$W^{1,\mathrm{\infty }}$ functions on convex domains have Lipschitz representatives

Prepare the $W^{1,1}$ class. If $\mathrm{\Omega }=\mathrm{\varnothing }$ both assertions hold vacuously; assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. Then $0<\lambda (\mathrm{\Omega })<\mathrm{\infty }$ by [F4]. Since $u\in W^{1,\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$, the class $u$ and every $D_{i}u$ lie in $L^{\mathrm{\infty }}$ by [F1], and $\mathrm{\mid }Du\mathrm{\mid }\le L$ almost everywhere. By [F5] applied to the finite measure space $\mathrm{\Omega }$, also $u\in L^1(\mathrm{\Omega };\mathbb{K})$ and $D_{i}u\in L^1(\mathrm{\Omega };\mathbb{K})$. The test-function identities of [F2] that define the weak derivatives $D_{i}u$ are the same identities with the same test functions, so they remain valid with the $L^1$ in place of the $L^{\mathrm{\infty }}$ classes: $u\in W^{1,1}(\mathrm{\Omega };\mathbb{K})$, its $W^{1,1}$ weak gradient is the given $Du$, and $\mathrm{\mid }Du\mathrm{\mid }\le L$ almost everywhere.

Choose smooth approximations and a single subsequence. Let $u_{\epsilon }={\rho }_{\epsilon }\ast \tilde{u}$ be the interior mollifications of [F6], where $\tilde{u}$ is the zero extension of a representative of $u$; then $u_{\epsilon }\in C^{\mathrm{\infty }}({\mathbb{R}}^n;\mathbb{K})$ and $u_{\epsilon }\to u$ in $W^{1,1}(U;\mathbb{K})$ for every open $U\subset \subset \mathrm{\Omega }$. Put $U_m:=\{x\in \mathrm{\Omega }:\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })>1\mathrm{/}m\}\cap B(0,m)$ for $m\ge 1$; each $U_m$ is open and convex, $U_m\subseteq U_{m+1}\subset \subset \mathrm{\Omega }$, and ${\bigcup }_m{U}_m=\mathrm{\Omega }$. Countable Choice, available from the Axiom of Choice by [F3], is what [F6] uses here. For every $m$ the family $(u_{\epsilon }{)}_{\epsilon >0}$ converges to $u$ in $W^{1,1}(U_m;\mathbb{K})$ as $\epsilon \to 0$, so by [F7] we may extract recursively for $m=1,2,\dots$ a subsequence converging almost everywhere on $U_m$; the resulting diagonal sequence, rewritten as ${\epsilon }_k\downarrow 0$, satisfies: $u_{{\epsilon }_k}\to u$ almost everywhere on $\mathrm{\Omega }$, and for every $m$, $u_{{\epsilon }_k}\to u$ in $W^{1,1}(U_m;\mathbb{K})$ and $D{u}_{{\epsilon }_k}\to Du$ in $L^1(U_m;\mathbb{K})$. Each $u_{{\epsilon }_k}$ is smooth on all of ${\mathbb{R}}^n$.

A multiplicity estimate. Let $U\subseteq \mathrm{\Omega }$ be open convex with $\lambda (U)<\mathrm{\infty }$ and let $g:{\mathbb{R}}^n\to [0,\mathrm{\infty }]$ be Borel measurable. Then ${\int }_U{\int }_U{\int }_0^{1}g((1-t)x+ty)dtdydx\le C(n)\lambda (U){\int }_{{\mathbb{R}}^n}g$, where $C(n):=2{\int }_{1\mathrm{/}2}^1{t}^{-n}dt$ is finite and depends only on $n$. Indeed, fix $t\in [1\mathrm{/}2,1]$ and $x\in U$. The map $y\mapsto z=(1-t)x+ty$ is an affine diffeomorphism of ${\mathbb{R}}^n$ with linear part $tI$ and $\mathrm{\mid }\det (tI)\mathrm{\mid }=t^n$, so change of variables [F10] gives ${\int }_{U}g((1-t)x+ty)dy=t^{-n}{\int }_{(1-t)x+tU}g(z)dz\le t^{-n}{\int }_{{\mathbb{R}}^n}g$; integrating over $x\in U$ and $t\in [1\mathrm{/}2,1]$ gives at most $({\int }_{1\mathrm{/}2}^1{t}^{-n}dt)\lambda (U)\int g$. Symmetrically, for fixed $t\in [0,1\mathrm{/}2]$ and $y\in U$ the substitution $x\mapsto z=(1-t)x+ty$ has linear part $(1-t)I$, so ${\int }_{U}g((1-t)x+ty)dx\le (1-t{)}^{-n}\int g$; integrating over $y\in U$ and $t\in [0,1\mathrm{/}2]$ and substituting $t\mapsto 1-s$ gives the same bound. Tonelli's theorem [F9] justifies all iterated integrals, and adding the two halves gives the estimate.

