Proof of Jensen's Lemma: the Contact Set of a Semiconvex Function at a Strict Maximum has Positive Measure

Step 0: the contact-set lemma applies. The ball $\bar{B}$ is contained in $U$, and by the continuity estimate for semiconvex functions, applied with the subset $S=\bar{B}$ of the open convex set $U$, the restriction of $\varphi$ to $\bar{B}$ satisfies hypothesis (H1) of Maximisers of Linearly Perturbed Continuous Functions on a Closed Ball: Existence, Localisation, and Compactness of the Contact Set; hypothesis (H2) there is exactly the strict maximum assumption made here. Consequently Maximisers of Linearly Perturbed Continuous Functions on a Closed Ball: Existence, Localisation, and Compactness of the Contact Set applies to $\hat{x}$, $r$ and the restriction of $\varphi$ to $\bar{B}$, and the sets $M(p)$ and $K_{\delta}$ of that lemma are the ones written here.

Step 1: the choice of $\delta_{0}$. Apply the localisation of maximisers with $\rho=r/2$ to obtain a positive real $\delta_{\ast}$ such that $K_{\delta}\subseteq B_{d_{E}}(\hat{x},r/2)$ whenever $0<\delta\le\delta_{\ast}$. Let $\delta_{0}$ be the smaller of $\delta_{\ast}$ and $\lambda r/4$; since $0<\lambda$ and $0<r$, this is a positive real number. Fix from now on a real $\delta$ with $0<\delta\le\delta_{0}$ where claims 1 to 3 are concerned.

Proof of claim 1. By the existence of maximisers the set $K_{\delta}$ is nonempty, and by the compactness of the contact set it is compact in $\mathbb{R}^{n}$; hence $K_{\delta}\in\mathcal{B}(\mathbb{R}^{n})$ by claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure. Since $\delta\le\delta_{0}\le\delta_{\ast}$, step 1 gives $K_{\delta}\subseteq B_{d_{E}}(\hat{x},r/2)$, and $B_{d_{E}}(\hat{x},r/2)\subseteq\bar{B}_{d_{E}}(\hat{x},r/2)$ by claim 1 of Elementary Properties of the Closed Ball in a Metric Space.

Proof of claim 2. For $x\in K_{\delta}$ put

$$P(x)=\bigl\{p\in\mathbb{R}^{n}:\lVert p\rVert\le\delta\ \text{ and }\ \varphi(y)+p\cdot y\le\varphi(x)+p\cdot x\ \text{ for every }y\in\bar{B}\bigr\},$$

which is nonempty by the definition of $K_{\delta}$.

(a) $P(x)$ has exactly one element. Let $p,p'\in P(x)$. By the triangle inequality and claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square,

$$\lVert p-p'\rVert\le\lVert p\rVert+\lVert p'\rVert\le 2\delta\le 2\delta_{0}\le\frac{\lambda r}{2}.$$

By claim 1 above, $x\in\bar{B}_{d_{E}}(\hat{x},r/2)$. Applying the Lipschitz estimate between maximisers with the convex set $U$, the constant $\lambda$, the ball $\bar{B}$, the vectors $p,p'$ and the points $x'=x$, we obtain $p=p'$. Write $T(x)$ for the unique element of $P(x)$; this defines a map $T:K_{\delta}\to\mathbb{R}^{n}$.

(b) $T$ is Lipschitz with constant $2\lambda$. Let $x,x'\in K_{\delta}$ and put $p=T(x)$, $p'=T(x')$. As in (a), $\lVert p-p'\rVert\le 2\delta\le\lambda r/2$, and by claim 1 both $x$ and $x'$ lie in $\bar{B}_{d_{E}}(\hat{x},r/2)$; each of $x,x'$ maximises the corresponding perturbation over $\bar{B}$. The Lipschitz estimate between maximisers therefore gives

$$\lVert T(x)-T(x')\rVert\le 2\lambda\,\lVert x-x'\rVert ,$$

so $T$ is Lipschitz with constant $2\lambda$ for the Euclidean distances.

(c) $T(K_{\delta})=\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)$. Every $x\in K_{\delta}$ satisfies $\lVert T(x)\rVert\le\delta$, that is, $d_{E}(0_{\mathbb{R}^{n}},T(x))\le\delta$, so $T(K_{\delta})\subseteq\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)$. Conversely let $p\in\mathbb{R}^{n}$ with $\lVert p\rVert\le\delta$. By the existence of maximisers there is $x\in M(p)$; then $x\in K_{\delta}$ and $p\in P(x)$, so $p=T(x)$. Hence $\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\subseteq T(K_{\delta})$.

(d) The measure estimate. The set $K_{\delta}$ is nonempty and compact by claim 1, and $T$ is Lipschitz with constant $2\lambda$ by (b). Applying The Lebesgue Measure of a Lipschitz Image of a Compact Subset of $\mathbb{R}^n$ and using (c),

$$\lambda_{n}\bigl(\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\bigr)=\lambda_{n}\bigl(T(K_{\delta})\bigr)\le\bigl(2\,\sigma_{n}\cdot 2\lambda\bigr)^{n}\lambda_{n}(K_{\delta})=\bigl(4\,\sigma_{n}\,\lambda\bigr)^{n}\lambda_{n}(K_{\delta}),$$

which is the displayed inequality of claim 2.

(e) Positivity. Since $0<\delta$, claim 1 of Elementary Properties of the Closed Ball in a Metric Space gives $B_{d_{E}}(0_{\mathbb{R}^{n}},\delta)\subseteq\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)$, and by claim 1 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure the open ball is a Borel set of positive $\lambda_{n}$-measure. The closed ball is compact, hence Borel, by claim 2 of A Closed Euclidean Ball is Convex and Compact and claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure, so monotonicity (claim 2 of Basic Properties of a Measure) gives $0<\lambda_{n}(\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta))$. Combining with (d), $0<(4\sigma_{n}\lambda)^{n}\lambda_{n}(K_{\delta})$; as $0<\sigma_{n}$ and $0<\lambda$ make $(4\sigma_{n}\lambda)^{n}$ positive, it follows that $0<\lambda_{n}(K_{\delta})$.

Proof of claim 3. Let $Z\in\mathcal{B}(\mathbb{R}^{n})$ with $\lambda_{n}(Z)=0$, and suppose that every $x\in K_{\delta}$ lay in $Z$, that is, $K_{\delta}\subseteq Z$. Then monotonicity (claim 2 of Basic Properties of a Measure) would give $\lambda_{n}(K_{\delta})\le\lambda_{n}(Z)=0$, contradicting claim 2. Hence some $x\in K_{\delta}$ satisfies $x\notin Z$.

Proof of claim 4. Let $\rho\in\mathbb{R}$ with $0<\rho$. By the localisation of maximisers there is a positive real $\delta_{\rho}$ with $K_{\delta}\subseteq B_{d_{E}}(\hat{x},\rho)$ for every real $\delta$ with $0<\delta\le\delta_{\rho}$. Let $\delta_{1}$ be the smaller of $\delta_{0}$ and $\delta_{\rho}$; it is positive and satisfies $\delta_{1}\le\delta_{0}$, and every real $\delta$ with $0<\delta\le\delta_{1}$ satisfies $\delta\le\delta_{\rho}$, so $K_{\delta}\subseteq B_{d_{E}}(\hat{x},\rho)$.