Proof of Maximisers of Linearly Perturbed Continuous Functions on a Closed Ball: Existence, Localisation, and Compactness of the Contact Set

The ball $\bar{B}$ is nonempty, since $\hat{x}\in\bar{B}$ by claim 1 of Elementary Properties of the Closed Ball in a Metric Space, and compact in $\mathbb{R}^{n}$ by claim 2 of A Closed Euclidean Ball is Convex and Compact. It is bounded and closed by claims 2 and 3 of Elementary Properties of the Closed Ball in a Metric Space. For $p\in\mathbb{R}^{n}$ let $\psi_{p}:\bar{B}\to\mathbb{R}$ be given by $\psi_{p}(x)=\varphi(x)+p\cdot x$; thus $M(p)$ is the set of points of $\bar{B}$ at which $\psi_{p}$ attains its greatest value on $\bar{B}$. Throughout we use the bilinearity and symmetry of the dot product (The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite) and the Cauchy-Schwarz inequality The Cauchy-Schwarz Inequality on $\mathbb{R}^n$.

Step 1 (the perturbed functions are continuous). Let $p\in\mathbb{R}^{n}$ and $S\subseteq\bar{B}$. We check that the restriction of $\psi_{p}$ to $S$ satisfies the continuity hypothesis of Extreme Value Theorem on a Compact Subset of a Metric Space. Let $x\in S$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. For $y\in\bar{B}$,

$$\bigl|\psi_{p}(y)-\psi_{p}(x)\bigr|\le\bigl|\varphi(y)-\varphi(x)\bigr|+\bigl|p\cdot(y-x)\bigr|\le\bigl|\varphi(y)-\varphi(x)\bigr|+\lVert p\rVert\,\lVert y-x\rVert .$$

By (H1) there is a positive real $\delta_{1}$ such that every $z\in\bar{B}$ with $\lVert z-x\rVert<\delta_{1}$ satisfies $|\varphi(z)-\varphi(x)|<\varepsilon/2$. Let $\delta$ be the smaller of $\delta_{1}$ and $\varepsilon/(2\lVert p\rVert+2)$, a positive real number. If $y\in S$ satisfies $d_{E}(x,y)=\lVert y-x\rVert<\delta$, then

$$\bigl|\psi_{p}(y)-\psi_{p}(x)\bigr|<\frac{\varepsilon}{2}+\lVert p\rVert\,\frac{\varepsilon}{2\lVert p\rVert+2}\le\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$

since $2\lVert p\rVert\le 2\lVert p\rVert+2$ gives $\lVert p\rVert/(2\lVert p\rVert+2)\le\tfrac12$.

Proof of claim 1. Let $p\in\mathbb{R}^{n}$. By step 1 with $S=\bar{B}$, the function $\psi_{p}$ satisfies the hypotheses of Extreme Value Theorem on a Compact Subset of a Metric Space on the nonempty compact set $\bar{B}$, so there is $x_{\max}\in\bar{B}$ with $\psi_{p}(y)\le\psi_{p}(x_{\max})$ for every $y\in\bar{B}$; that is, $x_{\max}\in M(p)$ and $M(p)\ne\varnothing$. Now let $\delta$ be a positive real number. The origin $0_{\mathbb{R}^{n}}$ satisfies $\lVert 0_{\mathbb{R}^{n}}\rVert=0\le\delta$ by claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so any element of $M(0_{\mathbb{R}^{n}})$ belongs to $K_{\delta}$, and $K_{\delta}\ne\varnothing$.

Proof of claim 2. Let $\rho\in\mathbb{R}$ with $0<\rho$.

Suppose first that $r<\rho$. Every $x\in\bar{B}$ satisfies $d_{E}(\hat{x},x)\le r<\rho$, so $\bar{B}\subseteq B_{d_{E}}(\hat{x},\rho)$; since $K_{\delta}\subseteq\bar{B}$ for every positive real $\delta$, the choice $\delta_{\rho}=1$ works.

