k L (U) k L (U) k and

sup kvkkW 1,p(U) ≤ 2C(n, p). k∈N By Remark 7.5 we have that

W 1,p(U) ⊂⊂ Lp(U), 1 ≤ p ≤ ∞.

∞ p By deﬁnition there exists a subsequence (vkj )j=1 and a v ∈ L (U) with kvkLp(U) = 1 and (v)U = 0 such that

lim vk − v p = 0. j→∞ j L (U)

∞ Let φ ∈ Cc (U). Then, using Lebesgue’s Theorem and the deﬁnition of the weak derivative, we have Z Z Z v φx dx = lim vk φx dx = − lim (vk )x φ dx = 0, i j→∞ j i j→∞ j i

p where the last equality follows from limj→∞ kDvkj kL (U) = 0. Hence, Dv = 0. Since U is connected, Proposition 8.2 implies that v is constant a.e on U. As (v)U = 0 we have v = 0 a.e. on U, which is a contradiction to kvkLp(U) = 1.

Proposition 8.2. Let U ⊆ Rn open and bounded and connected. Let u ∈ W 1,p(U) and Du = 0 a.e. in U. Then u is constant a.e. on U.

Proof. Step 1: Let ε > 0. We consider the smooth functions

ε ∞ u = ηε ∗ u ∈ C (Uε), where Uε = {x ∈ U : d(x, ∂U) > ε}. Corollary 3.8 yields

Dxi (uε) = ηε ∗ Dxi u.

Hence, by assumption Dxi uε = 0 a.e. on Uε. Consequently, uε is constant on each connected subset of Uε. Step 2: Choose x, y ∈ U. Since U is connected there exists a polygonal path Γ ⊆ U that connects x and y. Let δ = minz∈Γ d(z, ∂U) and ε < δ. Then Γ ⊆ Uε and x, y lie in the ε ε same connected subset of Uε. Hence, u (x) = u (y) = const. Step 3: u ∈ W 1,p(U). Theorem 3.9 yields that

uε −→ε→0 u a.e. on U.

Hence, u is constant a.e. on U. 72 8.2. Poincar´e’sinequality for a ball Theorem 8.3 (Poincar´e’sinequality for a ball). Let 1 ≤ p ≤ ∞. Then there exists a constant C, that depends only on n and p, such that

ku − (u)B(x,r)kLp(B(x,r)) ≤ C r kDukLp(B(x,r)), (8.5) for each ball B(x, r) ⊆ Rn and each function u ∈ W 1,p(B(x, r)). Proof. Let U = B(0, 1) and u ∈ W 1,p(U). Theorem 8.1 yields the estimate

ku − (u)B(0,1)kLp(B(0,1)) ≤ CkDukLp(B(0,1)). (8.6) Let now u ∈ W 1,p(B(x, r)). We deﬁne v(y) = u(x + ry), y ∈ B(0, 1). Then v ∈ W 1,p(B(0, 1)) and by equation (8.6) we have

kv − (v)B(0,1)kLp(B(0,1)) ≤ CkDvkLp(B(0,1)). (8.7) Changing variables, we recover equation (8.5). Remark 8.4. Let u ∈ W 1,n(Rn) and B(x, r) ⊆ Rn. Then Theorem 8.3 yields 1 1 Z n Z n n dy n u(y) − (u)B(x,r) ≤ Cr |Du(y)| dy B(x,r) |B(x, r)| B(x,r) Cr ≤ kDuk 1 Ln(Rn) |B(x, r)| n C = kDuk . 1 Ln(Rn) |B(0, 1)| n By H¨older’sinequality we obtain for the left-hand side 1 Z Z n dy n dy u(y) − (u)B(x,r) ≤ u(y) − (u)B(x,r) . B(x,r) |B(x, r)| B(x,r) |B(x, r)| Hence, Z dy u(y) − (u) ≤ CkDuk , (8.8) B(x,r) Ln(Rn) B(x,r) |B(x, r)| where C depends only on n. 1 n Space of bounded mean oscillation. A function f ∈ Lloc(R ) is called of bounded mean oscillation if Z dy sup f(y) − (f)B(x,r) < ∞. (8.9) B(x,r)⊆Rn B(x,r) |B(x, r)| The space of all such functions is called the space of functions of bounded mean oscillation (BMO( n)) and the left-hand side of equation (8.9) deﬁnes a norm k·k on this space. R BMO(Rn)

Therefore, we have that W 1,n(Rn) is continuously embedded into BMO(Rn) with kuk ≤ CkDuk ≤ Ckuk . (8.10) BMO(Rn) Ln(Rn) W 1,n(Rn)

73 8.3. Poincar´e’sinequality - an alternative proof Proof II of Theorem 8.3. Let B ⊆ Rn be a ball with radius R. We show that for u ∈ C∞(B) the following estimate holds Z |Du(y)| |u(x) − (u)B| ≤ C(n) n−1 dy, x ∈ B. (8.11) B |x − y|

y−x Let x, y ∈ B and ω = |x−y| . Then

Z |x−y| d u(x) − u(y) = − u(x + rω)dr 0 dr Z |x−y| = − Du(x + rω) · ω dr. 0 We integrate over B with respect to y and obtain Z Z |x−y| |B|(u(x) − (u)B) = − Du(x + rω) · ω drdy. B 0 By integration in polar coordinates (Theorem 11.16) on the right-hand side we obtain equation (8.11). Using Deﬁnition 6.9 we get the potential estimate

|u(x) − uB| ≤ C(n)I1(|∇u|)(x). (8.12) Case: n = 1: Equation (8.11) yields Z |u(x) − (u)B| ≤ C(n) |Du(y)|dy. (8.13) B Hence,

1 Z p p 1 p |u(x) − (u)B| ≤ C(n)|B| kDukL1 . B

1 1 We apply H¨older’sinequality to the right-hand side with exponents p + 1 − p = 1 and obtain

1 Z p p 1 1− 1 p p |u(x) − (u)B| ≤ C(n)|B| |B| kDukLp B

= C(n)|B|kDukLp = 2C(n) R kDukLp .

