Capacity

Abstract

We define the notion of abstract capacity in the sense of Choquet in Sect. 7.1 and the variational capacity of degree p in Sect. 7.2. The space of functions of bounded variation defined in Sect. 7.3 contains the Sobolev space W ^{1, 1}(Ω) and the characteristic functions of bounded sets of finite perimeter. The relation between the perimeter and the variational capacity of degree 1 is given in Sect. 7.4.

7.1 Capacity

The notion of capacity appears in potential theory. The abstract theory was formulated by Choquet in 1954. In this section, we denote by X a metric space, by $\mathcal{K}$ the class of compact subsets of X, and by $\mathcal{O}$ the class of open subsets of X.

Definition 7.1.1.

A capacity on X is a function

$$\displaystyle{ \mathrm{cap} : \mathcal{K}\rightarrow [0,+\infty ] : K \rightarrow \mathrm{ cap}(K) }$$

such that:

(C ₁) (monotonicity.) For every $A,B \in \mathcal{K}$ such that A ⊂ B, cap(A) ≤ cap(B).

(C ₂) (regularity.) For every $K \in \mathcal{K}$ and for every a > cap(K), there exists $U \in \mathcal{O}$ such that K ⊂ U, and for all $C \in \mathcal{K}$ satisfying C ⊂ U, cap(C) < a.

(C ₃) (strong subadditivity.) For every $A,B \in \mathcal{K}$,

$$\displaystyle{ \mathrm{cap}(A \cup B) +\mathrm{ cap}(A \cap B) \leq \mathrm{ cap}(A) +\mathrm{ cap}(B). }$$

The Lebesgue measure of a compact subset of ${\mathbb{R}}^{N}$ is a capacity.

We denote by cap a capacity on X. We extend the capacity to the open subsets of X.

Definition 7.1.2.

The capacity of $U \in \mathcal{O}$ is defined by

$$\displaystyle{ \mathrm{cap}(U) =\sup \{\mathrm{ cap}(K) : K \in \mathcal{K}\mbox{ and }K \subset U\}. }$$

Lemma 7.1.3.

Let $A,B \in \mathcal{O}$ and $K \in \mathcal{K}$ be such that K ⊂ A ∪ B. Then there exist $L,M \in \mathcal{K}$ such that L ⊂ A, M ⊂ B, and K = L ∪ M.

Proof.

The compact sets K ∖ A and K ∖ B are disjoint. Hence there exist disjoint open sets U and V such that K ∖ A ⊂ U and K ∖ B ⊂ V. It suffices to define L = K ∖ U and M = K ∖ V.

Proposition 7.1.4.

• (a) (monotonicity.) For every $A,B \in \mathcal{O}$ such that A ⊂ B, cap (A) ≤ cap (B).

• (b) (regularity.) For every $K \in \mathcal{K}$,$\mathrm{cap}(K) =\inf \{\mathrm{ cap}(U) : U \in \mathcal{O}$ and U ⊃ K}.

• (c) (strong subadditivity.) For every $A,B \in \mathcal{O}$,

  $$\displaystyle{ \mathrm{cap}(A \cup B) +\mathrm{ cap}\ A \cap B) \leq \mathrm{ cap}(A) +\mathrm{ cap}(B). }$$

Proof.

• (a) Monotonicity is clear.

• (b) Let us define Cap$(K) =\inf \{\mathrm{ cap}(U) : U \in \mathcal{O}$ and U ⊃ K}. By definition, cap(K) ≤ Cap(K). Let a > cap(K). There exists $U \in \mathcal{O}$ such that K ⊂ U and for every $C \in \mathcal{K}$ satisfying C ⊂ U, cap(C) < a. Hence Cap(K) ≤ cap(U) < a. Since a > cap(K) is arbitrary, we conclude that Cap(K) ≤ cap(K).

• (c) Let $A,B \in \mathcal{O}$, a < cap(A ∪B), and b < cap(A ∩ B). By definition, there exist $K,C \in \mathcal{K}$ such that K ⊂ A ∪B, C ⊂ A ∩ B, a < cap(K), and b ≤ cap(C). We can assume that C ⊂ K. The preceding lemma implies the existence of $L,M \in \mathcal{K}$ such that L ⊂ A, M ⊂ B, and K = L ∪M. We can assume that C ⊂ L ∩ M. We obtain by monotonicity and strong subadditivity that

  $$\displaystyle{ \begin{array}{ll} a + b \leq \mathrm{ cap}(K) +\mathrm{ cap}(C)& \leq \mathrm{ cap}(L \cup M) +\mathrm{ cap}(L \cap M) \\ & \leq \mathrm{ cap}(L) +\mathrm{ cap}(M) \leq \mathrm{ cap}(A) +\mathrm{ cap}(B).\end{array} }$$

Since a < cap(A ∪B) and b < cap(A ∩ B) are arbitrary, the proof is complete.

We extend the capacity to all subsets of X.

Definition 7.1.5.

The capacity of a subset A of X is defined by

$$\displaystyle{ \mathrm{cap}(A) =\inf \{\mathrm{ cap}(U) : U \in \mathcal{O}\mbox{ and }U \supset A\}. }$$

By regularity, the capacity of compact subsets is well defined.

Proposition 7.1.6.

• (a) (monotonicity.) For every A,B ⊂ X, cap (A) ≤ cap (B).

• (b) (strong subadditivity.) For every A,B ⊂ X,

  $$\displaystyle{ \mathrm{cap}(A \cup B) +\mathrm{ cap}(A \cap B) \leq \mathrm{ cap}(A) +\mathrm{ cap}(B). }$$

Proof.

• (a) Monotonicity is clear.

• (b) Let A, B ⊂ X and $U,V \in \mathcal{O}$ be such that A ⊂ U and B ⊂ V. We have

  $$\displaystyle{ \mathrm{cap}(A \cup B) +\mathrm{ cap}(A \cap B) \leq \mathrm{ cap}(U \cup V ) +\mathrm{ cap}(U \cap V ) \leq \mathrm{ cap}(U) +\mathrm{ cap}(V ). }$$

  It is easy to conclude the proof.

Proposition 7.1.7.

Let (K _n ) be a decreasing sequence in $\mathcal{K}$ . Then

$$\displaystyle{ \mathrm{cap}\left (\bigcap _{n=1}^{\infty }K_{ n}\right ) =\lim _{n\rightarrow \infty }\mathrm{cap}(K_{n}). }$$

Proof.

Let $K = \bigcap _{n=1}^{\infty }K_{ n}$ and $U \in \mathcal{O}$, U ⊃ K. By compactness, there exists m such that K _m  ⊂ U. We obtain, by monotonicity, cap(K) ≤ lim _{n → ∞} cap(K _n ) ≤ cap(U). It suffices then to take the infimum with respect to U.

Lemma 7.1.8.

Let (U _n ) be an increasing sequence in $\mathcal{O}$ . Then

$$\displaystyle{ \mathrm{cap}\left (\bigcup _{n=1}^{\infty }U_{ n}\right ) =\lim _{n\rightarrow \infty }\mathrm{cap}(U_{n}). }$$

Proof.

Let $U = \bigcup _{n=1}^{\infty }U_{ n}$ and $K \in \mathcal{K},K \subset U$. By compactness, there exists m such that K ⊂ U _m . We obtain by monotonicity cap(K) ≤ lim _{n → ∞} cap(U _n ) ≤ capU. It suffices then to take the supremum with respect to K.

Theorem 7.1.9.

