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

Let $M$, $\mathbb{R}$, $\mathbb{R}^{M}$ and $\lVert\,\cdot\,\rVert$ be as in Sup-Convolution of a Function on $\mathbb{R}^M$; regard $\mathbb{R}^{M}$ as a real vector space, with the sum of points, the scalar multiple and the difference $x-y$. Let $d$ be the Euclidean distance, a metric on $\mathbb{R}^{M}$; a subset of $\mathbb{R}^{M}$ is open if and only if it is open in $(\mathbb{R}^{M},d)$, by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$.

Let $v:\mathbb{R}^{M}\to\mathbb{R}$ be upper semicontinuous on $\mathbb{R}^{M}$ with respect to $d$, let $C\in\mathbb{R}$ be an upper bound for the set of values of $v$, let $\lambda\in\mathbb{R}$ satisfy $0<\lambda$, and let $v^{\lambda}$ be the sup-convolution of $v$ with parameter $\lambda$.

Let $U\subseteq\mathbb{R}^{M}$ be open, let $\eta\in U$, let $\varphi:U\to\mathbb{R}$ be of class $C^{2}$ on $U$, and suppose that the function $U\to\mathbb{R}$ whose value at $\xi$ is $v^{\lambda}(\xi)-\varphi(\xi)$ has a local maximum at $\eta$ relative to $U$. Write $D\varphi$ and $D^{2}\varphi$ for the gradient and the Hessian matrix of $\varphi$.

Let $y\in\mathbb{R}^{M}$ satisfy

at least one such point existing by claim 2 of The Sup-Convolution of an Upper Semicontinuous Function Attains its Supremum. Put $z=y-\eta$, let

and let $\varphi_{z}:U'\to\mathbb{R}$ be given by $\varphi_{z}(x)=\varphi(x-z)$. Then the following hold.

1. (Location of the maximiser) $\lambda\,z=D\varphi(\eta)$; equivalently $y=\eta+\lambda^{-1}D\varphi(\eta)$. In particular $y$ is the only point of $\mathbb{R}^{M}$ with the displayed property.

2. (Value at the maximiser) $v(y)=v^{\lambda}(\eta)+\tfrac{1}{2}\lambda^{-1}\lVert D\varphi(\eta)\rVert^{2}$.

3. (Transfer of the test function) $U'$ is open and contains $y$; the function $\varphi_{z}$ is of class $C^{2}$ on $U'$ and satisfies

and the function $U'\to\mathbb{R}$ whose value at $x$ is $v(x)-\varphi_{z}(x)$ has a local maximum at $y$ relative to $U'$.