Jensen maximum principle for locally semiconvex functions

Shrink to a ball ${\bar{B}}_R\subseteq U$ on which $g(x)=f(x)+\lambda \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}2$ is convex and $f(0)>f(x)$ for $0<\mathrm{\mid }x\mathrm{\mid }\le R$. By [F1], $f$ is continuous. Given $0<r<R\mathrm{/}4$, the compact annulus $r\le \mathrm{\mid }x\mathrm{\mid }\le R$ has a strictly positive gap $c=f(0)-\max f$. Fix $0<\eta \le \delta$ with $\eta R<c\mathrm{/}2$. For every $\mathrm{\mid }q\mathrm{\mid }\le \eta$, all global maximizers of $f(x)+q\cdot x$ on ${\bar{B}}_R$ lie in $B_r$, since their value at zero is $f(0)$ and annulus values are at most $f(0)-c+\eta R$. Their maximizing pairs form a closed subset of the compact product ${\bar{B}}_R\times {\bar{B}}_{\eta }$, so their set $K$ is compact and lies in $B_r$.

At a maximizing pair $(x,q)$, $f(z)\le f(x)-q\cdot (z-x)$ on ${\bar{B}}_R$. Put $v=-q+\lambda x$. Expanding the quadratic gives $g(x+h)\le g(x)+v\cdot h+\lambda \mathrm{\mid }h{\mathrm{\mid }}^2\mathrm{/}2$. Convexity gives the reverse support $g(z)\ge g(x)+v\cdot (z-x)$: for $h=z-x$ and small $t>0$, convexity on the segment from $x-th$ to $z$ yields $g(z)-g(x)\ge (g(x)-g(x-th))\mathrm{/}t\ge v\cdot h-\lambda t\mathrm{\mid }h{\mathrm{\mid }}^2\mathrm{/}2$, and let $t\downarrow 0$. Thus $f$ is differentiable at $x$, $p_x=Df(x)=-q$, and $f(x)+p_x\cdot (z-x)-\lambda \mathrm{\mid }z-x{\mathrm{\mid }}^2\mathrm{/}2\le f(z)\le f(x)+p_x\cdot (z-x)$ for $z\in {\bar{B}}_R$; in particular the slope at each $x\in K$ is unique.

For distinct $x,y\in K$, set $d=y-x$. The support bounds in step 2.1 and $f(y)\le f(x)+p_x\cdot d$ give $(p_x-p_y)\cdot h\le \lambda \mathrm{\mid }d+h{\mathrm{\mid }}^2\mathrm{/}2$ whenever $y+h\in B_R$. If $p_x\mathrm{\ne }{p}_y$, take $h=\mathrm{\mid }d\mathrm{\mid }(p_x-p_y)\mathrm{/}\mathrm{\mid }{p}_x-p_y\mathrm{\mid }$, which is admissible since $\mathrm{\mid }y+h\mathrm{\mid }<3r<R$. Then $\mathrm{\mid }{p}_x-p_y\mathrm{\mid }\mathrm{\mid }d\mathrm{\mid }\le 2\lambda \mathrm{\mid }d{\mathrm{\mid }}^2$. Equal slopes need no estimate, so $x\mapsto p_x$ is $2\lambda$-Lipschitz on $K$.

For each coordinate set $P_j(z)={\inf }_{x\in K}(p_x^j+2\lambda \mathrm{\mid }z-x\mathrm{\mid })$. Compactness makes this finite; the triangle inequality makes it $2\lambda$-Lipschitz, and step 3.1 gives $P_j(x)=p_x^j$ on $K$. Hence $P=(P_1,\dots ,P_m)$ is $2\lambda \sqrt{m}$-Lipschitz and, since each $\mathrm{\mid }q\mathrm{\mid }\le \eta$ has a maximizer in $K$ with $P(x)=-q$, $P(K)\supseteq {\bar{B}}_{\eta }$. If $K$ were null, [F2] would make $P(K)$ null, contradicting the positive outer measure of $B_{\eta }$; thus $K$ has positive outer measure. If $\lambda =0$, this same construction makes $P$ constant while its image contains ${\bar{B}}_{\eta }$, a contradiction, so no strict-maximum case is missed. Countable Choice required by [F2] follows from [A1].

Since $K$ has positive outer measure, it cannot be contained in the given null set $N$. Choose $x\in K\setminus N$ and its maximizing slope $q$; then $\mathrm{\mid }x\mathrm{\mid }<r$ and $\mathrm{\mid }q\mathrm{\mid }\le \eta \le \delta$. This proves both conclusions.