Let (A _n ) be an increasing sequence of subsets of X. Then

$$\displaystyle{ \mathrm{cap}\left (\bigcup _{n=1}^{\infty }A_{ n}\right ) =\lim _{n\rightarrow \infty }\mathrm{cap}(A_{n}). }$$

Proof.

Let $A = \bigcup _{n=1}^{\infty }A_{ n}$. By monotonicity, lim _{n → ∞} cap(A _n ) ≤ cap(A). We can assume that lim _{n → ∞} cap(A _n ) <  + ∞. Let ɛ > 0 and $a_{n} = 1 - 1/(n + 1)$. We construct, by induction, an increasing sequence $(U_{n}) \subset \mathcal{O}$ such that A _n  ⊂ U _n and

$$\displaystyle{ \mathrm{cap}(U_{n}) \leq \mathrm{ cap}(A_{n}) +\varepsilon \ a_{n}. } $$

(*)

When n = 1, ( ∗ ) holds by definition. Assume that ( ∗ ) holds for n. By definition, there exists $V \in \mathcal{O}$ such that A _n + 1 ⊂ V and

$$\displaystyle{ \mathrm{cap}(V ) \leq \mathrm{ cap}(A_{n+1}) +\varepsilon (a_{n+1} - a_{n}). }$$

We define $U_{n+1} = U_{n} \cup V$, so that $A_{n+1} \subset U_{n+1}$. We obtain, by strong subadditivity,

$$\displaystyle{ \begin{array}{ll} \mathrm{cap}(U_{n+1})& \leq \mathrm{ cap}(U_{n}) +\mathrm{ cap}(V ) -\mathrm{ cap}(U_{n} \cap V ) \\ & \leq \mathrm{ cap}(A_{n}) +\varepsilon \ a_{n} +\mathrm{ cap}(A_{n+1}) +\varepsilon (a_{n+1} - a_{n}) -\mathrm{ cap}(A_{n}) \\ & =\mathrm{ cap}(A_{n+1}) +\varepsilon \ a_{n+1}. \end{array} }$$

It follows from ( ∗ ) and the preceding lemma that

$$\displaystyle{ \mathrm{cap}(A) \leq \mathrm{ cap}\left (\bigcup _{n=1}^{\infty }U_{ n}\right ) =\lim _{n\rightarrow \infty }\mathrm{cap}(U_{n}) \leq \lim _{n\rightarrow \infty }\mathrm{cap}(A_{n}) +\varepsilon. }$$

Since ɛ > 0 is arbitrary, the proof is complete.

Corollary 7.1.10 (Countable subadditivity). 

Let (A _n ) be a sequence of subsets of X. Then $\mathrm{cap}\left (\bigcup _{n=1}^{\infty }A_{ n}\right ) \leq \sum _{n=1}^{\infty }\mathrm{cap}(A_{ n})$ .

Proof.

Let $B_{k} = \bigcup _{n=1}^{k}A_{ k}$. We have

$$\displaystyle{\mathrm{cap}\left (\bigcup _{n=1}^{\infty }A_{ n}\right ) =\mathrm{ cap}\left (\bigcup _{k=1}^{\infty }B_{ k}\right ) =\lim _{k\rightarrow \infty }\mathrm{cap}(B_{k}) \leq \sum _{n=1}^{\infty }\mathrm{cap}(A_{ n}).}$$

 □ 

Definition 7.1.11.

The outer Lebesgue measure of a subset of ${\mathbb{R}}^{N}$ is defined by

$$\displaystyle{ {m}^{{\ast}}(A) =\inf \{ m(U) : U\mbox{ is open and }U \supset A\}. }$$

7.2 Variational Capacity

In order to define variational capacity, we introduce the space ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$.

Definition 7.2.1.

Let 1 ≤ p < N. On the space

$$\displaystyle{ {\mathcal{D}}^{1,p}({\mathbb{R}}^{N}) =\{ u \in {L}^{{p}^{{\ast}} }({\mathbb{R}}^{N}) : \nabla u \in {L}^{p}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})\}, }$$

we define the norm

$$\displaystyle{ \vert \vert u\vert \vert _{{\mathcal{D}}^{1,p}({\mathbb{R}}^{N})} =\vert \vert \nabla u\vert \vert _{p}. }$$

Proposition 7.2.2.

Let 1 ≤ p < N.

• (a) The space $\mathcal{D}({\mathbb{R}}^{N})$ is dense in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ .

• (b) (Sobolev’s inequality.) There exists c = c(p,N) such that for every $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$,

  $$\displaystyle{ \vert \vert u\vert \vert _{{p}^{{\ast}}} \leq c\vert \vert \nabla u\vert \vert _{p}. }$$

• (c) The space ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ is complete.

Proof.

The space $\mathcal{D}({\mathbb{R}}^{N})$ is dense in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ with the norm $\vert \vert u\vert \vert _{{p}^{{\ast}}} +\vert \vert \nabla u\vert \vert _{p}$. The argument is similar to that of the proof of Theorem 6.1.10.

Sobolev’s inequality follows by density from Lemma 6.4.2. Hence for every $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$,

$$\displaystyle{ \vert \vert \nabla u\vert \vert _{p} \leq \vert \vert u\vert \vert _{{p}^{{\ast}}} +\vert \vert \nabla u\vert \vert _{p} \leq (c + 1)\vert \vert \nabla u\vert \vert _{p}. }$$

Let (u _n ) be a Cauchy sequence in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$. Then u _n  → u in ${L}^{{p}^{{\ast}} }({\mathbb{R}}^{N})$, and for 1 ≤ k ≤ N, ∂ _k u _n  → v _k in ${L}^{p}({\mathbb{R}}^{N})$. By the closing lemma, for 1 ≤ k ≤ N, ∂ _k u = v _k . We conclude that u _n  → u in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$.

Proposition 7.2.3.

Every bounded sequence in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ contains a subsequence converging in $L_{\mathrm{loc}}^{1}({\mathbb{R}}^{N})$ and almost everywhere on ${\mathbb{R}}^{N}$ .

Proof.

Cantor’s diagonal argument will be used. Let (u _n ) be bounded in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$. By Sobolev’s inequality, for every k ≥ 1, (u _n ) is bounded in W ^1, 1(B(0, k)). Rellich’s theorem and Proposition 4.2.10 imply the existence of a subsequence (u _1, n ) of (u _n ) converging in L ¹(B(0, 1)) and almost everywhere on B(0, 1). By induction, for every k, there exists a subsequence (u _k, n ) of (u _{k − 1, n} ) converging in L ¹(B(0, k)) and almost everywhere on B(0, k). The sequence v _n  = u _n, n converges in $L_{\mathrm{loc}}^{1}({\mathbb{R}}^{N})$ and almost everywhere on ${\mathbb{R}}^{N}$.

Definition 7.2.4.

Let 1 ≤ p < N and let K be a compact subset of ${\mathbb{R}}^{N}$. The capacity of degree p of K is defined by

$$\displaystyle{ \mathrm{cap}_{p}(K) =\inf \left \{\int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx : u \in \mathcal{D}_{ K}^{1,p}({\mathbb{R}}^{N})\right \}, }$$

where

$$\displaystyle{ \mathcal{D}_{K}^{1,p}({\mathbb{R}}^{N}) =\{ u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N}) : \mbox{ there exists}U\mbox{ open such that}K \subset U\mbox{ and}\chi _{ U} \leq u }$$

almost everywhere}.

Theorem 7.2.5.

The capacity of degree p is a capacity on ${\mathbb{R}}^{N}$ .

Proof.

• (a) Monotonicity is clear by definition.

