Proof of Transfer of a Test Function from the Sup-Convolution to the Original Function

Order arithmetic is taken from Elementary Arithmetic in an Ordered Field and Elementary Order Arithmetic in an Ordered Field, and vector identities in $\mathbb{R}^{M}$ from Euclidean Space $\mathbb{R}^n$ is a Real Vector Space.

Two standing facts. First, by Sup-Convolution of a Function on $\mathbb{R}^M$, for all $x,\xi\in\mathbb{R}^{M}$ the number $v(x)-\frac{\lambda}{2}\lVert x-\xi\rVert^{2}$ belongs to the set whose least upper bound is $v^{\lambda}(\xi)$, so

$$v(x)-\frac{\lambda}{2}\,\lVert x-\xi\rVert^{2}\le v^{\lambda}(\xi).$$

Second, 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, $d(a,b)=\lVert a-b\rVert=\lVert b-a\rVert$ for all $a,b\in\mathbb{R}^{M}$.

A radius. By the local maximum hypothesis there is $\delta_{0}\in\mathbb{R}$ with $0<\delta_{0}$ such that every $\xi\in U$ with $d(\eta,\xi)<\delta_{0}$ satisfies

$$v^{\lambda}(\xi)-\varphi(\xi)\le v^{\lambda}(\eta)-\varphi(\eta).$$

Since $U$ is open in $(\mathbb{R}^{M},d)$ by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$, there is $\sigma\in\mathbb{R}$ with $0<\sigma$ such that the open ball $B_{d}(\eta,\sigma)$ is contained in $U$. Let $\rho$ be the lesser of $\delta_{0}$ and $\sigma$, which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field; as $\rho$ equals one of them, $0<\rho$. Put $W=B_{d}(\eta,\rho)$, which contains $\eta$, is contained in $U$, and is open in $(\mathbb{R}^{M},d)$ by Open Ball in a Metric Space is Open, hence open by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$.

Claim 1. Let $h:W\to\mathbb{R}$ be given by

$$h(\xi)=v(y)+\Bigl(-\frac{\lambda}{2}\Bigr)d(\xi,y)^{2}-\varphi(\xi).$$

By claims 2 and 3 of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian, applied on the open set $W$ with the point $y$ and the constant $-\frac{\lambda}{2}$, the function $\xi\mapsto(-\frac{\lambda}{2})d(\xi,y)^{2}$ is of class $C^{2}$ on $W$ and its gradient at $\xi$ is $(-\lambda)(\xi-y)$. The restriction of $\varphi$ to $W$ is of class $C^{2}$ on $W$ by claim 3 of Restriction of a $C^k$ Map to an Open Subset, $W$ being an open subset of $U$, and its partial derivatives at points of $W$ agree with those of $\varphi$ by claim 1 of that lemma, so its gradient at $\eta$ is $D\varphi(\eta)$. Hence, by claim 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set applied to the sum of these two functions and the constant function with value $v(y)$, the function $h$ is of class $C^{2}$ on $W$; and by claim 1 of that lemma together with Gradient of a Real-Valued Function on a Euclidean Open Set, computing coordinatewise,

$$Dh(\eta)=(-\lambda)(\eta-y)-D\varphi(\eta)=\lambda\,(y-\eta)-D\varphi(\eta)=\lambda\,z-D\varphi(\eta).$$

Now let $\xi\in W$. The first standing fact with $x=y$, together with $d(\xi,y)^{2}=\lVert y-\xi\rVert^{2}$, gives $v(y)+(-\frac{\lambda}{2})d(\xi,y)^{2}\le v^{\lambda}(\xi)$, so

$$h(\xi)\le v^{\lambda}(\xi)-\varphi(\xi)\le v^{\lambda}(\eta)-\varphi(\eta)=h(\eta),$$

the middle inequality because $d(\eta,\xi)<\rho\le\delta_{0}$ and $\xi\in U$, and the final equality because $v^{\lambda}(\eta)=v(y)-\frac{\lambda}{2}\lVert y-\eta\rVert^{2}$ and $d(\eta,y)^{2}=\lVert y-\eta\rVert^{2}$. Thus $h$ has a local maximum at $\eta$ relative to $W$, any positive radius serving in Local Maximum of a Function Relative to a Subset of a Metric Space. By claim 1 of First- and Second-Order Conditions at a Local Extremum of a Function of Class $C^2$, $Dh(\eta)$ is the origin of $\mathbb{R}^{M}$, that is,

$$\lambda\,z=D\varphi(\eta).$$

Multiplying by $\lambda^{-1}$ and using the vector space axioms gives $z=\lambda^{-1}D\varphi(\eta)$, that is $y=\eta+\lambda^{-1}D\varphi(\eta)$. The argument used only the defining property of $y$, so any point with that property equals $\eta+\lambda^{-1}D\varphi(\eta)$; hence $y$ is unique.