Case n > 1: 1. Let 1 < p < n. We apply Proposition 6.10 to the potential estimate (8.12) and obtain

1 Z p∗ p∗ |u(x) − (u)B| ≤ C(n)kI1(|Du|)kLp∗(B) ≤ C(n)kDukLp(B). B

74 p p H¨older’sinequality applied to the left-hand side with exponents p∗ + 1 − p∗ = 1 yields

1 1 ∗ Z p 1 1 Z ∗ p p − ∗ p |u(x) − (u)B| ≤ |B| p p |u(x) − (u)B| B B 1 n ≤ |B| C(n)kDukLp(B)

= C(n) R kDukLp(B). 2. Let n ≤ p < ∞. Choose 1 < q < n so that q∗ ≥ p. (This is always possible!!) Then, by the above computations

1 Z p∗ q∗ |u(x) − (u)B| ≤ C(n)kDukLq(B). B p p Applying H¨older’sinequality to the left-hand side with exponents q∗ + 1 − q∗ = 1 yields

1 1 ∗ Z p 1 1 Z ∗ q p − ∗ q |u(x) − (u)B| ≤ |B| p q |u(x) − (u)B| B B 1 − 1 p q∗ ≤ |B| C(n)kDukLq(B). q q Now applying H¨older’sinequality to the right-hand side with exponents p + 1 − p = 1 yields

1 Z p p 1 − 1 1 − 1 p q∗ q p |u(x) − (u)B| ≤ |B| |B| C(n)kDukLp(B) B 1 − 1 q q∗ = |B| C(n)kDukLp(B) 1 n = |B| C(n)kDukLp(B)

= C(n) R kDukLp(B). 3. Let p = ∞. Then equation (8.11) yields Z 1 |u(x) − (u)B| ≤ C(n)kDuk∞ n−1 dy B |x − y| Integration in polar coordinates (Theorem 11.16) gives

|u(x) − (u)B| ≤ C(n)RkDuk∞. Hence,

ku − (u)Bk∞ ≤ C(n)RkDuk∞. 4. Let p = 1. We apply the same procedure as in the case p = 1 in the alternative proof of the Gagliardo-Nirenberg inequality (Proof II of Theorem 6.2 in Section 6.3):

Let h(x) = u(x) − (u)B, x ∈ B. We set h+(x) = max {h(x), 0} and h−(x) = max {−h(x), 0}. In the following let h = h+ or h = h−. The support of h can be written as union of the sets n j j+1 Aj := x ∈ R : 2 < h(x) ≤ 2 , j ∈ Z. 75 We consider the function  0, if h(x) ≤ 2j  j j j+1 vj(x) = h(x) − 2 , if 2 < h(x) ≤ 2 (8.14) 2j, if 2j+1 < h(x).

j−1 j j−1 Since vj(x) > 2 if and only if h(x) − 2 > 2 , we obtain

j+1 j+2 j+1 j−1 |Aj+1| = | 2 < h ≤ 2 | ≤ | h > 2 | = | h > 4 · 2 | (8.15) j−1 j−1 ≤ | h > 3 · 2 | = | vj > 2 |.

The function vj is continuous on B. Hence, by smoothing by convolution we can construct a sequence of smooth functions which converges by Theorem 3.9 uniformly to vj on B. This approximation argument allows us to apply the potential estimate (8.12) to vj and obtain

|vj(x) − (vj)B| ≤ C(n)I1(|Dvj|)(x). (8.16)

Equation (8.16) and (8.15) yield

j−1 |Aj+1| ≤ | vj > 2 | j−1 ≤ | vj − (vj)B > 2 | j−1 ≤ | |vj − (vj)B| > 2 | −1 j−1 ≤ | I1(|Dvj|) > C(n) 2 |.

Using the weak estimate (6.23) in Proposition 6.10 for λ = C(n)−12j−1 we get

n Z n−1 −j+1 |Aj+1| ≤ C1(n) C(n)2 |Dvj|dx . Rn

The deﬁnition of vj yields that the support of Dvj is contained in Aj and Dvj = Dh on Aj. Hence,

n Z ! n−1 −j |Aj+1| ≤ C(n) 2 |Dh|dx . (8.17) Aj

By the deﬁnition of Aj we obtain Z Z n X n |h(x)| n−1 dx = |h(x)| n−1 dx n A R j∈Z j n X j+1 n−1 ≤ 2 |Aj| (8.18) j∈Z n n X j+1 n−1 = 2 n−1 2 |Aj+1|. j∈Z 76 Equation (8.17) yields n ! n−1 n n Z X j+1 n−1 X 2 n−1 2 |Aj+1| ≤ C(n) |Dh(x)|dx A j∈Z j∈Z j n ! −1 X Z n ≤ C(n) |Dh(x)|dx (8.19) A j∈Z j n Z n−1 = C(n) |Dh(x)|dx . Rn Equation (8.18) and (8.19) give the estimate + + h n ≤ C(n) Dh . L n−1 L1 The same argument holds for h−. Hence, we have + − + − khk n = h − h n ≤ h n + h n L n−1 L n−1 L n−1 L n−1 Z + − ≤ C(n) Dh (x) + Dh (x) dx B Z = C(n) |Dh(x)|dx B

= C(n)kDhkL1 .

= C(n) R kDukL1(B).

CHAPTER 9

Fourier transform

Definition 9.1 (L1-Fourier transform). For f ∈ L1(Rn) we deﬁne its Fourier transform Z fb(ξ) = e−2πix·ξf(x)dx (9.1) Rn and its inverse Fourier transform Z f(ξ) = e2πix·ξf(x)dx. (9.2) Rn

Remark 9.2. Since e ±2πixξ q= 1 and u ∈ L1(Rn), these integrals converge absolutely for all ξ ∈ Rn and deﬁne bounded functions: Z Z ±2πixξ ±2πixξ | f(x)e dx| ≤ |e ||f(x)|dx = kfk1 .

Theorem 9.3 (Fourier Inversion). Let f, fb ∈ L1(Rn). Then Z 2πiξ·x n f(x) = fb(ξ)e dξ, for a.e. x ∈ R . (9.3)

Proof. See[5]. Remark 9.4. Let f, f,b f ∈ L1(Rn). Then we have

b n q fb = f = f a.e. in R . (9.4) q 2 The Fourier transform on L . We intendq now to extend the deﬁnition of the Fourier transform and its inverse to L2(Rn). Theorem 9.5 (Plancherel). Let f ∈ L1(Rn) ∩ L2(Rn). Then f,b f ∈ L2(Rn) and

f = f = kfk . (9.5) b L2(Rn) L2(Rn) L2(Rn) q Proof. See [4]. q In view of the equality (9.5) we can deﬁne the Fourier transform of f ∈ L2(Rn) as follows. 1 n 2 n 2 n ∞ n Choose a sequence (fk) ∈ L (R ) ∩ L (R ) with fk → f in L (R ). Note that Cc (R ) is dense in Lp(Rn), 1 ≤ p < ∞. According to (9.5) we have

fbk − fbj = f\k − fj = kfk − fjkL2 . L2 L2 2 n 2 Hence, (fbk) is Cauchy sequence in L (R ) with limit in L . We deﬁne

fb = lim fbk. k→∞ 79 Note that the limit does not depend on the particular choice of the sequence (fk). We similarly deﬁne f. This gives the following theorem.