• (b) Let K be compact and $a >\mathrm{ cap}_{p}({\mathbb{R}}^{N})$. There exist $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ and U open such that K ⊂ U, χ _U  ≤ u almost everywhere, and $\int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx < a$. For every compact set C ⊂ U, we have

  $$\displaystyle{ \mathrm{cap}_{p}(C) \leq \int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx < a, }$$

  so that cap _p is regular.

• (c) Let A and B be compact sets, a > cap _p (A), and b > cap _p (B). There exist $u,v \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ and U and V open sets such that A ⊂ U, B ⊂ V, χ _U  ≤ u, and χ _V  ≤ v almost everywhere and

  $$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx < a,\quad \int _{{ \mathbb{R}}^{N}}\vert \nabla v{\vert }^{p}dx < b. }$$

  Since $\max (u,v) \in \mathcal{D}_{A\cup B}^{1,p}({\mathbb{R}}^{N})$ and $\min (u,v) \in \mathcal{D}_{A\cap B}^{1,p}({\mathbb{R}}^{N})$, Corollary 6.1.14 implies that

  $$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla \max (u,v){\vert }^{p}dx +\int _{{ \mathbb{R}}^{N}}\vert \nabla \min (u,v){\vert }^{p} =\int _{{ \mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx +\int _{{ \mathbb{R}}^{N}}\vert \nabla v{\vert }^{p}dx \leq a + b. }$$

  We conclude that

  $$\displaystyle{ \mathrm{cap}_{p}(A \cup B) +\mathrm{ cap}_{p}(A \cap B) \leq a + b. }$$

  Since a > cap _p (A) and b > cap _p (B) are arbitrary, cap _p is strongly subadditive.

The variational capacity is finer than the Lebesgue measure.

Proposition 7.2.6.

There exists a constant c = c(p,N) such that for every $A \subset {\mathbb{R}}^{N}$,

$$\displaystyle{ {m}^{{\ast}}(A) \leq c\ \mathrm{cap}_{ p}{(A)}^{N/(N-p)}. }$$

Proof.

Let K be a compact set and $u \in \mathcal{D}_{K}^{1,p}({\mathbb{R}}^{N})$. It follows from Sobolev’s inequality that

$$\displaystyle{ m(K) \leq \int _{{\mathbb{R}}^{N}}\vert {u\vert }^{{p}^{{\ast}} }dx \leq c{\left (\int _{{\mathbb{R}}^{N}}\vert \nabla {u\vert }^{p}dx\right )}^{{p}^{{\ast}}/p }. }$$

By definition,

$$\displaystyle{ m(K) \leq c\ \mathrm{cap}_{p,{\mathbb{R}}^{N}}{(K)}^{N/(N-p)}. }$$

To conclude, it suffices to extend this inequality to open subsets of ${\mathbb{R}}^{N}$ and to arbitrary subsets of ${\mathbb{R}}^{N}$.

The variational capacity differs essentially from the Lebesgue measure.

Proposition 7.2.7.

Let K be a compact set. Then

$$\displaystyle{ \mathrm{cap}_{p}(\partial K) =\mathrm{ cap}_{p}(K). }$$

Proof.

Let a > cap _p (∂K). There exist $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ and an open set U such that ∂K ⊂ U, χ _U  ≤ u almost everywhere, and

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx < a. }$$

Let us define V = U ∪K and v = max(u, χ _V ). Then $v \in \mathcal{D}_{K}^{1,p}({\mathbb{R}}^{N})$ and

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla v{\vert }^{p}dx \leq \int _{{ \mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx, }$$

so that cap _p (K) < a. Since a > cap _p (∂K) is arbitrary, we obtain

$$\displaystyle{\mathrm{cap}_{p}(K) \leq \mathrm{ cap}_{p}(\partial K) \leq \mathrm{ cap}_{p}(K).}$$

 □ 

Example.

Let 1 ≤ p < N and let B be a closed ball in ${\mathbb{R}}^{N}$. We deduce from the preceding propositions that

$$\displaystyle{ 0 <\mathrm{ cap}_{p}(B) =\mathrm{ cap}_{p}(\partial B). }$$

Theorem 7.2.8.

Let 1 < p < N and U an open set. Then

$$\displaystyle{ \mathrm{cap}_{p}(U) =\inf \left \{\int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx : u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N}),\chi _{ U} \leq u\mbox{ almost everywhere}\right \}. }$$

Proof.

Let us denote by Cap _p (U) the second member of the preceding equality. It is clear by definition that cap _p (U) ≤ Cap _p (U).

Assume that cap _p (U) < ∞. Let (K _n ) be an increasing sequence of compact subsets of U such that $U = \bigcup _{n=1}^{\infty }K_{ n}$, and let $(u_{n}) \subset {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ be such that for every n, $\chi _{K_{n}} \leq u_{n}$ almost everywhere and

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla u_{n}{\vert }^{p}dx \leq \mathrm{ cap}_{ p}(K_{n}) + 1/n. }$$

The sequence (u _n ) is bounded in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$. By Proposition 7.2.3, we can assume that u _n  → u in $L_{\mathrm{loc}}^{1}({\mathbb{R}}^{N})$ and almost everywhere. It follows from Sobolev’s inequality that $u \in {L}^{{p}^{{\ast}} }({\mathbb{R}}^{N})$. Theorem 6.1.7 implies that

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx \leq \begin{array}{c} \underline{\lim } \\ n\rightarrow \infty \end{array} \int _{{\mathbb{R}}^{N}}\vert \nabla u_{n}{\vert }^{p}dx \leq \lim _{ n\rightarrow \infty }\mathrm{cap}_{p}(K_{n}) \leq \mathrm{ cap}_{p}(U). }$$

(By Theorem 7.1.9, lim _{n → ∞} cap _p (K _n ) = cap _p (U).) Since almost everywhere, χ _U  ≤ u, we conclude that Cap _p (U) ≤ cap _p (U).

Corollary 7.2.9.

Let 1 < p < N, and let U and V be open sets such that U ⊂ V and m(V ∖ U) = 0. Then cap _p (U) = cap _p (V ).

Proof.

Let $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ be such that χ _U  ≤ u almost everywhere. Then χ _V  ≤ u almost everywhere.

Corollary 7.2.10 (Capacity inequality). 

Let 1 < p < N and $u \in \mathcal{D}({\mathbb{R}}^{N})$ . Then for every t > 0,

$$\displaystyle{ \mathrm{cap}_{p}(\{\vert u\vert > t\}) \leq {t}^{-p}\int _{{ \mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx. }$$

Proof.

By Corollary 6.1.14, $\vert u\vert /t \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$.

Definition 7.2.11.

Let 1 ≤ p < N. A function $v : {\mathbb{R}}^{N} \rightarrow \mathbb{R}$ is quasicontinuous of degree p if for every ɛ > 0, there exists an ω-open set such that cap _p (ω) ≤ ɛ and $v\big\vert _{{\mathbb{R}}^{N}\setminus \omega }$ is continuous. Two quasicontinuous functions of degrees p, v, and w are equal quasi-everywhere if $\mathrm{cap}_{p}(\{\vert v - w\vert > 0\}) = 0$.

Proposition 7.2.12.

Let 1 < p < N and let v and w be quasicontinuous functions of degree p and almost everywhere equal. Then v and w are quasi-everywhere equal.

Proof.