Suppose now that $\rho\le r$, and put

$$A=\bigl\{x\in\bar{B}:\rho\le\lVert x-\hat{x}\rVert\bigr\}.$$

The set $A$ is nonempty: let $e_{1}\in\mathbb{R}^{n}$ have first component $1$ and all other components $0$, so that $\lVert e_{1}\rVert^{2}=1$ and hence $\lVert e_{1}\rVert=1$ by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square; then $z=\hat{x}+\rho\,e_{1}$ satisfies $\lVert z-\hat{x}\rVert=\lVert\rho\,e_{1}\rVert=\rho$ by claim 5 there, so $d_{E}(\hat{x},z)=\rho\le r$ and $z\in A$.

The set $A$ is compact. Indeed $A=\bar{B}\cap C$ where $C=\mathbb{R}^{n}\setminus B_{d_{E}}(\hat{x},\rho)$, because $d_{E}(\hat{x},x)=\lVert x-\hat{x}\rVert$ and $B_{d_{E}}(\hat{x},\rho)$ consists of the points at distance less than $\rho$ from $\hat{x}$. The open ball is open by Open Ball in a Metric Space is Open, so $C$ is closed; $\bar{B}$ is closed; and the intersection of two closed sets is closed by Complements, Unions and Intersections of Closed Sets in a Topological Space. Moreover $A\subseteq\bar{B}$ is bounded, directly from Bounded Subset of a Metric Space. By The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere, $A$ is compact.

By step 1 with $p=0_{\mathbb{R}^{n}}$ and $S=A$, and by Extreme Value Theorem on a Compact Subset of a Metric Space, there is $x_{A}\in A$ with $\varphi(x)\le\varphi(x_{A})$ for every $x\in A$. Since $0<\rho\le\lVert x_{A}-\hat{x}\rVert$ we have $x_{A}\ne\hat{x}$, so (H2) gives $\varphi(x_{A})<\varphi(\hat{x})$. Put

$$\eta=\varphi(\hat{x})-\varphi(x_{A})>0,\qquad \delta_{\rho}=\frac{\eta}{2r}>0 .$$

Let $\delta$ be real with $0<\delta\le\delta_{\rho}$ and let $x\in K_{\delta}$, say $x\in M(p)$ with $\lVert p\rVert\le\delta$. Since $\hat{x}\in\bar{B}$ we have $\varphi(\hat{x})+p\cdot\hat{x}\le\varphi(x)+p\cdot x$, hence, using Cauchy-Schwarz and $\lVert x-\hat{x}\rVert=d_{E}(\hat{x},x)\le r$,

$$\varphi(x)\ \ge\ \varphi(\hat{x})-p\cdot(x-\hat{x})\ \ge\ \varphi(\hat{x})-\lVert p\rVert\,\lVert x-\hat{x}\rVert\ \ge\ \varphi(\hat{x})-\delta\,r\ \ge\ \varphi(\hat{x})-\frac{\eta}{2}\ >\ \varphi(\hat{x})-\eta=\varphi(x_{A}).$$

So $\varphi(x_{A})<\varphi(x)$, which rules out $x\in A$; as $x\in\bar{B}$, this forces $\lVert x-\hat{x}\rVert<\rho$, that is, $x\in B_{d_{E}}(\hat{x},\rho)$. Hence $K_{\delta}\subseteq B_{d_{E}}(\hat{x},\rho)$, proving claim 2.

Proof of claim 3. Let $\delta$ be a positive real number. Since $K_{\delta}\subseteq\bar{B}$, the set $K_{\delta}$ is bounded. We show that it is closed, using Sequential Characterization of Closed Subsets of a Metric Space.

Let $(x^{(q)})_{q\in\mathbb{N}}$ be a sequence with $x^{(q)}\in K_{\delta}$ for every $q$, converging to a point $x\in\mathbb{R}^{n}$, and for each $q$ choose $p^{(q)}\in\mathbb{R}^{n}$ with $\lVert p^{(q)}\rVert\le\delta$ and $x^{(q)}\in M(p^{(q)})$. As $\bar{B}$ is closed, $x\in\bar{B}$.