2 2 Theorem 9.6q . There exists a unique linear and bounded operator F : L → L , such that 2 (1) kFfk2 = kfk2 for f ∈ L , (2) Ff = fb for f ∈ L1 ∩ L2.

Theorem 9.7 (Properties). Let f ∈ L2(Rn) and τ ∈ Rn.

(1) Let fτ (x) = f(x − τ). Then

−2πiτ ξ fbτ (ξ) = e fb(ξ). (9.6)

2πix τ (2) Let eτ (x) = e . Then

edτ f(ξ) = fb(ξ − τ). (9.7)

ε −n 1 (3) Let f (x) = ε f( ε x). Then

fbε(ξ) = fb(εξ) (9.8)

(4) Let Dαf ∈ L2(Rn) for some multiindex α. Then

Ddαf(ξ) = (2πi)|α|ξαfb(ξ). (9.9)

(5) Let g ∈ L2(Rn). Then

f[∗ g(ξ) = fb(ξ) gb(ξ). (9.10) (6) Let g ∈ L2(Rn). Then Z Z f(x)g(x)dx = fb(x)gb(x)dx. Rn Rn ∞ n Proof. We prove (4) and (6) only. Let f ∈ Cc (R ). Then Z Ddαu(ξ) = e−2πix·ξDαf(x)dx Rn Z = (−1)|α| Dαe−2πix·ξf(x)dx Rn Z = (−1)|α|(−2πiξ)α e−2πix·ξf(x)dx Rn = (2πi)|α|ξαfb(ξ).

2 n ∞ n 2 n ∞ Let now f ∈ L (R ). Since Cc (R ) is dense in L (R ), there exists a sequence (fk)k=1 ⊆ ∞ n 2 α 2 α ∞ ∞ Cc (R ) that converges to f in L . Let f be the L -limit of the sequence (D fk)k=1 ⊆ Cc . 80 ∞ n Then, for every ϕ ∈ Cc (R ) Z Z α α f (x)ϕ(x)dx = lim D fk(x)ϕ(x)dx k→∞ Rn Rn Z |α| α = (−1) lim fk(x)D ϕ(x)dx k→∞ Rn Z = (−1)|α| f(x)Dαϕ(x)dx Rn Z = Dαf(x)ϕ(x)dx. Rn α α Hence, D f = f . Note that by Theorem 9.5 (Plancherel) we have that fbk converges to fb in L2. By the above we obtain

α α |α| α |α| α Ddf = lim D[fk = lim (2πi) ξ fbk = (2πi) ξ f.b k→∞ k→∞ This gives (4). Let α ∈ C. Then, by Theorem 9.5 2 2 kf + αgkL2 = fb+ αg . (9.11) c L2 Expanding we deduce Z |f(x)|2 + αg(x)f(x) + αg(x)f(x) + |αg(x)|2dx Rn Z 2 2 = fb(x) + αgc(x)fb(x) + αgc(x)fb(x) + |αgc(x)| dx. Rn Again, by Theorem 9.5 we obtain Z Z αg(x)f(x) + αg(x)f(x)dx = αgc(x)fb(x) + αgc(x)fb(x)dx. Rn Rn If we take α = 1 we obtain Z Z g(x)f(x) + g(x)f(x)dx = gb(x)fb(x) + gb(x)fb(x)dx. (9.12) Rn Rn If we take α = i we obtain Z Z ig(x)f(x) − ig(x)f(x)dx = i gb(x)fb(x) − i gb(x)fb(x)dx. (9.13) Rn Rn We multiply equation (9.13) with i and obtain Z Z −g(x)f(x) + g(x)f(x)dx = − gb(x)fb(x) + gb(x)fb(x)dx. (9.14) Rn Rn Combining equation (9.12) and (9.14) yields the statement. Theorem 9.8 (Characterization of Hk by the Fourier Transform). Let k ∈ N. A function u ∈ L2(Rn) belongs to Hk(Rn) if and only if k 2 n (1 + | · | )ub ∈ L (R ). (9.15) 81 In addition, there exists a positive constant C such that

1 k kuk k n ≤ 1 + | · | u ≤ Ckuk k n . (9.16) H (R ) b 2 H (R ) C L (Rn) Proof. 1. Assume ﬁrst that u ∈ Hk(Rn). Then for each multiindex |α| ≤ k, we have Dαu ∈ L2(Rn). Theorem 9.7 (4) asserts that α |α| α n Ddu(ξ) = (2πi) ξ ub(ξ) for a.e. ξ ∈ R . (9.17) Thus, by Theorem 9.5, we have that (2πiξ)|α|u(ξ) ∈ L2(Rn) for each |α| ≤ k. b th Let 1 ≤ j ≤ n and αj = (0,...,k,..., 0), where k is at the j position of the multiindex αj. Then we have

αj k k D[u(ξ) = (2πi) ξj ub(ξ). Hence, by Theorem 9.5, 2 X α 2 X α kD ukL2 = Ddu L2 1≤|α|≤k 1≤|α|≤k n X 2 ≥ D[αj u L2 j=1 n Z X k k 2 = (2πi) yj u(y) dy n b j=1 R Z n 2k 2 X 2k = (2π) |u(y)| |yj| dy n b R j=1 Z 2k 2 2k ≥ c(n, k)(2π) |ub(y)| |y| dy. Rn Summarizing we have Z 2 2k X α 2 |u(y)| |y| dy ≤ C(n, k) kD ukL2 . (9.18) n b R 1≤|α|≤k Since |y|k ≤ 1, if |y| ≤ 1 and |y|k ≤ |y|2k, if |y| ≥ 1, the following estimate holds: Z 2 Z k 2 k 2 ub(y)(1 + |y| ) dy = |ub(y)| (1 + |y| ) dy Rn Rn Z Z Z 2 2 2k 2 k = |ub(y)| dy + |ub(y)| |y| dy + 2 |ub(y)| |y| dy Rn Rn Rn Z Z 2 2 2k ≤ 3 |ub(y)| dy + |ub(y)| |y| dy . Rn Rn We apply Theorem 9.5 and equation (9.18) to the right-hand side and obtain   Z 2 k 2 X α 2 2 u(y)(1 + |y| ) dy ≤ C(n, k) kukL2 + kD ukL2  = C(n, k)kukHk . n b R 1≤|α|≤k