By assumption, m(A) = 0, where $A =\{ \vert v - w\vert > 0\}$, and for every n, there exists an ω _n -open set such that cap _p (ω _n ) ≤ 1 ∕ n and $\vert v - w\vert \big\vert _{{\mathbb{R}}^{N}\setminus \omega _{n}}$ is continuous. It follows that A ∪ω _n is open. We conclude, using Corollary 7.2.9, that

$$\displaystyle{\mathrm{cap}_{p}(A) \leq \mathrm{ cap}_{p}(A \cup \omega _{n}) =\mathrm{ cap}_{p}(\omega _{n}) \rightarrow 0,\quad n \rightarrow \infty.}$$

 □ 

Proposition 7.2.13.

Let 1 < p < N and $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ . Then there exists a function v quasicontinuous of degree p and almost everywhere equal to u.

Proof.

By Proposition 7.2.2, there exists $(u_{n}) \subset \mathcal{D}({\mathbb{R}}^{N})$ such that u _n  → u in ${\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$. Using Proposition 7.2.3, we can assume that u _n  → u almost everywhere and

$$\displaystyle{ \sum _{k=1}^{\infty }{2}^{kp}\int _{{ \mathbb{R}}^{N}}\vert \nabla (u_{k+1} - u_{k}){\vert }^{p}dx < \infty. }$$

We define

$$\displaystyle{ U_{k} =\{ \vert u_{k+1} - u_{k}\vert > {2}^{-k}\},\quad \omega _{ m} =\bigcup _{ k=m}^{\infty }U_{ k}. }$$

Corollary 7.2.10 implies that for every k,

$$\displaystyle{ \mathrm{cap}_{p}(U_{k}) \leq {2}^{kp}\int _{{ \mathbb{R}}^{N}}\vert \nabla (u_{k+1} - u_{k}){\vert }^{p}dx. }$$

It follows from Corollary 7.1.10 that for every m,

$$\displaystyle{ \mathrm{cap}_{p}(\omega _{m}) \leq \sum _{k=m}^{\infty }{2}^{kp}\int _{{ \mathbb{R}}^{N}}\vert \nabla (u_{k+1} - u_{k}){\vert }^{p}dx \rightarrow 0,\quad m \rightarrow \infty. }$$

We obtain, for every x ∈ ℝ ^N ∖ ω _m and every k ≥ j ≥ m,

$$\displaystyle{ \vert u_{j}(x) - u_{k}(x)\vert \leq {2}^{1-j}, }$$

so that (u _n ) converges simply to v on ${\mathbb{R}}^{N} \setminus \bigcap _{m=1}^{\infty }\omega _{ m}$. Moreover, $v\big\vert _{{\mathbb{R}}^{N}\setminus \omega _{m}}$ is continuous, since the convergence of (u _n ) on ${\mathbb{R}}^{N} \setminus \omega _{m}$ is uniform. For x ∈ ⋂ _m = 1 ^∞ ω _m , we define v(x) = 0. Since by Proposition 7.2.6, m(ω _m ) → 0, we conclude that u = v almost everywhere.

7.3 Functions of Bounded Variations

A function is of bounded variation if its first-order derivatives, in the sense of distributions, are bounded measures.

Definition 7.3.1.

Let Ω be an open subset of ${\mathbb{R}}^{N}$. The divergence of $v \in {\mathcal{C}}^{1}(\Omega ; {\mathbb{R}}^{N})$ is defined by

$$\displaystyle{ \mathrm{div\ }v =\sum _{ k=1}^{N}\partial _{ k}v_{k}. }$$

The total variation of u ∈ L _loc ¹(Ω) is defined by

$$\displaystyle{ \vert \vert Du\vert \vert _{\Omega } =\sup \left \{\int _{\Omega }u\mathrm{\ div\ }v\ dx : v \in \mathcal{D}(\Omega ; {\mathbb{R}}^{N}),\vert \vert v\vert \vert _{ \infty }\leq 1\right \}, }$$

where

$$\displaystyle{ \vert \vert v\vert \vert _{\infty } =\sup _{x\in \Omega }{\left (\sum _{k=1}^{N}{(v_{ k}(x))}^{2}\right )}^{1/2}. }$$

Theorem 7.3.2.

Let (u _n ) be such that u _n → u in L _loc ¹ (Ω). Then

$$\displaystyle{ \vert \vert Du\vert \vert _{\Omega } \leq \begin{array}{c} \underline{\lim } \\ n\rightarrow \infty \end{array} \vert \vert Du_{n}\vert \vert _{\Omega }. }$$

Proof.

Let $v \in \mathcal{D}(\Omega ; {\mathbb{R}}^{N})$ be such that $\vert \vert v\vert \vert _{\infty }\leq 1$. We have, by definition,

$$\displaystyle{ \int _{\Omega }u\ \mathrm{div\ }v\ dx =\lim _{n\rightarrow \infty }\int _{\Omega }u_{n}\ \mathrm{div}\ v\ dx \leq \begin{array}{c} \underline{\lim } \\ n\rightarrow \infty \end{array} \vert \vert Du_{n}\vert \vert _{\Omega }. }$$

It suffices then to take the supremum with respect to v.

Theorem 7.3.3.

Let u ∈ W _loc ^1,1 (Ω). Then the following properties are equivalent:

• (a) $\nabla u \in {L}^{1}(\Omega ; {\mathbb{R}}^{N})$

• (b) $\vert \vert Du\vert \vert _{\Omega } < \infty$ .

In this case,

$$\displaystyle{ \vert \vert Du\vert \vert _{\Omega } =\vert \vert \nabla u\vert \vert _{{L}^{1}(\Omega )}. }$$

Proof.

• (a) Assume that $\nabla u \in {L}^{1}(\Omega ; {\mathbb{R}}^{N})$. Let $v \in \mathcal{D}(\Omega ; {\mathbb{R}}^{N})$ be such that $\vert \vert v\vert \vert _{\infty }\leq 1$. It follows from the Cauchy–Schwarz inequality that

  $$\displaystyle{ \int _{\Omega }u\mathrm{\ div\ }v\ dx = -\int _{\Omega }\sum _{k=1}^{N}v_{ k}\partial _{k}u\ dx \leq \int _{\Omega }\vert \nabla u\vert dx. }$$

  Hence $\vert \vert Du\vert \vert _{\Omega } \leq \vert \vert \nabla u\vert \vert _{{L}^{1}(\Omega )}$.

  Theorem 4.3.11 implies the existence of $(w_{n}) \subset \mathcal{D}(\Omega ; {\mathbb{R}}^{N})$ converging to ∇ u in ${L}^{1}(\Omega ; {\mathbb{R}}^{N})$. We can assume that w _n  → ∇ u almost everywhere on Ω. Let us define

  $$\displaystyle{ v_{n} = w_{n}/\sqrt{\vert w_{n } {\vert }^{2 } + 1/n}. }$$

  We infer from Lebesgue’s dominated convergence theorem that

  $$\displaystyle{ \vert \vert \nabla u\vert \vert _{{L}^{1}(\Omega )} =\int _{\Omega }\vert \nabla u\vert dx =\lim _{n\rightarrow \infty }\int _{\Omega }v_{n} \cdot \nabla u\ dx \leq \vert \vert Du\vert \vert _{\Omega }. }$$

• (b) Assume that $\vert \vert Du\vert \vert _{\Omega } < \infty$ and define

  $$\displaystyle{ \omega _{n} =\{ x \in \Omega : d(x,\partial \Omega ) > 1/n\mbox{ and }\vert x\vert < n\}. }$$

  Then by the preceding step, we obtain

  $$\displaystyle{ \vert \vert \nabla u\vert \vert _{{L}^{1}(\omega _{n})} =\vert \vert Du\vert \vert _{\omega _{n}} \leq \vert \vert Du\vert \vert _{\Omega } < \infty. }$$

  Levi’s theorem ensures that $\nabla u \in {L}^{1}(\Omega ; {\mathbb{R}}^{N})$.

