Proof of The Sup-Convolution of an Upper Semicontinuous Function Attains its Supremum

Let $\Phi:\mathbb{R}^{M}\to\mathbb{R}$ be given by $\Phi(x)=v(x)-\frac{\lambda}{2}\lVert x-\xi\rVert^{2}$, so that the set $S_{\lambda,v}(\xi)$ of Sup-Convolution of a Function on $\mathbb{R}^M$ is the set of values of $\Phi$ and $v^{\lambda}(\xi)$ is its least upper bound. Since $\xi-\xi$ is the origin of $\mathbb{R}^{M}$, of norm $0$ 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, claim 1 of Zero Products and Elementary Identities in a Field gives $\Phi(\xi)=v(\xi)$. Order arithmetic is taken from Elementary Arithmetic in an Ordered Field and Elementary Order Arithmetic in an Ordered Field. Put $K=C-v(\xi)$; as $v(\xi)\le C$, claim 3 of Elementary Arithmetic in an Ordered Field gives $0\le K$.

Claim 1. Let $y\in\mathbb{R}^{M}$ satisfy $v(\xi)\le v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}$. Adding $\frac{\lambda}{2}\lVert y-\xi\rVert^{2}$ to both sides (claim 3 of Elementary Arithmetic in an Ordered Field) gives $v(\xi)+\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le v(y)$, and $v(y)\le C$, so $v(\xi)+\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le C$; subtracting $v(\xi)$ by the same claim gives $\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le K$, which is the assertion.

Claim 2, Step 1 (a radius beyond which $\Phi$ drops below $\Phi(\xi)$). From $0<\lambda$ and $0<2^{-1}$ (claim 7 of Elementary Order Arithmetic in an Ordered Field applied to $0<2$) we get $0<\frac{\lambda}{2}$ by claim 5 of that result, so $(\frac{\lambda}{2})^{-1}$ exists and is positive by claim 7. Because $\le$ is a total order, one of the two real numbers $1$ and $(K+1)(\frac{\lambda}{2})^{-1}$ is greater than or equal to the other; let $r$ be such a one, so that $1\le r$ and $(K+1)(\frac{\lambda}{2})^{-1}\le r$. Multiplying the second inequality by the nonnegative number $\frac{\lambda}{2}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $K+1\le\frac{\lambda}{2}r$, while $K<K+1$ by claims 6 and 1 of Elementary Order Arithmetic in an Ordered Field; hence $K<\frac{\lambda}{2}r$.

Let $x\in\mathbb{R}^{M}$ with $r<\lVert x-\xi\rVert$. Both numbers are nonnegative, so claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $r^{2}<\lVert x-\xi\rVert^{2}$, and $r\le r^{2}$ follows from $1\le r$ by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative factor $r$. Multiplying by $\frac{\lambda}{2}$, positive, using claim 10 of Elementary Order Arithmetic in an Ordered Field for the strict inequality and claim 5 of Elementary Arithmetic in an Ordered Field for the other, gives

$$K<\frac{\lambda}{2}\,r\le\frac{\lambda}{2}\,r^{2}<\frac{\lambda}{2}\,\lVert x-\xi\rVert^{2}.$$

Reversing signs (claim 4 of Elementary Order Arithmetic in an Ordered Field) and adding $v(x)$ gives $\Phi(x)<v(x)-K$, and $v(x)\le C$ gives $v(x)-K\le C-K=v(\xi)$; by mixed transitivity (claim 2 of Elementary Order Arithmetic in an Ordered Field),

$$\Phi(x)<\Phi(\xi)\qquad\text{whenever }r<\lVert x-\xi\rVert .$$

Step 2 (upper semicontinuity of $\Phi$). Let $q:\mathbb{R}^{M}\to\mathbb{R}$ be given by $q(x)=-\frac{\lambda}{2}\lVert x-\xi\rVert^{2}$. The map $y\mapsto\lVert y\rVert^{2}$ is smooth on $\mathbb{R}^{M}$ by The Squared Euclidean Norm is Smooth, and $\mathbb{R}^{M}$ is open in itself by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous. Apply Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution to this map with translation vector $-\xi$, scaling factor $1$ and multiplier $-\frac{\lambda}{2}$, both of the latter nonzero: since $(-\xi)+x=x-\xi$ by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space together with commutativity of addition, the substituted function is $q$ and its domain is all of $\mathbb{R}^{M}$, so claim 5 of that lemma shows $q$ is smooth on $\mathbb{R}^{M}$. By claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous, $q$ is continuous on $\mathbb{R}^{M}$ as a map from $(\mathbb{R}^{M},d)$ into $\mathbb{R}$ with the metric of The Absolute Value Metric on the Real Numbers; hence $q$ is upper semicontinuous on $\mathbb{R}^{M}$ by claim 2 of Semicontinuity Under Negation and Characterization of Continuity. Since $v$ is upper semicontinuous on $\mathbb{R}^{M}$ as well, claim 1 of Sums and Nonnegative Multiples of Semicontinuous Functions shows that $\Phi=v+q$ is upper semicontinuous on $\mathbb{R}^{M}$.

Step 3 (a maximum on a closed ball). Let $\bar B(\xi,r)$ be the closed ball of centre $\xi$ and radius $r$ in $(\mathbb{R}^{M},d)$. It contains $\xi$, because $d(\xi,\xi)=0\le r$, and it is compact by claim 2 of A Closed Euclidean Ball is Convex and Compact. By claim 2 of Negation, Restriction, and Separated Differences of Semicontinuous Functions the restriction of $\Phi$ to $\bar B(\xi,r)$ is upper semicontinuous on $\bar B(\xi,r)$, so claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set provides $y\in\bar B(\xi,r)$ with $\Phi(x)\le\Phi(y)$ for every $x\in\bar B(\xi,r)$.

Step 4 (the maximum is the supremum). Let $x\in\mathbb{R}^{M}$. If $d(x,\xi)\le r$ then $x\in\bar B(\xi,r)$ and $\Phi(x)\le\Phi(y)$ by Step 3. Otherwise $r<d(x,\xi)$, and $d(x,\xi)=\lVert x-\xi\rVert$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so Step 1 gives $\Phi(x)<\Phi(\xi)$; since $\xi\in\bar B(\xi,r)$ we have $\Phi(\xi)\le\Phi(y)$, and mixed transitivity gives $\Phi(x)\le\Phi(y)$. Thus $\Phi(y)$ is an upper bound for the set of values of $\Phi$, that is, for $S_{\lambda,v}(\xi)$; and $\Phi(y)$ is itself an element of that set, so it is its least upper bound. Therefore

$$v^{\lambda}(\xi)=\Phi(y)=v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}.$$

Finally, let $y\in\mathbb{R}^{M}$ be any point with $v^{\lambda}(\xi)=v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}$. By claim 1 of Domination, Monotonicity and Semiconvexity of the Sup-Convolution we have $v(\xi)\le v^{\lambda}(\xi)$, so the hypothesis of claim 1 above holds for $y$ and the stated bound follows.