82 k 2 n k n 2. Assume (1 + | · | )ub ∈ L (R ). We show that u ∈ H (R ). Let |α| ≤ k. Then Z Z |α| α 2 2|α| α 2 2 (2πi) y ub(y) dy = (2π) |y | |ub(y)| dy Rn Rn Z 2k 2k 2 ≤ (2π) |y| |ub(y)| dy Rn Z (9.19) 2k k 2 2 ≤ (2π) (1 + |y| ) |ub(y)| dy Rn 2 = (2π)2k (1 + |·|k)u . b 2 L (Rn) |α| α 2 n Hence, (2πi) y ub(y) ∈ L (R ). Let |α| α uα = (2πi) y ub(y) . q ∞ n We show that uα is the weak derivative of u. Let ϕ ∈ Cc (R ). We use Theorem 9.7 to obtain Z Z α α D ϕ(x)u(x)dx = D[ϕ(x)ub(x)dx Rn Rn Z |α| α = (2πi) x ϕb(x)ub(x)dx Rn Z |α| |α| α = (−1) (2πi) x ub(x)ϕb(x)dx Rn Z |α| = (−1) uα(x)ϕ(x)dx, Rn where u denotes the complex conjugate of the function u : Rn → C. Hence, Z Z α |α| ∞ D ϕ(x)u(x)dx = (−1) uα(x)ϕ(x)dx, for all ϕ ∈ Cc . Rn Rn If we take the complex conjugate on both sides of the equation we obtain Z Z α |α| ∞ D ϕ(x)u(x)dx = (−1) uα(x)ϕ(x)dx, for all ϕ ∈ Cc . Rn Rn 2 n Hence, uα is the weak derivative of u and by equation (9.19) the weak derivative is in L (R ). It remains to show the left-hand side inequality of (9.16). By Theorem 9.5 and equation (9.19) we have that for every |α| ≤ k the following estimate holds Z kDαuk2 ≤ (2π)2k |y|2k|u(y)|2dy. L2(Rn) b Rn Hence, Z X 2 2k 2 kDαuk ≤ C(n, k) |y| |u(y)| dy. (9.20) L2(Rn) n b 1≤|α|≤k R Theorem 9.5 yields Z kuk2 = |u(y)|2dy. (9.21) L2(Rn) b Rn 83 By combining equation (9.20) and (9.21) we obtain Z 2 2k 2 kukHk ≤ C(n, k) (1 + |y| )|ub(y)| dy Rn Z k 2 2 ≤ C(n, k) (1 + |y| ) |ub(y)| dy Rn

= C(n, k) (1 + | · |k)u . b 2 L (Rn) We can define the fractional Sobolev spaces. 2 n s n s Definition 9.9. Let 0 < s < ∞ and u ∈ L (R ). We say u ∈ H (R ) if (1 + |·| )ub ∈ L2(Rn) and define for s∈ / N the norm kuk = k(1 + | · |s)uk . Hs(Rn) b L2(Rn) Remark 9.10. Note that k 2 k n (1 + |y| ) ≈ (1 + |y| ) 2 , for all y ∈ R . Hence, we have the following equivalent characterization: a function u ∈ L2(Rn) belongs to Hk(Rn) if and only if 2 k 2 n (1 + | · | ) 2 ub ∈ L (R ). (9.22) Alternatively, the following definition is common as well

2 n s n 2 s Definition 9.11. Let 0 < s < ∞ and u ∈ L (R ). We say u ∈ H (R ) if (1 + | · | ) 2 ub ∈ L2(Rn) and deﬁne for s∈ / N the norm 2 s 2 kuk s n = (1 + | · | ) u . H (R ) b L2(Rn)

84 CHAPTER 10

Exercises

Chapter 2 1 Exercise 1. Let u ∈ Lloc(R) and Z ∞ Tu : Cc (R) → R,Tu(φ) = u(x)φ(x)dx. R

Show that the integral exists and that Tu is linear and continuous. ∞ ∞ Remark: Note that a sequence ϕn ∈ Cc (R) converges to ϕ in Cc (R), if (1) there exists a compact interval [a, b] such that supp ϕn ⊆ [a, b] for all n ∈ N.

(l) (l) (2) ∀ > 0 ∀l ∈ N0 ∃ Nl ∀n ≥ Nl : supx ϕn (x) − ϕ (x) < . Exercise 2. Consider the function u : R → R, given by ( 0, if x ≤ 0, u(x) = x, if x > 0.

Determine the distributional and the weak derivative (if it exists) of u.

Exercise 3. The Heaviside function H : R → R is deﬁned by ( 0, if x ≤ 0, H(x) = 1, if x > 0. Determine the distributional derivative of H. Prove or disprove that H does not have a weak derivative. Exercise 4. Prove Lemma 2.6. Exercise 5. Prove Lemma 2.7. Exercise 6. Show that W k,p(U) is a normed vector space. Exercise 7. Show that X Z hu, vi = Dαu(x)Dαv(x)dx |α|≤k U

k p deﬁnes an inner product on H and hu, ui = k·kHk . Exercise 8. Prove Lemma 2.16

85 Chapter 3 ε Exercise 9. Let Uε and f be as in Deﬁnition 3.4. Show that for x ∈ Uε and 1 ≤ i ≤ n ε ε Z ∂ ε f (x + eih) − f (x) ε f (x) := lim = ∂x η (x − y)f(y)dy . h→0 i ∂xi h U Exercise 10. Let u be absolutely continuous (cf. Section 11.5) on an interval U = (a, b) ⊆ R. Show that u admits a weak derivative v ∈ L1(U) and that the weak derivative coincides with the classical derivative almost everywhere.