Example.

There exists a function everywhere differentiable on [ − 1, 1] such that $\vert \vert Du\vert \vert _{]-1,1[} = +\infty$. We define

$$\displaystyle\begin{array}{rcl} \begin{array}{lll} u(x)& = 0, &\quad x = 0, \\ & = {x}^{2}\sin \frac{1} {{x}^{2}},&\quad 0 < \vert x\vert \leq 1.\end{array} & & {}\\ \end{array}$$

We obtain

$$\displaystyle\begin{array}{rcl} \begin{array}{lll} u^{\prime}(x)& = 0, &\quad x = 0, \\ & = 2x\sin \frac{1} {{x}^{2}} -\frac{2} {x}\cos \frac{1} {{x}^{2}},&\quad 0 < \vert x\vert \leq 1.\end{array} & & {}\\ \end{array}$$

The preceding theorem implies that

$$\displaystyle{ +\infty =\lim _{n\rightarrow \infty }\vert \vert u^{\prime}\vert \vert _{{L}^{1}(]1/n,1[)} \leq \vert \vert Du\vert \vert _{]-1,1[}. }$$

Indeed,

$$\displaystyle{ 2\int _{0}^{1}\vert \cos \frac{1} {{x}^{2}}\vert \frac{dx} {x} =\int _{ 1}^{\infty }\vert \cos t\vert \frac{dt} {t} = +\infty. }$$

The function u has no weak derivative!

Example (Cantor function). 

There exists a continuous nondecreasing function with almost everywhere zero derivative and positive total variation. We use the notation of the last example of Sect. 2.2. We consider the Cantor set C corresponding to $\ell_{n} = 1/{3}^{n+1}$. Observe that

$$\displaystyle{ m(C) = 1 -\sum _{j=0}^{\infty }{2}^{j}/{3}^{j+1} = 0. }$$

We define on $\mathbb{R}$,

$$\displaystyle{ u_{n}(x) ={ \left (\frac{3} {2}\right )}^{n}\int _{ 0}^{x}\chi _{ C_{n}}(t)dt. }$$

It is easy to verify by symmetry that

$$\displaystyle{ \vert \vert u_{n+1} - u_{n}\vert \vert _{\infty }\leq \frac{1} {3} \frac{1} {{2}^{n+1}}. }$$

By the Weierstrass test, (u _n ) converges uniformly to the Cantor function $u \in \mathcal{C}(\mathbb{R})$. For n ≥ m, u′ _n  = 0 on $\mathbb{R} \setminus C_{m}$. The closing lemma implies that u′ = 0 on $\mathbb{R} \setminus C_{m}$. Since m is arbitrary, u′ = 0 on $\mathbb{R} \setminus C$. Theorems 7.3.2 and 7.3.3 ensure that

$$\displaystyle{ \vert \vert Du\vert \vert _{\mathbb{R}} \leq \begin{array}{c} \underline{\lim } \\ n\rightarrow \infty \end{array} \vert \vert u^{\prime}_{n}\vert \vert _{{L}^{1}(\mathbb{R})} = 1. }$$

Let $v \in \mathcal{D}(\mathbb{R})$ be such that $\vert \vert v\vert \vert _{\infty } = 1$ and $v = -1$ on [0, 1] and integrate by parts:

$$\displaystyle{ \int _{\mathbb{R}}v^{\prime}u\ dx =\lim _{n\rightarrow \infty }\int _{\mathbb{R}}v^{\prime}u_{n}\ dx = -\lim _{n\rightarrow \infty }\int _{\mathbb{R}}vu^{\prime}_{n}dx =\lim _{n\rightarrow \infty }{\left (\frac{3} {2}\right )}^{n}m(C_{ n}) = 1. }$$

We conclude that $\vert \vert Du\vert \vert _{\mathbb{R}} = 1$. The function u has no weak derivative.

Definition 7.3.4.

Let Ω be an open subset of ${\mathbb{R}}^{N}$. On the space

$$\displaystyle{ BV (\Omega ) =\{ u \in {L}^{1}(\Omega ) :\vert \vert Du\vert \vert _{ \Omega } < \infty \}, }$$

we define the norm

$$\displaystyle{ \vert \vert u\vert \vert _{BV (\Omega )} =\vert \vert u\vert \vert _{{L}^{1}(\Omega )} +\vert \vert Du\vert \vert _{\Omega } }$$

and the distance of strict convergence

$$\displaystyle{ d_{S}(u,v) =\vert \vert u - v\vert \vert _{{L}^{1}(\Omega )} +\big\vert \vert \vert Du\vert \vert _{\Omega } -\vert \vert Dv\vert \vert _{\Omega }\big\vert. }$$

Remark.

It is clear that convergence in norm implies strict convergence.

Example.

The space BV (]0, π[), with the distance of strict convergence, is not complete. We define on ]0, π[,

$$\displaystyle{ u_{n}(x) = \frac{1} {n}\cos nx, }$$

so that u _n  → 0 in L ¹(]0, π[). By Theorem 7.3.3, for every n,

$$\displaystyle{ \vert \vert Du_{n}\vert \vert _{]0,\pi [} =\int _{ 0}^{\pi }\vert \sin nx\vert dx = 2. }$$

Hence $\lim _{j,k\rightarrow \infty }d_{S}(u_{j},u_{k}) =\lim _{j,k\rightarrow \infty }\vert \vert u_{j} - u_{k}\vert \vert _{{L}^{1}(]0,\pi [)} = 0$. If lim _{n → ∞} d _S (u _n , v) = 0, then v = 0. But lim _{n → ∞} d _S (u _n , 0) = 2. This is a contradiction.

Proposition 7.3.5.

The normed space BV (Ω) is complete.

Proof.

Let (u _n ) be a Cauchy sequence on the normed space BV (Ω). Then (u _n ) is a Cauchy sequence1pc]“Theorem 27.2 and Proposition 15.15” is not given. Please check and provide. in L ¹(Ω), so that u _n  → u in L ¹(Ω).

Let ɛ > 0. There exists m such that for j, k ≥ m, $\vert \vert D(u_{j} - u_{k})\vert \vert _{\Omega } \leq \varepsilon$. Theorem 7.3.2 implies that for k ≥ m, $\vert \vert D(u_{k}-u)\vert \vert \leq \begin{array}{c} \underline{\lim } \\ j\rightarrow \infty \end{array} \vert \vert D(u_{j}-u_{k})\vert \vert _{\Omega } \leq \varepsilon$. Since ɛ > 0 is arbitrary, $\vert \vert D(u_{k} - u)\vert \vert _{\Omega } \rightarrow 0$, k → ∞.

Lemma 7.3.6.

Let $u \in L_{\mathrm{loc}}^{1}({\mathbb{R}}^{N})$ be such that $\vert \vert Du\vert \vert _{{\mathbb{R}}^{N}} < \infty$ . Then

$$\displaystyle{ \vert \vert \nabla (\rho _{n} {\ast} u)\vert \vert _{{L}^{1}({\mathbb{R}}^{N})} \leq \vert \vert Du\vert \vert _{{\mathbb{R}}^{N}}\mbox{ and }\vert \vert Du\vert \vert _{{\mathbb{R}}^{N}} =\lim _{n\rightarrow \infty }\vert \vert \nabla (\rho _{n} {\ast} u)\vert \vert _{{L}^{1}({\mathbb{R}}^{N})}. }$$

Proof.