The closed ball $\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)$ is bounded and closed by claims 2 and 3 of Elementary Properties of the Closed Ball in a Metric Space, and $\lVert p^{(q)}\rVert=d_{E}(0_{\mathbb{R}^{n}},p^{(q)})\le\delta$ places every $p^{(q)}$ in it. By The Bolzano-Weierstrass Theorem in $\mathbb{R}^n$ there are $p\in\mathbb{R}^{n}$ and a strictly increasing sequence $(q_{t})_{t\in\mathbb{N}}$ in $\mathbb{N}$ such that $(p^{(q_{t})})_{t\in\mathbb{N}}$ converges to $p$; by Sequential Characterization of Closed Subsets of a Metric Space, $p\in\bar{B}_{d_{E}}(0_{\mathbb{R}^{n}},\delta)$, that is, $\lVert p\rVert\le\delta$. The subsequence $(x^{(q_{t})})_{t\in\mathbb{N}}$ converges to $x$.

Fix $y\in\bar{B}$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By (H1) there is a positive real $\delta_{2}$ such that every $z\in\bar{B}$ with $\lVert z-x\rVert<\delta_{2}$ satisfies $|\varphi(z)-\varphi(x)|<\varepsilon/4$. By convergence, choose $t\in\mathbb{N}$ so large that

$$\lVert x^{(q_{t})}-x\rVert<\delta_{2},\qquad \lVert x^{(q_{t})}-x\rVert<\frac{\varepsilon}{4\delta+4},\qquad \lVert p^{(q_{t})}-p\rVert<\frac{\varepsilon}{4\lVert x\rVert+4\lVert y\rVert+4}.$$

Writing $p^{(q_{t})}\cdot x^{(q_{t})}-p\cdot x=p^{(q_{t})}\cdot(x^{(q_{t})}-x)+(p^{(q_{t})}-p)\cdot x$ and using Cauchy-Schwarz,

$$\bigl|\psi_{p^{(q_{t})}}(x^{(q_{t})})-\psi_{p}(x)\bigr|\le\bigl|\varphi(x^{(q_{t})})-\varphi(x)\bigr|+\delta\,\lVert x^{(q_{t})}-x\rVert+\lVert p^{(q_{t})}-p\rVert\,\lVert x\rVert<\frac{3\varepsilon}{4},$$

while $\bigl|p^{(q_{t})}\cdot y-p\cdot y\bigr|\le\lVert p^{(q_{t})}-p\rVert\,\lVert y\rVert<\varepsilon/4$. Since $x^{(q_{t})}\in M(p^{(q_{t})})$ and $y\in\bar{B}$,

$$\varphi(y)+p^{(q_{t})}\cdot y\ \le\ \varphi(x^{(q_{t})})+p^{(q_{t})}\cdot x^{(q_{t})} ,$$

and combining the three displays,

$$\varphi(y)+p\cdot y\ <\ \varphi(y)+p^{(q_{t})}\cdot y+\frac{\varepsilon}{4}\ \le\ \varphi(x^{(q_{t})})+p^{(q_{t})}\cdot x^{(q_{t})}+\frac{\varepsilon}{4}\ <\ \varphi(x)+p\cdot x+\varepsilon .$$

As $\varepsilon$ was an arbitrary positive real number, $\varphi(y)+p\cdot y\le\varphi(x)+p\cdot x$: were the reverse strict inequality to hold, taking $\varepsilon$ to be half of the positive difference would contradict the display. Since $y\in\bar{B}$ was arbitrary, $x\in M(p)$ with $\lVert p\rVert\le\delta$, so $x\in K_{\delta}$.

By Sequential Characterization of Closed Subsets of a Metric Space the set $K_{\delta}$ is closed, and being bounded it is compact by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere. This proves claim 3.