The converse: $f$ is bounded and its a.e. partial derivatives are bounded functions. Assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$ and fix $x_0\in \mathrm{\Omega }$. For every $x\in \mathrm{\Omega }$ the Lipschitz condition [F15] gives $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }f(x_0)\mathrm{\mid }+L\mathrm{\mid }x-x_0\mathrm{\mid }\le \mathrm{\mid }f(x_0)\mathrm{\mid }+L\mathrm{diam}(\mathrm{\Omega })<\mathrm{\infty }$, so $f$ is bounded, hence $f\in L^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ and $f\in L^1(\mathrm{\Omega };\mathbb{K})$ by [F5]. For $i\in \{1,\dots ,n\}$ define $g_i(x)$ to be the limit of $(f(x+e_i\mathrm{/}k)-f(x))\mathrm{/}(1\mathrm{/}k)$ where that limit exists, and $0$ elsewhere, the quotient being declared $0$ when $x+e_i\mathrm{/}k\mathrm{\notin }\mathrm{\Omega }$. Each quotient is continuous on its open domain and has modulus at most $L$ there, so $g_i$ is measurable with $\mathrm{\mid }{g}_i\mathrm{\mid }\le L$ everywhere (use componentwise convergence for complex values) on $\mathrm{\Omega }$. Moreover $g_i={\mathrm{\partial }}_{i}f$ almost everywhere: for each fixed value of the other $n-1$ coordinates the section $t\mapsto f(\dots ,t,\dots )$ is $L$-Lipschitz on an interval, hence absolutely continuous and differentiable at almost every $t$ by [F12], and by Fubini [F9] the set of $x\in \mathrm{\Omega }$ at which the classical partial derivative fails to exist has measure zero.

Segment identities and the integrated error. Each $u_{{\epsilon }_k}$ is smooth on ${\mathbb{R}}^n$, so for all $x,y\in {\mathbb{R}}^n$ the chain rule and the fundamental theorem of calculus [F11] give $u_{{\epsilon }_k}(y)-u_{{\epsilon }_k}(x)={\int }_0^{1}D{u}_{{\epsilon }_k}(x+t(y-x))\cdot (y-x)dt$. Choose finite-valued Borel representatives of the components of $Du$ by [F16]. Fix $m$ and apply step 1.3 with $U=U_m$ and $g={\mathbf{1}}_{U_m}\mathrm{\mid }D{u}_{{\epsilon }_k}-Du\mathrm{\mid }$, extended by zero: by convexity $x+t(y-x)\in U_m$ whenever $x,y\in U_m$, so ${\int }_{U_m}{\int }_{U_m}{\int }_0^1\mathrm{\mid }D{u}_{{\epsilon }_k}-Du\mathrm{\mid }(x+t(y-x))dtdydx\le C(n)\lambda (U_m)\mathrm{\parallel }D{u}_{{\epsilon }_k}-Du{\mathrm{\parallel }}_{L^1(U_m;\mathbb{K})}$, and the right-hand side tends to $0$ as $k\to \mathrm{\infty }$ by step 1.2. Passing to a further subsequence, once more chosen diagonally over $m$ and relabelled, we may suppose that for every $m$ and almost every $(x,y)\in U_m\times U_m$ the integral ${\int }_0^1\mathrm{\mid }D{u}_{{\epsilon }_k}-Du\mathrm{\mid }(x+t(y-x))dt$ tends to $0$.