Let $v \in \mathcal{D}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})$ be such that $\vert \vert v\vert \vert _{\infty }\leq 1$. It follows from Proposition 4.3.15 that

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}(\rho _{n} {\ast} u)\mathrm{\ div\ }v\ dx =\int _{{\mathbb{R}}^{N}}u\sum _{k=1}^{N}\rho _{ n} {\ast} \partial _{k}v_{k}dx =\int _{{\mathbb{R}}^{N}}u\sum _{k=1}^{N}\partial _{ k}(\rho _{n} {\ast} v_{k})dx. }$$

The Cauchy–Schwarz inequality implies that for every $x \in {\mathbb{R}}^{N}$,

$$\displaystyle{ \sum _{k=1}^{N}{(\rho _{ n}{\ast}v_{k}(x))}^{2} =\sum _{ k=1}^{N}{\left (\int _{{ \mathbb{R}}^{N}}\rho _{n}(x - y)v_{k}(y)dy\right )}^{2} \leq \sum _{ k=1}^{N}\int _{{ \mathbb{R}}^{N}}\rho _{n}(x-y){(v_{k}(y))}^{2}dy \leq 1. }$$

Hence we obtain

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}(\rho _{n} {\ast} u)\mathrm{\ div\ }v\ dx \leq \vert \vert Du\vert \vert _{{\mathbb{R}}^{N}}, }$$

and by Theorem 7.3.3, $\vert \vert \nabla (\rho _{n} {\ast} u)\vert \vert _{{L}^{1}({\mathbb{R}}^{N})} \leq \vert \vert Du\vert \vert _{{\mathbb{R}}^{N}}$.

By the regularization theorem, ρ _n  ∗ u → u in $L_{\mathrm{loc}}^{1}({\mathbb{R}}^{N})$. Theorems 7.3.2 and 7.3.3 ensure that

$$\displaystyle{\vert \vert Du\vert \vert _{{\mathbb{R}}^{N}} \leq \begin{array}{c} \underline{\lim } \\ n\rightarrow \infty \end{array} \vert \vert \nabla (\rho _{n}{\ast}u)\vert \vert _{{L}^{1}({\mathbb{R}}^{N})}.}$$

 □ 

Theorem 7.3.7.

• (a) For every $u \in BV ({\mathbb{R}}^{N})$ , (ρ _n ∗ u) converges strictly to u.

• (b) (Gagliardo–Nirenberg inequality.) Let N ≥ 2. There exists c _​N > 0 such that for every $u \in BV ({\mathbb{R}}^{N})$,

  $$\displaystyle{ \vert \vert u\vert \vert _{{L}^{N/(N-1)}({\mathbb{R}}^{N})} \leq c_{N}\vert \vert Du\vert \vert _{{\mathbb{R}}^{N}}. }$$

Proof.

• (a) Proposition 4.3.14 and the preceding lemma imply the strict convergence of (ρ _n  ∗ u) to u.

• (b) Let N ≥ 2. We can assume that $\rho _{n_{k}} {\ast} u \rightarrow u$ almost everywhere on ${\mathbb{R}}^{N}$. It follows from Fatou’s lemma and Sobolev’s inequality in ${\mathcal{D}}^{1,1}({\mathbb{R}}^{N})$ that

  $$\displaystyle{\vert \vert u\vert \vert _{N/(N-1)} \leq \begin{array}{c} \underline{\lim } \\ k\rightarrow \infty \end{array} \vert \vert \rho _{n_{k}}{\ast}u\vert \vert _{N/(N-1)} \leq c_{N}\lim _{n\rightarrow \infty }\vert \vert \nabla (\rho _{n_{k}}{\ast}u)\vert \vert _{1} = c_{N}\vert \vert Du\vert \vert _{{\mathbb{R}}^{N}}.}$$

   □ 

7.4 Perimeter

The perimeter of a smooth domain is the total variation of its characteristic function.

Theorem 7.4.1.

Let Ω be an open subset of ${\mathbb{R}}^{N}$ of class ${\mathcal{C}}^{1}$ with a bounded boundary Γ. Then

$$\displaystyle{ \int _{\Gamma }d\gamma =\vert \vert D\chi _{\Omega }\vert \vert _{{\mathbb{R}}^{N}}. }$$

Proof.

Let $v \in \mathcal{D}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})$ be such that $\vert \vert v\vert \vert _{\infty }\leq 1$. The divergence theorem and the Cauchy–Schwarz inequality imply that

$$\displaystyle{ \int _{\Omega }\mathrm{div}\ v\ dx =\int _{\Gamma }v \cdot n\ d\gamma \leq \int _{\Gamma }\vert v\vert \ \vert n\vert d\gamma \leq \int _{\Gamma }d\gamma. }$$

Taking the supremum with respect to v, we obtain $\vert \vert D\chi _{\Omega }\vert \vert _{{\mathbb{R}}^{N}} \leq \int _{\Gamma }d\gamma$.

We use the notation of Definition 9.2.1 and define

$$\displaystyle{ U =\{ x \in {\mathbb{R}}^{N} : \nabla \varphi (x)\neq 0\}, }$$

so that Γ ⊂ U. The theorem of partitions of unity ensures the existence of $\psi \in \mathcal{D}(U)$ such that 0 ≤ ψ ≤ 1 and ψ = 1 on Γ. We define

$$\displaystyle\begin{array}{rcl} \begin{array}{lll} v(x)& =\psi (x)\nabla \varphi (x)/\vert \nabla \varphi (x)\vert,&x \in U \\ & = 0, &x \in {\mathbb{R}}^{N} \setminus U.\end{array} & & {}\\ \end{array}$$

It is clear that $v \in \mathcal{K}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})$ and for every γ ∈ Γ, v(γ) = n(γ). For every m ≥ 1, $w_{m} =\rho _{m} {\ast} v \in \mathcal{D}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})$. We infer from the divergence and regularization theorems that

$$\displaystyle{ \lim _{m\rightarrow \infty }\int _{\Omega }\mathrm{div}\ w_{m}\ dx =\lim _{m\rightarrow \infty }\int _{\Gamma }w_{m} \cdot n\ d\gamma =\int _{\Gamma }n \cdot n\ d\gamma =\int _{\Gamma }d\gamma. }$$

By definition, $\vert \vert v\vert \vert _{\infty }\leq 1$, and by the Cauchy–Schwarz inequality,

$$\displaystyle{ \sum _{k=1}^{N}{(\rho _{ m}{\ast}v_{k}(x))}^{2} =\sum _{ k=1}^{N}{\left (\int _{{ \mathbb{R}}^{N}}\rho _{m}(x - y)v_{k}(y)dy\right )}^{2} \leq \sum _{ k=1}^{N}\int _{{ \mathbb{R}}^{N}}\rho _{m}(x-y){(v_{k}(y))}^{2}dy \leq 1. }$$

We conclude that $\int _{\Gamma }d\gamma \leq \vert \vert D\chi _{\Omega }\vert \vert _{{\mathbb{R}}^{N}}$.

The preceding theorem suggests a functional definition of the perimeter due to De Giorgi.

Definition 7.4.2.

Let A be a measurable subset of ${\mathbb{R}}^{N}$. The perimeter of A is defined by $p(A) =\vert \vert D\chi _{A}\vert \vert _{{\mathbb{R}}^{N}}$.

The proof of the Morse–Sard theorem is given in Sect. 9.3.

Theorem 7.4.3.

Let Ω be an open subset of ${\mathbb{R}}^{N}$ and $u \in {\mathcal{C}}^{\infty }(\Omega )$ . Then the Lebesgue measure of

$$\displaystyle{ \{t \in \mathbb{R} : \mbox{ there exists }x \in \Omega \mbox{ such that }u(x) = t\mbox{ and }\nabla u(x) = 0\} }$$

is equal to zero.

