Lemma 21.4.39AEE54E3B1E1

Proof. If $n=0$, then $A*\rho _\varepsilon =A$ and there is nothing to prove; let $n\geq 1$. Recall that $\rho $ is smooth, nonnegative, even, supported in the unit ball, and has integral one. Since $A$ is continuous, being locally Lipschitz by proposition 21.2.4, the functions $A*\rho _\varepsilon $ are smooth by section 2.5, and they converge to $A$ locally uniformly by lemma 2.5.1(1). They are convex, because $A*\rho _\varepsilon (x)=\int A(x-y)\rho _\varepsilon (y)\, \mathrm dy$ and the convexity inequalities of the functions $x\mapsto A(x-y)$ can be integrated against $\rho _\varepsilon (y)\, \mathrm dy\geq 0$. For fixed $x$, consider the function

which is finite because $A$ is continuous and $\rho $ has compact support. The substitution $y\mapsto \varepsilon y$ shows that $g(\varepsilon )=A*\rho _\varepsilon (x)$ for $\varepsilon >0$, and $g(0)=A(x)$. For each $y$, the function $\varepsilon \mapsto A(x-\varepsilon y)$ is convex, so $g$ is convex; and $g$ is even, because $\rho $ is even. A convex even function on $\mathbb R$ is nondecreasing on $[0,+\infty )$: if $0\leq a<b$, then $a=\lambda b+(1-\lambda )(-b)$ with $\lambda =(a+b)/(2b)\in (0,1]$, so that $g(a)\leq \lambda g(b)+(1-\lambda )g(-b)=g(b)$. Hence $A*\rho _\varepsilon (x)$ is nondecreasing in $\varepsilon >0$ and bounded below by $g(0)=A(x)$. $\square$