The converse: the functions $g_i$ are the weak derivatives and $f\in W^{1,\mathrm{\infty }}$. Let $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$. Fix $i$ and write $h=1\mathrm{/}k$. Since $\phi$ is smooth with compact support, the difference quotients $(\phi (x)-\phi (x-h{e}_i))\mathrm{/}h$ converge uniformly to ${\mathrm{\partial }}_i\phi$ as $h\to 0$, and $f\in L^1(\mathrm{\Omega };\mathbb{K})$ by step 1.4, so ${\int }_{\mathrm{\Omega }}f(x)(\phi (x)-\phi (x-h{e}_i))\mathrm{/}hdx\to {\int }_{\mathrm{\Omega }}f{\mathrm{\partial }}_i\phi$. Substituting $x=y+h{e}_i$ and using the translation case of the change-of-variables identity [F10] (determinant one) turns that integral into ${\int }_{\mathrm{\Omega }}(f(x)-f(x+h{e}_i))\mathrm{/}h\phi (x)dx$. On the compact set $\mathrm{supp}\phi \subset \subset \mathrm{\Omega }$ and for $k$ large both $x$ and $x+e_i\mathrm{/}k$ lie in $\mathrm{\Omega }$, so there the quotient $q_k(x):=(f(x+e_i\mathrm{/}k)-f(x))\mathrm{/}(1\mathrm{/}k)$ is a Lipschitz difference quotient with $\mathrm{\mid }{q}_k\mathrm{\mid }\le L$, and $q_k\to {\mathrm{\partial }}_{i}f$ almost everywhere by step 1.4; dominated convergence [F13] with dominating function $L\mathrm{\mid }\phi \mathrm{\mid }$ on the finite measure set $\mathrm{supp}\phi$ therefore gives ${\int }_{\mathrm{\Omega }}f{\mathrm{\partial }}_i\phi =-{\int }_{\mathrm{\Omega }}{g}_i\phi$. By the defining identity [F2], $g_i=D_{i}f$ for every $i$, and with $f\in L^{\mathrm{\infty }}$ from step 1.4 and $g_i\in L^{\mathrm{\infty }}$ this shows $f\in W^{1,\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$.

The almost-everywhere pair bound. Fix $m$. By Fubini [F9], for almost every $(x,y)\in U_m\times U_m$ both $x$ and $y$ are points of almost-everywhere convergence of $u_{{\epsilon }_k}$ toward $u$ and at the same time the integral convergence of step 2.1 holds. For such a pair define $u^{\ast }(x):={\lim }_k{u}_{{\epsilon }_k}(x)$ and $u^{\ast }(y)$ similarly, and extend $u^{\ast }$ arbitrarily elsewhere. Passing to the limit in the identity of step 2.1, the left-hand side tends to $u^{\ast }(y)-u^{\ast }(x)$, while the difference of the right-hand sides obeys $\mid {\int }_0^1[D{u}_{{\epsilon }_k}-Du](x+t(y-x))\cdot (y-x)dt\mid \le \mathrm{\mid }y-x\mathrm{\mid }{\int }_0^1\mathrm{\mid }D{u}_{{\epsilon }_k}-Du\mathrm{\mid }(x+t(y-x))dt\to 0$; hence $u^{\ast }(y)-u^{\ast }(x)={\int }_0^{1}Du(x+t(y-x))\cdot (y-x)dt$, where $\mathrm{\mid }Du\mathrm{\mid }\le L$ for almost every parameter on almost every pair segment: apply step 1.3 to the Borel null set where the chosen gradient exceeds $L$, whose indicator has zero integral. Thus the integral is finite for these pairs. Therefore $\mathrm{\mid }{u}^{\ast }(y)-u^{\ast }(x)\mathrm{\mid }\le L\mathrm{\mid }y-x\mathrm{\mid }$. Since $\mathrm{\Omega }\times \mathrm{\Omega }={\bigcup }_m(U_m\times U_m)$ and a countable union of null sets is null, there is one measurable representative $u^{\ast }$ of $u$ (the almost-everywhere limit of the subsequence of step 1.2) and one null set $N\subseteq \mathrm{\Omega }\times \mathrm{\Omega }$ such that $\mathrm{\mid }{u}^{\ast }(x)-u^{\ast }(y)\mathrm{\mid }\le L\mathrm{\mid }x-y\mathrm{\mid }$ whenever $(x,y)\mathrm{\notin }N$.

