Proposition 2.8.5B412D7DC28A3

Proof. Steps 1 and 2 show by mollification that $f\in C^{1,1}_{\mathrm{loc}}(\Omega )$, Step 3 proves the Taylor expansion at the common Lebesgue points of the $A_{ab}$ by means of lemma 2.8.4, and Step 4 proves the converse.

Step 1: $f\in C^1(\Omega )$. Let $B=B(x_0,R)$ and $B'=B(x_0,R+s)$, where $R,s>0$ are such that $\overline{B(x_0,R+2s)}\subset \Omega $, and let $C$ bound the functions $|A_{ab}|$ on $B(x_0,R+2s)$. For $\varepsilon <s$, the closed ball $\overline{B'}$ lies in $\Omega _\varepsilon $, the mollification $f_\varepsilon $ is smooth near $\overline{B'}$, and $\partial _ a\partial _ bf_\varepsilon =(\partial _ a\partial _ bf)_\varepsilon =A_{ab}*\rho _\varepsilon $ there by proposition 2.5.3(1). At points of $\overline{B'}$ this average involves only values of $A_{ab}$ on $B(x_0,R+2s)$, so $|\partial _ a\partial _ bf_\varepsilon |\leq C$ on $\overline{B'}$. By lemma 2.5.1(1), $f_\varepsilon \to f$ uniformly on $\overline{B'}$, so $|f_\varepsilon |\leq M$ on $\overline{B'}$ for some $M$ and all small $\varepsilon $. For $x\in \overline B$ and a unit vector $e$, Taylor's formula

in which $|\partial _ e^2f_\varepsilon |=|\sum _{a,b}e_ ae_ b\partial _ a\partial _ bf_\varepsilon |\leq dC$, gives $|\partial _ ef_\varepsilon (x)|\leq 2M/s+dCs/2$. Thus the gradients $\nabla f_\varepsilon $ are bounded on $\overline B$ uniformly in small $\varepsilon $. They are also Lipschitz on the convex set $\overline B$ with constant $dC$. Indeed,

where $D^2f_\varepsilon $ is the Hessian matrix, and a $d\times d$ matrix $H$ with entries bounded by $C$ satisfies $|Hv|\leq \sqrt d\, C\sum _ b|v_ b|\leq dC|v|$ for $v\in \mathbb R^ d$, by the Cauchy–Schwarz inequality.

Applying proposition 2.7.4 successively to the components of the maps $\nabla f_{1/k}$, $k>1/s$, we find $\varepsilon _ k\to 0$ such that $\nabla f_{\varepsilon _ k}$ converges uniformly on $\overline B$ to a continuous map $g=(g_1,\ldots ,g_ d)$. Since $\partial _ af_\varepsilon =(\partial _ af)_\varepsilon \to \partial _ af$ in the sense of distributions by proposition 2.5.3(1) and (2), the distribution $\partial _ af$ on $B$ is the continuous function $g_ a$. Letting $\varepsilon =\varepsilon _ k\to 0$ in the identity

we obtain the same identity for $f$ and $g$. Since $g$ is continuous, it follows that $f$ is differentiable on $B$ with $\nabla f=g$. Covering $\Omega $ by such balls $B$, we conclude that $f\in C^1(\Omega )$ and that its distributional first derivatives coincide with the continuous functions $\partial _ af$ on each such ball, hence on $\Omega $ by the last assertion of lemma 2.5.2. Consequently, $\partial _ af_\varepsilon =(\partial _ af)_\varepsilon \to \partial _ af$ uniformly on compact subsets of $\Omega $, by lemma 2.5.1(1).

Step 2: the Lipschitz bound. Let $V\subset \Omega $ be convex and open with $|A_{ab}|\leq C$ almost everywhere on $V$, and let $x,y\in V$. The segment $[x,y]$ is a compact subset of $V$; choose $\delta >0$ smaller than its distance to $\mathbb R^ d\setminus V$. For $\varepsilon <\delta $ and $z\in [x,y]$, the closed ball $\overline{B(z,\varepsilon )}$ lies in $V$, so $z\in \Omega _\varepsilon $ and $|\partial _ a\partial _ bf_\varepsilon (z)|=|(A_{ab}*\rho _\varepsilon )(z)|\leq C$. As in Step 1, $|\nabla f_\varepsilon (y)-\nabla f_\varepsilon (x)|\leq dC|y-x|$, and letting $\varepsilon \to 0$ gives $|\nabla f(y)-\nabla f(x)|\leq dC|y-x|$. Every point of $\Omega $ lies in a ball $V\Subset \Omega $, on which the $A_{ab}$ are bounded, so $\nabla f$ is locally Lipschitz and $f\in C^{1,1}_{\mathrm{loc}}(\Omega )$.