1 Exercise 11. Let U = (a, b) ⊆ R be an open interval. Assume that u ∈ Lloc(U) admits a weak derivative v ∈ L1(U). Show that there exists an absolutely continuous functionu ˜ such that u˜(x) = u(x), for a.e. x ∈ U, (10.1) u˜(x + h) − u˜(x) v(x) = lim , for a.e. x ∈ U. (10.2) h→0 h

Hint: Let x0 ∈ U be a Lebesgue point of u, i.e. 1 Z lim |u(y) − u(x0)| dy = 0. r→0+ |B(x0, r)| B(x0,r) R x Deﬁneu ˜(x) = u(x0) + v(y)dy and use Theorem 3.9. x0 Exercise 12. Let U = (a, b) ⊆ R. Show that u ∈ W 1,p(U) if and only if u coincides a.e. with an absolutely continuous functionu ˜ : U → R withu ˜ ∈ Lp(U) and its derivative u˜0 ∈ Lp(U). Hint for exercises 10-12: The following facts for absolutely continuous functions follow from the fundamental theorem (Theorem 11.23) (1) u: I = (a, b) → R is absolutely continuous if and only if there exists a function v ∈ L1(I) such that Z x u(x) = u(a) + v(t)dt, x ∈ I. a (2) u: I → R is absolutely continuous if and only if its classical derivative u0 exists a.e. in I and belongs to L1(I).

p n n Exercise 13. Let 1 ≤ p < ∞ and f ∈ L (R ). Let fh(t) = f(t − h), h ∈ R . Prove

kfh − fkLp → 0. n p Hint. Use that Cc(R ) is dense in L , 1 ≤ p < ∞. Exercise 14. Give an example to show that the result in Exercise 13 is not true when p = ∞. Exercise 15. Reconsider the estimate in Remark 3.18

86 Chapter 4 Exercise 16. Show that

kukW 1,p(B) ≤ Cp kukW 1,p(B+), where B , B+, u and u are as in the proof of Theorem 4.1 (Extension Theorem). Exercise 17. Let u ∈ W 2,p(U) ∩ C2(U). Show that u∈ / C2(B), but

kukW 2,p(B) ≤ Cp kukW 2,p(B+), where B , B+ and u are as in the proof of Theorem 4.1 (Extension Theorem). Exercise 18. Let u ∈ W 1,∞(U). Let ( u(x), if x ∈ B+ u(x) = − u(x1, . . . , xn−1, −xn), if x ∈ B . Show that the weak derivatives of u are given by

( + ∂u uxi , on B = − ∂xi uxi (x1, . . . , xn−1, −xn), on B if 1 ≤ i < n and

( + ∂u uxn , on B = − ∂xn −uxn (x1, . . . , xn−1, −xn), on B .

Exercise 19. Show that the operator E : W 1,p(U) → W 1,p(Rn) deﬁned by

Eu = lim Eum m→∞ ∞ 1 does not depend on the particular choice of the sequence (um)m=1 ⊆ C (U) and satisﬁes the properties of Theorem 4.1.

Chapter 5 Exercise 20. Verify Step 2 in the proof of Theorem 5.1 analogously to Step 4 in the proof of Theorem 4.1.

Exercise 21. Let U = (a, b) ⊆ R and u ∈ W 1,p(U), 1 ≤ p < ∞. Then T u ≡ 0 ⇔ u˜(a) =u ˜(b) = 0, where T : W 1,p(U) → Lp(∂U) is the trace operator andu ˜ is the absolutely continuous representation of u, cf. Exercise 12.

k,p Exercise 22. Let u ∈ W (U) and x0 ∈ U. Show that Z ∞ x0 ∈ supp u ⇔ ∀ V ⊆ U open with x0 ∈ V ∃ ϕ ∈ Cc (V ): u(x)ϕ(x) 6= 0, V where supp u = U \ S{V ⊆ U open : u = 0 a.e. on V }.

87 Exercise 23. Let u ∈ C(U). Let supp(u) = {x ∈ U : u(x) 6= 0} and [ ess supp(u) = U \ {V ⊆ U open : u = 0 a.e. on V }. Show that supp u = ess supp u.

2 2 2 ∞ Exercise 24. Let U = {(x, y) ∈ R : x + y < 1}. Construct a sequence (un)n=1 in C(U) ∩ W 1,1(U) such that C ku k ≤ and ku k = C, n L1(U) n n L1(∂U) where C is some constant independent from n.

∞ Exercise 25. Show that the sequence (un)n=1 constructed in Exercise 24 satisﬁes

kunkL1(∂U) ≤ CkunkW 1,1(U), where C is some constant independent from n.

Chapter 6

Exercise 26. Prove the general H¨olderinequality (Theorem 11.6): Let 1 ≤ p1, . . . , pm ≤ 1 1 p ∞, with + ··· + = 1. Assume uk ∈ L k for k = 1 . . . , m. Then p1 pm m Z Y |u1 ··· um|dx ≤ kuikLpk (U) . U k=1 ∞ n n Exercise 27. Let η ∈ Cc (R ) such that η(0) = 1, 0 ≤ η(x) ≤ 1 for all x ∈ R , and ∞ n k,p n supp η ⊆ B(0, 1). Let f ∈ C (R ) ∩ W (R ). Show that fR(x) := f(x)η(x/R) converges k,p n k,p n k,p n to f in W (R ) for R → ∞. As a consequence show that W0 (R ) = W (R ) for all 1 ≤ p < ∞. Exercise 28. Show that the statement of Theorem 6.2 is true for p = n, if n = 1. Exercise 29. Recalculate the proof of Theorem 6.2 for p = 1 and n = 2. n 1,p Exercise 30. Let U ⊆ R open and bounded. Let 1 ≤ p < n. Show that for u ∈ W0 we have that kukW 1,p(U) is equivalent to kDukLp (U). Exercise 31. Let U ⊆ Rn be open and bounded and suppose ∂U is C1. Assume 1 ≤ p < n and u ∈ W 2,p(U). Show that

kukW 1,q(U) ≤ CkukW 2,p(U) for all q ∈ [1, p∗]. Exercise 32. Show that f(x) = xα, x ∈ [0, 1], is H¨oldercontinuous with exponent α0, 0 < α0 ≤ α < 1, but not for α0 > α.

k,γ Exercise 33. Show that (C (U), k · kk,γ) is a Banach space.

88 α Exercise 34. Let U = {x ∈ Rn : |x| < 1}. Let u(x) = |x| , α ∈ (0, 1). Show that 1,p n u ∈ W (U), p > n, if and only if α > 1 − p .

