Pome59B8A4C8DA05Lemma 14.21.5.59B8A4C8DA05 In Euclidean space $\mathbb R^ m$, let $B(p,r)$ be the ball of center $p$ and radius $r>0$, and let $I$ be the identity matrix. Let $g\in C^{1,1}$ near $\overline{B(p,r)}\subset \mathbb R^ m$, suppose $g\leq g(p)$ on the ball, and let $E$ be any subset of full measure in the ball at whose points $g$ has a second-order Taylor expansion. For every $\delta >0$ and $0<\rho <r$, there is $x\in E\cap B(p,\rho )$ such that
Proof. Put $g_\delta (x)=g(x)-\delta |x-p|^2$. The gradient map $F=\nabla g_\delta $ is Lipschitz on the ball, so it maps the null set $N=\overline{B(p,r)}\setminus E$ to a null set. Choose $s\in B(0,\delta \rho )\setminus F(N)$ and maximize $g_\delta (x)-s\cdot (x-p)$ on the closed ball. At any point with $|x-p|\geq \rho $, its value is strictly smaller than its value at $p$, because
Thus a maximizer $x$ lies in $B(p,\rho )$. The first-order condition at $x$ gives $F(x)=s$, so $x\notin N$. The second-order Taylor expansion at this upper contact point gives $D^2g(x)\leq 2\delta I$. Finally, sets $E$ as in the statement exist: by Rademacher's theorem, applied to the Lipschitz gradient of $g$, the function $g$ has a second-order Taylor expansion almost everywhere. $\square$
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