Theorem 7.4.4.

Let 1 < p < ∞, u ∈ L ^p (Ω), u ≥ 0, and g ∈ L ^p′ (Ω). Then

• (a) ∫ _Ω g u dx = ∫ ₀ ^∞ dt∫ _u>t g dx;

• (b) $\vert \vert u\vert \vert _{p} \leq \int _{0}^{\infty }m{(\{u > t\})}^{1/p}dt$

• (c) $\vert \vert u\vert \vert _{p}^{p} =\int _{ 0}^{\infty }m(\{u > t\})p{t}^{p-1}dt$ .

Proof.

• (a) We deduce from Fubini’s theorem that

  $$\displaystyle{ \begin{array}{ll} \int _{\Omega }g\ u\ dx& = \int _{\Omega }dx\int _{0}^{\infty }g\ \chi _{ u>t}\ dt\\ \\ & = \int _{0}^{\infty }dt\int _{ \Omega }g\ \chi _{u>t}\ dx\\ \\ & = \int _{0}^{\infty }dt\int _{ u>t}g\ dx.\end{array} }$$

• (b) If $\vert \vert g\vert \vert _{p^{\prime}} = 1$, we obtain from Hölder’s inequality that

  $$\displaystyle{ \int _{\Omega }g\ u\ dx \leq \int _{0}^{\infty }m{(\{u > t\})}^{1/p}dt. }$$

  It suffices then to take the supremum.

• (c) Define f(t) = t ^p. It follows from Fubini’s theorem that

  $$\displaystyle\begin{array}{rcl} \vert \vert u\vert \vert _{p}^{p}& = \int _{ \Omega }dx\int _{0}^{u}f^{\prime}(t)dt & {}\\ & & {}\\ & = \int _{\Omega }dx\int _{0}^{\infty }\chi _{ u>t}f^{\prime}(t)dt & {}\\ & & {}\\ & = \int _{0}^{\infty }dt\int _{ \Omega }\chi _{u>t}f^{\prime}(t)dx & {}\\ & & {}\\ & = \int _{0}^{\infty }m(\{u > t\})f^{\prime}(t)dt.& \sqcap \sqcup {}\\ \end{array}$$

Theorem 7.4.5 (Coarea formula). 

Let $u \in \mathcal{D}({\mathbb{R}}^{N})$ and $f \in {\mathcal{C}}^{1}({\mathbb{R}}^{N})$ . Then

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}f\ \vert \nabla u\vert \ dx =\int _{ 0}^{\infty }dt\int _{ \vert u\vert =t}f\ d\gamma. }$$

Moreover, for every open subset Ω of ${\mathbb{R}}^{N}$,

$$\displaystyle{\int _{\Omega }\vert \nabla u\vert \ dx =\int _{ 0}^{\infty }dt\int _{ \vert u\vert =t}\chi _{\Omega }\ d\gamma.}$$

Proof.

By the Morse–Sard theorem, for almost every $t \in \mathbb{R}$,

$$\displaystyle{ u(x) = t\Longrightarrow\nabla u(x)\neq 0. }$$

Hence for almost every t > 0, the open sets {u > t} and {u <  − t} are smooth.

We infer from Lemma 6.1.1, Theorem 7.4.4, and the divergence theorem that for every $v \in {\mathcal{C}}^{1}({\mathbb{R}}^{N}; {\mathbb{R}}^{N})$,

$$\displaystyle{ \begin{array}{ll} \int _{{\mathbb{R}}^{N}}\nabla u \cdot v\ dx& = -\int _{{\mathbb{R}}^{N}}u\ \mathrm{div\ }v\ dx\\ \\ & = -\int _{0}^{\infty }dt\int _{ u>t}\mathrm{div\ }v\ dx +\int _{ 0}^{\infty }dt\int _{ u<-t}\mathrm{div\ }v\ dx\\ \\ & = \int _{0}^{\infty }dt\int _{ \vert u\vert =t}v \cdot \frac{\nabla u} {\vert \nabla u\vert }d\gamma. \end{array} }$$

Define

$$\displaystyle{ v_{n} = f\ \nabla u/\sqrt{\vert \nabla u{\vert }^{2 } + 1/n}. }$$

Lebesgue’s dominated convergence theorem implies that

$$\displaystyle{\int _{{\mathbb{R}}^{N}}f\ \vert \nabla u\vert \ dx =\lim _{n\rightarrow \infty }\int _{{\mathbb{R}}^{N}}\nabla u \cdot v_{n}\ dx =\lim _{n\rightarrow \infty }\int _{0}^{\infty }dt\int _{ \vert u\vert =t}v_{n} \cdot \frac{\nabla u} {\vert \nabla u\vert }d\gamma =\int _{ 0}^{\infty }dt\int _{ \vert u\vert =t}f\ d\gamma.}$$

Define

$$\displaystyle{\omega _{n} =\{ x \in \Omega \ :\ d(x,\partial \Omega ) > 1/n\ \text{and}\ \vert x\vert < n\}.}$$

For all n, there exists $\varphi _{n} \in \mathcal{D}(\omega _{n+1})$ such that 0 ≤ φ _n  ≤ 1 and φ _n  = 1 on ω _n . Levi’s monotone convergence theorem implies that

$$\displaystyle{\int _{\Omega }\vert \nabla u\vert dx =\lim _{n\rightarrow \infty }\int _{{\mathbb{R}}^{N}}\varphi _{n}\vert \nabla u\vert dx =\lim _{n\rightarrow \infty }\int _{0}^{\infty }dt\int _{ \vert u\vert =t}\varphi _{n}d\gamma =\int _{ 0}^{\infty }dt\int _{ \vert u\vert =t}\chi _{\Omega }\ d\gamma.}$$

 □ 

Lemma 7.4.6.

Let 1 ≤ p < N, let K be a compact subset of ${\mathbb{R}}^{N}$ , and a > cap _p (K). Then there exist V open and $v \in \mathcal{D}({\mathbb{R}}^{N})$ such that K ⊂ V, χ _V ≤ v, and ∫ _Ω |∇v| dx < a.

Proof.

By assumption, there exist $u \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N})$ and U open such that K ⊂ U, χ _U  ≤ u, and

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla u{\vert }^{p}dx < a. }$$

There exists V open such that K ⊂ V ⊂ ⊂ U. For m large enough, χ _V  ≤ w = ρ _m  ∗ u and

$$\displaystyle{ \int _{{\mathbb{R}}^{N}}\vert \nabla w{\vert }^{p}dx < a. }$$

Let $\theta _{n}(x) =\theta (\vert x\vert /n)$ be a truncating sequence. For n large enough, χ _V  ≤ v = θ _n w and

$$\displaystyle{\int _{{\mathbb{R}}^{N}}\vert \nabla v{\vert }^{p}dx < a.}$$

 □ 

Theorem 7.4.7.

Let N ≥ 2 and let K be a compact subset of ${\mathbb{R}}^{N}$ . Then

$$\displaystyle{ \mathrm{cap}_{1}(K) =\inf \{ p(U) : U\mbox{ is open and bounded, and }U \supset K\}. }$$

Proof.

We denote by Cap₁(K) the second member of the preceding equality. Let U be open, bounded, and such that U ⊃ K. Define u _n  = ρ _n  ∗ χ _U . For n large enough, $u \in \mathcal{D}_{K}^{1,1}({\mathbb{R}}^{N})$. Lemma 7.3.6 implies that for n large enough,

$$\displaystyle{ \mathrm{cap}_{1}(K) \leq \int _{{\mathbb{R}}^{N}}\vert \nabla u_{n}\vert dx \leq \vert \vert D\chi _{U}\vert \vert _{{\mathbb{R}}^{N}} = p(U). }$$

Taking the infimum with respect to U, we obtain cap₁(K) ≤ Cap₁(K).