Claim 2. From $v^{\lambda}(\eta)=v(y)-\frac{\lambda}{2}\lVert y-\eta\rVert^{2}$ and claim 3 of Elementary Arithmetic in an Ordered Field we get $v(y)=v^{\lambda}(\eta)+\frac{\lambda}{2}\lVert y-\eta\rVert^{2}$. By claim 1, $y-\eta=\lambda^{-1}D\varphi(\eta)$, so 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 gives $\lVert y-\eta\rVert=|\lambda^{-1}|\,\lVert D\varphi(\eta)\rVert$. Since $0<\lambda$ we have $0<\lambda^{-1}$ by claim 7 of Elementary Order Arithmetic in an Ordered Field, and then $|\lambda^{-1}|=\lambda^{-1}$: by claim 1 of Properties of the Absolute Value in an Ordered Field the absolute value $|\lambda^{-1}|$ is nonnegative and equals $\lambda^{-1}$ or $-\lambda^{-1}$, and $-\lambda^{-1}<0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, which excludes the second alternative. Squaring and multiplying by $\frac{\lambda}{2}$ in the field of real numbers, where $\frac{\lambda}{2}(\lambda^{-1})^{2}=\frac{1}{2}\lambda^{-1}$, gives

$$\frac{\lambda}{2}\lVert y-\eta\rVert^{2}=\tfrac{1}{2}\lambda^{-1}\lVert D\varphi(\eta)\rVert^{2},$$

which is the assertion.

Claim 3. Apply Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution to $\varphi$ on $U$ with translation vector $-z$, scaling factor $1$ and multiplier $1$, both of the latter nonzero. Since $(-z)+x=x-z$ by claim 3 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and commutativity of addition, the substituted domain of claim 1 of that lemma is exactly $U'$, which is therefore open, and the substituted function is $\varphi_{z}$. Moreover $y-z=\eta\in U$, so $y\in U'$. Claim 4 of the same lemma, with $k=2$, shows that $\varphi_{z}$ is of class $C^{2}$ on $U'$, and claim 2 gives, for every $x\in U'$ and every $i\in\{1,\dots,M\}$,

$$\partial_{i}\varphi_{z}(x)=\partial_{i}\varphi(x-z);$$

at $x=y$ this reads $\partial_{i}\varphi_{z}(y)=\partial_{i}\varphi(\eta)$, so $D\varphi_{z}(y)=D\varphi(\eta)$ by Gradient of a Real-Valued Function on a Euclidean Open Set. The displayed identity says that $\partial_{j}\varphi_{z}$ is obtained from $\partial_{j}\varphi:U\to\mathbb{R}$ by the same substitution, so a second application of claim 2 of Partial Derivatives, Continuity and $C^k$ Regularity under a Scaling Substitution, now to $\partial_{j}\varphi$, gives $\partial_{i}\partial_{j}\varphi_{z}(x)=\partial_{i}\partial_{j}\varphi(x-z)$ for $x\in U'$; at $x=y$ and for all $i,j$ this is $D^{2}\varphi_{z}(y)=D^{2}\varphi(\eta)$, by Hessian Matrix of a C^2 Function.

Finally, let $x\in U'$ satisfy $d(y,x)<\rho$, and put $\xi=x-z$. Then $\xi-\eta=x-(z+\eta)=x-y$, so $d(\eta,\xi)=\lVert\xi-\eta\rVert=\lVert x-y\rVert=d(y,x)<\rho$; in particular $\xi\in W\subseteq U$ and $d(\eta,\xi)<\delta_{0}$. Since $x-\xi=z$, the first standing fact gives

$$v(x)-\frac{\lambda}{2}\lVert z\rVert^{2}\le v^{\lambda}(\xi)\le v^{\lambda}(\eta)-\varphi(\eta)+\varphi(\xi),$$

the second inequality being the local maximum hypothesis. As $\lVert z\rVert=\lVert y-\eta\rVert$ and $v^{\lambda}(\eta)+\frac{\lambda}{2}\lVert y-\eta\rVert^{2}=v(y)$, rearranging by claim 3 of Elementary Arithmetic in an Ordered Field yields

$$v(x)-\varphi_{z}(x)=v(x)-\varphi(\xi)\le v(y)-\varphi(\eta)=v(y)-\varphi_{z}(y),$$

using $\varphi_{z}(y)=\varphi(\eta)$. Hence the function $x\mapsto v(x)-\varphi_{z}(x)$ has a local maximum at $y$ relative to $U'$, with radius $\rho$.