Exercise 35. Let U ⊆ Rn be open, bounded and suppose ∂U is C1. Assume u ∈ 1,p n W (U), n < p ≤ ∞. Show that for all γ ∈ (0, 1 − p ] ∗ ku kC0,γ (U) ≤ CkukW 1,p(U). (10.3) where u∗ is a version of u. Use Exercise 34 to show that equation (10.3) does not hold for n γ ∈ (1 − p , 1]. Then show that ku∗k ≤ Ckuk C0,γ (Rn) W 1,p(Rn) n does not hold for γ ∈ (1 − p , 1]. Exercise 36. Prove Theorem 6.20.

n 1,p Exercise 37. Let U ⊆ R open and bounded. Assume u ∈ W0 (U), n < p ≤ ∞. Then ∗ 0,γ n u has a version u ∈ C (U) for γ = 1 − p , with the estimate ∗ ku kC0,γ (U) ≤ CkDukLp . Exercise 38. Present the proof of Proposition 6.10. Introduce ﬁrst all deﬁnitions needed. Exercise 39. Present the alternative proof of Theorem 6.2 (given in Section 6.3) in the case 1 < p < n. Exercise 40. Present the alternative proof of Theorem 6.2 (given in Section 6.3) in the case p = 1.

Exercise 41. Let U = {x ∈ Rn : |x| < 1}, where |x| is the euclidean norm on Rn, i.e.

1 n ! 2 X 2 |x| = xi . i=1

1 n Let u(x) = ln ln 1 + |x| on U \{0}. Show that u ∈ L (U). Hint: (1) ln(1 + x) ≤ x for all x ≥ 0. R ∞ n−1 −t (2) 0 t e dt = Γ(n). Exercise 42. Let U and u be as in Exercise 41. Show that the classical derivatives ∂u , ∂xi which exist on U \{0}, are the weak derivatives of u on U and ∂u ∈ Ln(U). ∂xi

89 Chapter 7 Exercise 43. Show that the estimate (7.1) in the proof of Theorem 7.2 is valid for ∞ 1,p (um)m=1 in W (V ). Exercise 44. Verify equation 7.6, by applying Lemma 3.7 and Minkowski’s inequality for integrals (Theorem 11.4). Exercise 45. Prove Theorem 7.3. Exercise 46. What do we have to change in the proof of Theorem 7.2 to obtain a proof for Theorem 7.4?

Chapter 8 2 dt Exercise 47. Let f ∈ L ([0, 2π], 2π ) be given by its Fourier series X int f(t) = ane , n∈Z where Z 2π in· 1 −int an = f, e = f(t)e dt ∈ C. 2π 0 Let P n2|a |2 < ∞. Show that the weak derivative of f is given by n∈Z n X int g(t) = inane n∈Z 2 dt and is in L ([0, 2π], 2π ).

Remark: The partial sums N X int SN = ane n=−N 2 dt converge to f in L ([0, 2π], 2π ). 2 dt Exercise 48. Let f ∈ L ([0, 2π], 2π ) be given by its Fourier series X int f(t) = ane n∈Z with P n2|a |2 < ∞. Show that Poincar´e’sinequality (Theorem 8.1) holds with constant n∈Z n one.

90 CHAPTER 11

Appendix

11.1. Notation 11.1.1. Geometric notation. (1) Rn is the n-dimensional real euclidean space equipped with the euclidean norm 1 n ! 2 X 2 kxk = |x| = |x | . Rn i i=1 n n (2) R+ = {x = (x1, . . . , xn) ∈ R : xn > 0}. 11.1.2. Notation for functions. Let U ⊆ Rn. (1) If u : U −→ R, we write

u(x) = u(x1, . . . xn), x ∈ U. We say u is smooth if u is inﬁnitely diﬀerentiable. (2) If u : U −→ Rm, we write u(x) = (u1(x), . . . , um(x)), x ∈ U. The function uk is the kth component of u, k = 1, . . . , m.

11.1.3. Function Spaces. Let U ⊆ Rn. (1) k C (U) = {u : U → R | u is continuous}, C(U) = {u ∈ C(U) | u is uniformly continuous on bounded subsets of U}, k C (U) = {u : U → R | u is k-times continuously differentiable}, α k k D u is uniformly continuous on bounded subsets of U C (U) = u ∈ C (U) . for all multiindex α with |α| ≤ k Thus, if u ∈ Ck(U), then Dαu continuously extends to U for each multiindex α, |α| ≤ k. On the other hand, if V ⊂ U is compact, then every continuous function on V is uniformly continuous (Heine-Cantor). (2) ∞ ∞ \ k C (U) = {u : U → R | u is infinitely differentiable } = C (U) k=0 ∞ \ C∞(U) = Ck(U) k=0

91 k k (3) Cc(U),Cc (U), etc. denote the functions in C(U),C (U), etc. with compact support. ∞ Cc (U), cf. page 7. (4) Let 1 ≤ p < ∞. The space of p-times integrable functions, denoted by Lp(U), consists of equivalence classes of measurable functions u : U → Rn such that 1 Z p p kukLp = |u| < ∞, U where two such functions are equivalent if they are equal a.e. (w.r.t. the Lebesgue measure). The space of essentially bounded functions, denoted by L∞(U), consists of equivalence (a.e. equivalence) classes of measurable functions u : U → Rn such that

kuk∞ = ess supRn |u| < ∞, (11.1) n where ess supRn u = inf{a ∈ R : |{x ∈ R : u(x) > a}| = 0}. p p (5) Lloc(U) = {u : U → R | u ∈ L (V ) for each V ⊂⊂ U}, cf. page 7. Note that V ⊂⊂ U means V ⊂ K ⊂ U, where K is compact (compactly contained), cf. page 12. (6) W k,p(U), cf. page 10, Hk(U), cf. page 11, k,p W0 (U), cf. page 11.