Step 3: the Taylor expansion. By theorem 2.8.3, almost every point of $\Omega $ is a common Lebesgue point of the functions $A_{ab}$. Let $x$ be such a point, let $Q=(-1,1)^ d$, and choose $r_0>0$ with $x+3r_0\overline Q\subset \Omega $. The functions $|A_{ab}|$ are bounded on the compact set $x+3r_0\overline Q$ by some constant $C$, so by Step 2, $\nabla f$ is Lipschitz with constant $\lambda =dC$ on the convex open set $x+3r_0Q$, which contains $x+2r_0\overline Q$. For $0<r\leq r_0$, define on $2Q$

We show that $G_ r\to 0$ uniformly on $\overline Q$ as $r\downarrow 0$; the expansion then follows by integrating $\nabla f$ along segments. The maps $G_ r$ vanish at $0$ and are Lipschitz with the common constant $\lambda +|A(x)|$, where $|A(x)|$ is the operator norm, and hence also with $\Lambda =\lambda +|A(x)|+1>0$.

To compute the derivatives of $G_ r$, let $\psi $ be a test function on $2Q$. The function $z\mapsto \psi ((z-x)/r)$ is a test function on $\Omega $, and the distributional derivative $\partial _ b$ of the continuous function $\partial _ af$ is $\partial _ b\partial _ af=\partial _ a\partial _ bf=A_{ab}$, since partial derivatives of distributions commute. Hence the change of variables $z=x+ry$ gives

Thus $\partial _{y_ b}(G_ r)_ a(y)=A_{ab}(x+ry)-A_{ab}(x)$ in the sense of distributions on $2Q$, and since $x$ is a Lebesgue point,

Lemma 2.8.4, applied on the cube $\overline Q$ of side length $2$ to each component of $G_ r$, gives

We deduce that $\sup _{\overline Q}|G_ r-(G_ r)_ Q|\to 0$. Otherwise there would be $\kappa >0$, radii $r_ k\to 0$ and points $y_ k\in \overline Q$ with $|G_{r_ k}(y_ k)-(G_{r_ k})_ Q|\geq \kappa $, and by the common Lipschitz bound, $|G_{r_ k}-(G_{r_ k})_ Q|\geq \kappa /2$ on $Q\cap B(y_ k,\kappa /(2\Lambda ))$. For $y\in \overline Q$ and $0<\sigma \leq 1$, the open cube $\prod _ lI_ l$, with $I_ l=(y_ l-\sigma ,y_ l)$ if $y_ l>0$ and $I_ l=(y_ l,y_ l+\sigma )$ if $y_ l\leq 0$, lies in $Q$ and within distance $\sqrt d\sigma $ of $y$. With $\sigma =\min \{ 1,\kappa /(2\Lambda \sqrt d)\} $, this shows that $Q\cap B(y_ k,\kappa /(2\Lambda ))$ has measure at least $\sigma ^ d$ for every $k$, so that $\int _ Q|G_{r_ k}-(G_{r_ k})_ Q|\, \mathrm dy\geq \sigma ^ d\kappa /2$, contradicting the convergence of these integrals to zero. Since $G_ r(0)=0$, also $|(G_ r)_ Q|=|G_ r(0)-(G_ r)_ Q|\to 0$, and hence $\sup _{\overline Q}|G_ r|\to 0$ as $r\downarrow 0$.

Now let $0<|h|\leq r_0$, and put $r=|h|$ and $y=h/|h|\in \overline Q$. For $0<t\leq 1$ we have $\nabla f(x+th)-\nabla f(x)=rt\, G_{rt}(y)+tA(x)h$, so, as $f$ is continuously differentiable,

The absolute value of the right side is at most $|h|^2\sup _{0<\varrho \leq |h|}\sup _{\overline Q}|G_\varrho |=o(|h|^2)$.

Step 4: the converse. Let $B\Subset \Omega $ be a ball, and let $f_ j$ be as in the statement, with $|\partial _ a\partial _ bf_ j|\leq C_ B$ on $B$ for all $j$. For every nonnegative test function $\chi $ supported in $B$, two integrations by parts give

so $|\langle \partial _ a\partial _ bf,\chi \rangle |\leq C_ B\int \chi \, \mathrm dx$. The distribution $\partial _ a\partial _ bf$ is real, because $f$ is real, and every point of $\Omega $ lies in such a ball $B$. By corollary 2.8.2(2), applied with $\nu $ equal to Lebesgue measure, $\partial _ a\partial _ bf$ is given by a Borel function $A_{ab}$ with $|A_{ab}|\leq C_ B$ almost everywhere on every such ball $B$. Each set of a compact exhaustion of $\Omega $ is covered by finitely many such balls, so countably many such balls $B_ k$ cover $\Omega $. Redefine $A_{ab}$ to be $0$ on the null set $\bigcup _ k\{ x\in B_ k:|A_{ab}(x)|>C_{B_ k}\} $. The new function still represents $\partial _ a\partial _ bf$, and it is bounded on every compact set, because a compact set is covered by finitely many of the balls $B_ k$. Finally, since $f_\varepsilon \to f$ in the sense of distributions (proposition 2.5.3(2)), the mollifications $f_ j=f_{\varepsilon _ j}$ with $\varepsilon _ j\to 0$ satisfy the hypothesis of the converse whenever their second derivatives are bounded on each ball $B\Subset \Omega $ uniformly in small $\varepsilon $. $\square$