The converse: the derivative operator norm bound. Let $v$ be a unit vector. Choose $j$ with $v_j\mathrm{\ne }0$ and define the invertible linear map $T$ by $T{e}_j=v$ and $T{e}_i=e_i$ for $i\mathrm{\ne }j$; then $\mathrm{\mid }\det T\mathrm{\mid }=\mathrm{\mid }{v}_j\mathrm{\mid }>0$. The function $g:=f\circ T$ is Lipschitz on the open set $T^{-1}\mathrm{\Omega }$, and for each fixed value of the other coordinates its sections $t\mapsto g(\dots ,t,\dots )$ are Lipschitz on intervals, hence differentiable at almost every $t$ by [F12]; Fubini [F9] shows that the set of $w\in T^{-1}\mathrm{\Omega }$ at which the $e_j$-directional derivative of $g$ fails to exist is null, and therefore, by the linear change-of-variables identity [F10], the set of $x\in \mathrm{\Omega }$ at which the $v$-directional derivative of $f$ fails to exist is null. Define $q_k(x):=(f(x+v\mathrm{/}k)-f(x))\mathrm{/}(1\mathrm{/}k)$ when $x\in \mathrm{\Omega }$ and $x+v\mathrm{/}k\in \mathrm{\Omega }$, and $q_k(x):=0$ otherwise; then $q_k$ is measurable, $\mathrm{\mid }{q}_k\mathrm{\mid }\le L$ wherever the quotient is defined, and $q_k\to {\mathrm{\partial }}_{v}f$ almost everywhere on $\mathrm{\Omega }$. For $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ the same difference-quotient computation as in step 2.2, now in the direction $v$, gives ${\int }_{\mathrm{\Omega }}f{\mathrm{\partial }}_v\phi =-{\lim }_k{\int }_{\mathrm{\Omega }}{q}_k\phi =-{\int }_{\mathrm{\Omega }}{h}_v\phi$, where $h_v$ is the limit of $q_k$ where it exists, and $0$ elsewhere satisfies $\mathrm{\mid }{h}_v\mathrm{\mid }\le L$ everywhere and equals ${\mathrm{\partial }}_{v}f$ almost everywhere. Hence $h_v$ is a representative of the weak directional derivative $D_{v}f:={\sum }_i{v}_i{D}_{i}f$ of [F2], and $\mid {\sum }_i{v}_i{D}_{i}f\mid \le L$ almost everywhere for every unit vector $v$. Now let $V$ be a countable dense subset of the unit sphere. Choosing representatives of $D_{1}f,\dots ,D_{n}f$, for each $v\in V$ the identity ${\sum }_i{v}_i{D}_{i}f=h_v$ holds almost everywhere, so on a set of full measure all the countably many inequalities $\mid {\sum }_i{v}_i{D}_{i}f\mid \le L$ hold simultaneously. At every such $x$ one has $\mathrm{\parallel }Df(x){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}={\sup }_{\mathrm{\mid }w\mathrm{\mid }=1}\mid {\sum }_i{w}_i{D}_{i}f(x)\mid ={\sup }_{v\in V}\mid {\sum }_i{v}_i{D}_{i}f(x)\mid \le L$, because $w\mapsto {\sum }_i{w}_i{D}_{i}f(x)$ is continuous and $V$ is dense in the unit sphere. For real $f$ this operator norm is $\mathrm{\mid }Df\mathrm{\mid }$. For $f=a+ib$, each row gradient satisfies $\mathrm{\mid }Da\mathrm{\mid },\mathrm{\mid }Db\mathrm{\mid }\le L$, so $\mathrm{\mid }Df{\mathrm{\mid }}^2=\mathrm{\mid }Da{\mathrm{\mid }}^2+\mathrm{\mid }Db{\mathrm{\mid }}^2\le 2L^2$. This gives the asserted real and complex Euclidean bounds and completes the converse.