11.1.4. Notation for derivatives. Assume u : U −→ R, x ∈ U. ∂u u(x+hei−u(x) (1) (x) = limh→∞ , provided the limit exists. ∂xi h ∂u ∂2u (2) We write ux for , for ux x , etc. i ∂xi ∂xi∂xj i j (3) Multiindex Notation n (a) A vector of the form α = (α1, . . . , αn) ∈ N0 is called a multiindex of order

|α| = α1 + ··· + αn. (b) Let x ∈ Rn, then α α1 αn x = x1 ··· xn (c) We deﬁne for a given multiindex α ∂|α|u(x) α αn αn D u(x) := = ∂x ··· ∂x u ∂α1 ··· ∂αn 1 n x1 xn (d) Let k ∈ N0. The set of all partial derivatives of order k is denoted by Dku(x) := {Dαu(x)||α| = k}. (e) Let α, β be multiindices, then

β ≤ α ⇐⇒ β1 ≤ α1, . . . , βn ≤ αn. (f) Let α, β be multiindices with β ≤ α. Then α α! = , β β!(α − β)!

where α! = α1! ··· αn!. 92 ∞ (g) Leibniz formula. If u, v ∈ Cc (U), then X α Dα(uv) = DαuDα−βu β β≤α

11.2. Inequalities Theorem 11.1. Let 1 ≤ p < ∞. Then (a + b)p ≤ 2p−1(ap + bp), a, b > 0. (11.2) Proof. The function t 7→ tp is convex for t ≥ 0. Therefore, a + bp 1 ≤ (ap + bp). 2 2 Corollary 11.2. Let 1 ≤ p < ∞. Then

p p−1 p p kf + gkLp(U) ≤ 2 kfkLp(U) + kgkLp(U) .

Proof. Use the triangle inequality and apply Theorem 11.1 with a = |f| and b = |g|. 1 1 Theorem 11.3 (Young’s inequality). Let 1 < p, q ≤ ∞ and p + q = 1. Then ap bq ab ≤ + , a, b > 0. p q Proof. log (ab) log a+log b 1 log ap+ 1 log bq ab = e = e = e p q .

The function x 7→ ex is convex for all x ∈ R. Therefore, p q 1 log ap+ 1 log bq 1 log ap 1 log bq a b e p q ≤ e + e = + . p q p q

Theorem 11.4 (Minkowski’s inequality for integrals). Let (Ω1, dx) and (Ω2, dy) be measure spaces and F :Ω1 × Ω2 → R be measurable. Let r ≥ 1. Then r 1 1 Z Z r Z Z r |F (x, y)|dy dx ≤ |F (x, y)|rdx dy. Ω1 Ω2 Ω2 Ω1 1 1 Theorem 11.5 (H¨older’sinequality). Let 1 ≤ p, q ≤ ∞, p + q = 1. Then, if u ∈ Lp(U), v ∈ Lq(U), we have Z |uv|dx ≤ kukLp(U)kvkLq(U) . U

93 1 1 Theorem 11.6 (General HölderInequality). Let 1 ≤ p1, . . . , pm ≤ ∞, with +···+ = p1 pm p 1. Assume uk ∈ L k (U) for k = 1 . . . , m. Then m Z Y |u1 ··· um|dx ≤ kuikLpk (U) . U k=1 Proof. Induction, and using Höldersinequality. Let m = 2. This is clear from Höldersinequality. Induction step: 1 1 1 + ··· + + = 1 p1 pm pm+1 1 1 p + p p p + = m m+1 ; α := m m+1 pm pm+1 pmpm+1 pm + pm+1 ⇒ Z |u1(x) ··· um(x)um+1(x)|dx ≤ ku1kLp1 · · · kum−1kLpm−1 .kumum+1kLα U It remains to show that kumum+1kLα ≤ kumkLpm .kum+1kLpm+1 .

Note that α + α = 1. Hence,we can use H¨older’sinequality to obtain pm pm+1 α α ! pm ! p +1 Z Z Z p m α α α pm α m+1 |um|.|um+1| dx ≤ |um| α dx |um+1| α dx . U U U Corollary 11.7. Let 1 ≤ p ≤ q ≤ ∞. Let U ⊆ Rn be bounded. Then 1 − 1 p q kfkLp(U) ≤ |U| kfkLq(U). p p p Proof. Apply Theorem 11.5 with u = |f| , v = 1 and exponents q + 1 − q = 1. Similar proofs establish the following discrete versions of the above inequalities 1 1 n Theorem 11.8. Let 1 ≤ p < ∞ and p + q = 1. Let x = (x1, . . . , xn) ∈ R and n y = (y1, . . . , yn) ∈ R . Then 1 1 n n ! p n ! q X X p X q |xiyi| ≤ |xi| |yi| . i=1 i=1 i=1 n Corollary 11.9. Let 1 ≤ p ≤ q ≤ ∞. Let x = (x1, . . . , xn) ∈ R . Then 1 1 1 n ! q n ! p n ! q X q X p 1 − 1 X q |xi| ≤ |xi| ≤ n p q |xi| . (11.3) i=1 i=1 i=1

1 Pn p p Proof. Assume that ( i=1 |xi| ) = 1. Then, |xi| ≤ 1 for every 1 ≤ i ≤ n. Hence, for all q ≥ p the following inequality holds q n n n ! p X q X p X p |xi| ≤ |xi| = 1 = |xi| . i=1 i=1 i=1 94 Therefore, 1 1 n ! q n ! p X q X p |xi| ≤ |xi| . (11.4) i=1 i=1 1 Pn p p If ( i=1 |xi| ) 6= 1. Set xi zi = 1 . Pn p p ( i=1 |xi| ) 1 Pn p p Then ( i=1 |zi| ) = 1. Therefore, by the above we obtain 1 1 n ! q n ! p X q X p |zi| ≤ |zi| . (11.5) i=1 i=1 This is equivalent to 1 1 Pn q q Pn p p ( i=1 |xi| ) ( i=1 |xi| ) 1 ≤ 1 . (11.6) Pn p p Pn p p ( i=1 |xi| ) ( i=1 |xi| ) Hence, 1 1 n ! q n ! p X q X p |xi| ≤ |xi| . (11.7) i=1 i=1 This gives the left-hand side of inequality (11.3) p p Applying Theorem 11.8 with y = (1,..., 1) and q + 1 − q = 1 yields the right-hand side of equation (11.3). p Theorem 11.10 (Interpolation inequality for L -norms). Let 1 ≤ p0 ≤ q0 ≤ ∞ and 0 < θ < 1. We deﬁne 1 1 − θ θ = + . pθ p0 p1 Let f ∈ Lp0 ∩ Lp1 . Then f ∈ Lpθ and 1−θ θ kfkLpθ ≤ kfkLp0 kfkLp1 . (11.8) Proof. Apply H¨older’sinequality with θpθ + (1−θ)pθ = 1. p1 p0 Z Z |f|pθ dx = |f|pθθ|f|pθ(1−θ)dx

pθθ pθ(1−θ) Z p1 p1 Z p0 p0 pθθ pθ(1−θ) ≤ |f| pθθ |f| pθ(1−θ)

θ 1−θ = kfkLp1 kfkLp0 . 1 1 1 Theorem 11.11 (Young’s inequality). Let 1 ≤ p, q, r ≤ ∞ such that 1 + r = p + q . Let f ∈ Lp(Rn) and g ∈ Lq(Rn). Then f ∗ g exists a.e. and lies in Lr(Rn) with

kf ∗ gkr ≤ kfkp kgkq.