Let a > cap₁(K). By the preceding lemma, there exist V open and $v \in \mathcal{D}({\mathbb{R}}^{N})$ such that K ⊂ V, χ _V  ≤ v and $\int _{{\mathbb{R}}^{N}}\vert \nabla v\vert dx < a$. We deduce from the Morse–Sard theorem and from the coarea formula that

$$\displaystyle{ \mathrm{Cap}_{1}(K) \leq \int _{0}^{1}dt\int _{ v=t}d\gamma \leq \int _{0}^{\infty }dt\int _{ v=t}d\gamma =\int _{{\mathbb{R}}^{N}}\vert \nabla v\vert dx < a. }$$

Since a > cap₁(K) is arbitrary, we conclude that Cap₁(K) ≤ cap₁(K).

7.5 Comments

The book by Maz’ya ([51]) is the main reference on functions of bounded variations and on capacity theory. The beautiful proof of the coarea formula (Theorem 7.4.5) is due to Maz’ya. The derivative of the function of unbounded variation in Sect. 7.3 is Denjoy–Perron integrable (since it is a derivative); see Analyse, fondements techniques, évolution by J. Mawhin ([49]).

7.6 Exercises for Chap. 7

• 1. Let 1 ≤ p < N. Then

  $$\displaystyle\begin{array}{rcl} & \qquad \lambda p + N < 0 \Leftrightarrow {(1 + \vert x{\vert }^{2})}^{\lambda /2} \in {W}^{1,p}({\mathbb{R}}^{N}), & {}\\ & (\lambda -1)p + N < 0 \Leftrightarrow {(1 + \vert x{\vert }^{2})}^{\lambda /2} \in {\mathcal{D}}^{1,p}({\mathbb{R}}^{N}).& {}\\ \end{array}$$

• 2. What are the interior and the closure of W ^1, 1(Ω) in BV (Ω)?

• 3. Let u ∈ L _loc ¹(Ω). The following properties are equivalent:

  • (a) $\vert \vert Du\vert \vert _{\Omega } < \infty$;

  • (b) there exists c > 0 such that for every ω ⊂ ⊂ Ω and every $y \in {\mathbb{R}}^{N}$ such that | y |  < d(ω, ∂Ω)

    $$\displaystyle{ \vert \vert \tau _{y}u - u\vert \vert _{{L}^{1}(\omega )} \leq c\vert y\vert. }$$

• 4. (Relative variational capacity.) Let Ω be an open bounded subset of ${\mathbb{R}}^{N}$ (or more generally, an open subset bounded in one direction). Let 1 ≤ p < ∞ and let K be a compact subset of Ω. The capacity of degree p of K relative to Ω is defined by

  $$\displaystyle{ \mbox{ cap}_{p,\Omega }(K) =\inf \left \{\int _{\Omega }\vert \nabla u{\vert }^{p}dx : u \in W_{ K}^{1,p}(\Omega )\right \}, }$$

  where

  $$\displaystyle\begin{array}{rcl} W_{K}^{1,p}(\Omega ) =& \;\{u \in W_{ 0}^{1,p}(\Omega ) : \mbox{ there exists}\omega \mbox{ such that}K \subset \omega \subset \subset \Omega & {}\\ & \;\mbox{ and}\;\chi _{\omega } \leq u\;\mbox{ a.e. on}\;\Omega \}. & {}\\ \end{array}$$

  Prove that the capacity of degree p relative to Ω is a capacity on Ω.

• 5. Verify that

  $$\displaystyle{ \mbox{ cap}_{p,\Omega }(K) =\inf \left \{\int _{\Omega }\vert \nabla u{\vert }^{p}dx : u \in \mathcal{D}_{ K}(\Omega )\right \}, }$$

  where

  $$\displaystyle{ \mathcal{D}_{K}(\Omega ) =\{ u \in \mathcal{D}(\Omega ) : \mbox{ there exists}\omega \mbox{ such that}K \subset \omega \subset \subset \Omega \mbox{ and}\chi _{\omega } \leq u\}. }$$
• 6.

  • (a) If cap _p, Ω (K) = 0, then m(K) = 0. Hint: Use Poincaré’s inequality.

  • (b) If p > N and if cap _p, Ω (K) = 0, then K = ϕ. Hint: Use the Morrey inequalities.

• 7. Assume that cap _p, Ω (K) = 0. Then for every $u \in \mathcal{D}(\Omega )$, there exists $(u_{n}) \subset \mathcal{D}(\Omega \setminus K)$ such that | u _n  | ≤ | u | and u _n  → u in W ^1, p(Ω).

• 8. (Dupaigne–Ponce, 2004.) Assume that cap_1, Ω (K) = 0. Then W ^1, p(Ω ∖ K) = W ^1, p(Ω). Hint: Consider first the bounded functions in W ^1, p(Ω ∖ K).

• 9. For every $u \in BV ({\mathbb{R}}^{N})$,

  $$\displaystyle{ \vert \vert D\vert u\vert \vert \vert _{{\mathbb{R}}^{N}} \leq \vert \vert D{u}^{+}\vert \vert _{{ \mathbb{R}}^{N}} +\vert \vert D{u}^{-}\vert \vert _{{ \mathbb{R}}^{N}} =\| Du\|_{{\mathbb{R}}^{N}}. }$$

  Hint: Consider a sequence $(u_{n}) \subset {W}^{1,1}({\mathbb{R}}^{N})$ such that u _n  → u strictly in $BV ({\mathbb{R}}^{N})$.

• 10. Let u ∈ L ¹(Ω) and $f \in \mathcal{B}{\mathcal{C}}^{1}(\Omega )$. Then

  $$\displaystyle{ \vert \vert D(fu)\vert \vert _{\Omega } \leq \vert \vert f\vert \vert _{\infty }\vert \vert Du\vert \vert _{\Omega } +\vert \vert \nabla f\vert \vert _{\infty }\vert \vert u\vert \vert _{{L}^{1}(\Omega )}. }$$

• 11. (Cheeger constant.) Let Ω be an open bounded domain in ${\mathbb{R}}^{N}$ and define

  $$\displaystyle{ h(\Omega ) =\inf \{ p(\omega )/m(\omega ) :\omega \subset \subset \Omega \mbox{ and}\omega \mbox{ is of class}{\mathcal{C}}^{1}\}. }$$

  Then for 1 ≤ p < ∞ and every u ∈ W ₀ ^1, p(Ω),

  $$\displaystyle{{ \left (\frac{h(\Omega )} {p} \right )}^{p}\int _{ \Omega }\vert u{\vert }^{p}dx \leq \int _{ \Omega }\vert \nabla u{\vert }^{p}dx. }$$

  Hint: Assume first that p = 1 and apply the coarea formula to $u \in \mathcal{D}(\Omega )$.

• 12. Let u ∈ W ^1, 1(Ω). Then

  $$\displaystyle{ \int _{\Omega }\sqrt{1+ \vert \nabla u{\vert }^{2}}dx =\sup \left \{\int _{\Omega }(v_{N+1} + u\sum _{k=1}^{N}\partial _{ k}u_{k})dx : u \in \mathcal{D}(\Omega ; {\mathbb{R}}^{N+1}),\|u\|_{ \infty }\leq 1\right \}. }$$