Lebesgue points and the doubling comparison. Let $A_{r}f$ be the ball average of [F8], applied to the zero extension of a representative of $u$; by [F8] the set $E$ of $x\in \mathrm{\Omega }$ with $A_{r}u(x)\to u(x)$ as $r\downarrow 0$ has full measure in $\mathrm{\Omega }$, and on $E$ the limit equals $u^{\ast }(x)$ because $u^{\ast }=u$ almost everywhere. Replace $E$ by its intersection with the full-measure set where $u^{\ast }=u$. For $x,y\in E$ and $0<r<\min (\mathrm{dist}(x,\mathrm{\partial }\mathrm{\Omega }),\mathrm{dist}(y,\mathrm{\partial }\mathrm{\Omega }))$ one has $A_{r}u(x)-A_{r}u(y)=\mathrm{\mid }B(x,r){\mathrm{\mid }}^{-1}\mathrm{\mid }B(y,r){\mathrm{\mid }}^{-1}{\int }_{B(x,r)}{\int }_{B(y,r)}(u^{\ast }(z)-u^{\ast }(z^{\mathrm{\prime }}))d{z}^{\mathrm{\prime }}dz$, so using that the exceptional set $N$ of step 3.1 is null, $\mathrm{\mid }{A}_{r}u(x)-A_{r}u(y)\mathrm{\mid }\le L\mathrm{\mid }B(x,r){\mathrm{\mid }}^{-1}\mathrm{\mid }B(y,r){\mathrm{\mid }}^{-1}{\int }_{B(x,r)}{\int }_{B(y,r)}\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }d{z}^{\mathrm{\prime }}dz\le L(\mathrm{\mid }x-y\mathrm{\mid }+2r)$, because $\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }z-x\mathrm{\mid }+\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }y-z^{\mathrm{\prime }}\mathrm{\mid }\le 2r+\mathrm{\mid }x-y\mathrm{\mid }$ on the two balls. Letting $r\downarrow 0$ yields $\mathrm{\mid }{u}^{\ast }(x)-u^{\ast }(y)\mathrm{\mid }\le L\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y\in E$.

Extending to a Lipschitz representative and completing the first assertion. The full-measure set $E$ is dense in $\mathrm{\Omega }$: otherwise $E$ would be disjoint from some ball $B\subseteq \mathrm{\Omega }$, and $\lambda (B)>0$ by [F4], contradicting $\lambda (\mathrm{\Omega }\setminus E)=0$. On $E$ the restriction of $u^{\ast }$ is $L$-Lipschitz, hence uniformly continuous. By [F14] applied with $X:=\bar{\mathrm{\Omega }}$, the dense subset $E$ and the complete target $\mathbb{K}$, there is a continuous $g:\bar{\mathrm{\Omega }}\to \mathbb{K}$ with $g{\mathrm{\mid }}_E=u^{\ast }{\mathrm{\mid }}_E$. For arbitrary $x,y\in \bar{\mathrm{\Omega }}$ choose $x_j,y_j\in E$ with $x_j\to x$ and $y_j\to y$; by continuity $g(x)={\lim }_{j}g(x_j)$ and $g(y)={\lim }_{j}g(y_j)$, so step 4.1 gives $\mathrm{\mid }g(x)-g(y)\mathrm{\mid }={\lim }_j\mathrm{\mid }{u}^{\ast }(x_j)-u^{\ast }(y_j)\mathrm{\mid }\le L{\lim }_j\mathrm{\mid }{x}_j-y_j\mathrm{\mid }=L\mathrm{\mid }x-y\mathrm{\mid }$. Thus $g{\mathrm{\mid }}_{\mathrm{\Omega }}$ is an $L$-Lipschitz representative of $u$ on $\mathrm{\Omega }$ that equals $u$ almost everywhere, and the first assertion of the theorem is proved.