95 11.3. Calculus Facts Let U ⊂ Rn be open and bounded. k Definition 11.12. We say ∂U is C if for each point x0 ∈ ∂U there exists r > 0 and a Ck-function γ : Rn−1 → R such that we have

U ∩ B(x0, r) = {x ∈ B(x0, r): xn > γ(x1, ..., xn−1)}. Note that ∂U ∩ B(x0, r) = {x ∈ B(x0, r): xn = γ(x1, ..., xn−1)}. ∂U is C∞ if it is Ck for k = 1, 2,.... Definition 11.13. (1) If ∂U is C1, then along ∂U is deﬁned the outward pointing unit normal vector ﬁeld ν = (ν1, . . . , νn).

(2) The unit normal at any point x0 ∈ ∂U is ν(x0) = ν = (ν1, . . . , νn). (3) Let u ∈ C1(U). We call ∂u = ν · Du. ∂ν Theorem 11.14 (Gauss-Green Theorem). Suppose u ∈ C1(U). Then Z Z i uxi dx = uν dS, (i = 1, . . . , n). (11.9) U ∂U Theorem 11.15 (Integration by parts formula). Let u, v ∈ C1(U). Then Z Z Z i uxi v dx = − uvxi dx + uν dS, (i = 1, . . . , n). (11.10) U U ∂U Polarcoordinates. For x ∈ Rn \{0} the polar coordinates are given by x r = |x| and ω = ∈ Sn−1 = {ω ∈ n : |ω| = 1}. |x| R Theorem 11.16 (Integration in polar coordinates). Let f ∈ L1(Rn). Then Z Z ∞ Z f(x)dx = f(rω)rn−1dσ(ω)dr. Rn 0 Sn−1 where σ is the Borel measure on the unit sphere Sn−1 in Rn. Corollary 11.17. If f is a measurable function on Rn, nonnegative or integrable, such that f(x) = g(|x|) for some function g on (0, ∞), then Z Z ∞ f(x)dx = σ(Sn−1) g(r)rn−1dr. Rn 0 ∞ Theorem 11.18 (Arzela-Ascoli Compactness Criterion). Let {fk}k=1 be a sequence of real-valued functions on Rn such that n |fk(x)| ≤ M (k = 1, . . . , x ∈ R ) ∞ for some constant M and the {fk}k=1 are uniformly equicontinuous, i.e. n ∀η > 0 ∃δ > 0 ∀k ∈ N ∀x, y ∈ R : |x − y| < δ ⇒ |fk(x) − fk(y)| < η. 96 ∞ ∞ Then there exists a subsequence {fkj }j=1 ⊆ {fk}k=1 and a continuous function f, such that n fkj → f uniformly on compact subsets of R .

11.4. Convergence theorems for integrals Definition 11.19 (Average of f over a ball B(x, r)). We denote by Z 1 Z − f dy = f dy B(x,r) |B(x, r)| B(x,r) the average of f over the ball B(x, r) and by Z 1 Z (f)U = − f dx = f dx U | U | U the average of f over U ⊂ Rn. Theorem 11.20 (Lebesgue Diﬀerentiation Theorem). Let f : Rn → R be locally integrable. n (i) Then for a.e. point x0 ∈ R , Z − f dx −→ f(x0), as r → 0 B(x0,r) n (ii) In fact, for a.e. point x0 ∈ R , Z − | f(x) − f(x0) | dx −→ 0, as r → 0. (11.11) B(x0,r)

A point x0 at which (11.11) holds, is called a Lebesgue point of f. p n Remark 11.21. More generally, if f ∈ Lloc(R ) for some 1 ≤ p < ∞, then for a.e. point n x0 ∈ R we have Z p − | f(x) − f(x0) | dx −→ 0, as r → 0 B(x0,r)

11.5. Absolutely continuous functions Definition 11.22. Let I be an interval in R. A function u : I → R is absolutely continuous on I if and only if for every ε > 0 there exists a δ > 0 such that for every ﬁnite sequence of pairwise disjoint subintervals ((xk, yk))k of I we have that X X (yk − xk) < δ implies |u(yk) − u(xk)| < ε. k k Theorem 11.23 (Fundamental theorem of calculus). (1) Let f : I → R be Lebesgue integrable. The function Z x F (x) = f(t)dt, x ∈ I x0

is for every x0 ∈ I absolutely continuous. In particular F is diﬀerentiable a.e. in I and F 0(x) = f(x) for a.e. x ∈ I. 97 (2) Let F : I = (a, b) → R be absolutely continuous. Then F is diﬀerentiable a.e. in I and F 0 is Lebesgue integrable with Z b F (b) − F (a) = F 0(t)dt. a

98 Bibliography

[1] R. A. Adams. Sobolev spaces. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, 1975. Pure and Applied Mathematics, Vol. 65. [2] R. A. Adams and J. J. F. Fournier. Sobolev spaces, volume 140 of Pure and Applied Mathematics (Amsterdam). Elsevier/Academic Press, Amsterdam, second edition, 2003. [3] C. Bennett and R. Sharpley. Interpolation of operators, volume 129 of Pure and Applied Mathematics. Academic Press, Inc., Boston, MA, 1988. [4] L. C. Evans. Partial differential equations, volume 19 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 1998. [5] G. B. Folland. Fourier analysis and its applications. The Wadsworth & Brooks/Cole Mathematics Series. Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1992. [6] D. Gilbarg and N. S. Trudinger. Elliptic partial differential equations of second order. Springer-Verlag, Berlin-New York, 1977. Grundlehren der Mathematischen Wissenschaften, Vol. 224. [7] J. Heinonen. Lectures on analysis on metric spaces. Universitext. Springer-Verlag, New York, 2001. [8] S. Kindermann. Vorlesungsskriptum Partielle Differentialgleichungen. JKU, 2013. Institut fürIndus- triemathematik. [9] K. Königsberger. Analysis. 2. Springer-Lehrbuch. [Springer Textbook]. Springer-Verlag, Berlin, 1993. [10] E. M. Stein. Singular integrals: the roles of Calderónand Zygmund. Notices Amer. Math. Soc., 45(9):1130–1140, 1998.