Proof of The Sup-Convolution Converges Pointwise to an Upper Semicontinuous Function

Order arithmetic is taken from Elementary Arithmetic in an Ordered Field and Elementary Order Arithmetic in an Ordered Field. Fix $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$.

Step 1 (a modulus from upper semicontinuity). Since $v$ is upper semicontinuous at $\xi$ relative to $\mathbb{R}^{M}$, there is $\delta\in\mathbb{R}$ with $0<\delta$ such that every $y\in\mathbb{R}^{M}$ with $d(\xi,y)<\delta$ satisfies $v(y)<v(\xi)+\varepsilon$.

Step 2 (choice of $\lambda_{0}$). Put $K=C-v(\xi)$, so that $0\le K$ by claim 3 of Elementary Arithmetic in an Ordered Field, since $v(\xi)\le C$. Put $b=\frac{\delta^{2}}{2}$. From $0<\delta$ we get $0<\delta^{2}$ by claim 5 of Elementary Order Arithmetic in an Ordered Field, and $0<2^{-1}$ by claim 7 of the same result, so $0<b$ by claim 5 again; hence $b^{-1}$ exists and $0<b^{-1}$ by claim 7. Also $1\le K+1$ by claim 3 of Elementary Arithmetic in an Ordered Field and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so $0<K+1$ by mixed transitivity (claim 2). Put

$$\lambda_{0}=(K+1)\,b^{-1},$$

so that $0<\lambda_{0}$ by claim 5 of Elementary Order Arithmetic in an Ordered Field and $\lambda_{0}\,b=K+1$.

Step 3 (the estimate). Let $\lambda\in\mathbb{R}$ with $\lambda_{0}\le\lambda$; then $0<\lambda$ by mixed transitivity, so the sup-convolution $v^{\lambda}$ is defined. By claim 1 of Domination, Monotonicity and Semiconvexity of the Sup-Convolution we have $v(\xi)\le v^{\lambda}(\xi)$, so only the upper estimate remains.

By claim 2 of The Sup-Convolution of an Upper Semicontinuous Function Attains its Supremum there is $y\in\mathbb{R}^{M}$ with

$$v^{\lambda}(\xi)=v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\qquad\text{and}\qquad\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le K .$$

Suppose $\delta\le\lVert y-\xi\rVert$. Both numbers are nonnegative, so claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\delta^{2}\le\lVert y-\xi\rVert^{2}$, and multiplying by the nonnegative number $\frac{\lambda}{2}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives

$$\lambda\,b=\frac{\lambda}{2}\,\delta^{2}\le\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le K .$$

On the other hand $\lambda_{0}\le\lambda$ and $0\le b$ give $K+1=\lambda_{0}b\le\lambda b$ by claim 5 of Elementary Arithmetic in an Ordered Field, so $K+1\le K$; this contradicts $K<K+1$, which holds by claims 6 and 1 of Elementary Order Arithmetic in an Ordered Field. Hence $\lVert y-\xi\rVert<\delta$.

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 and the symmetry clause 3 of a metric, $d(\xi,y)=\lVert y-\xi\rVert<\delta$, so Step 1 gives $v(y)<v(\xi)+\varepsilon$. Since $\frac{\lambda}{2}\lVert y-\xi\rVert^{2}$ is nonnegative, as observed in Sup-Convolution of a Function on $\mathbb{R}^M$, claim 3 of Elementary Arithmetic in an Ordered Field gives $v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le v(y)$, and mixed transitivity yields

$$v^{\lambda}(\xi)=v(y)-\frac{\lambda}{2}\lVert y-\xi\rVert^{2}\le v(y)<v(\xi)+\varepsilon .$$

As $\lambda\ge\lambda_{0}$ was arbitrary, $\lambda_{0}$ has the required property.