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

Throughout, the notation is that of the statement; $\mathbb{R}^{M}$ is open, so it is an admissible domain for Quadruple Approximable by Test-Function Data and for Transfer of a Test Function from the Sup-Convolution to the Original Function. Put

$$L=v^{\lambda}(\eta_{0})+\tfrac{1}{2}\,\lambda^{-1}\lVert q_{0}\rVert^{2},\qquad \Theta=2+\lambda^{-1}+\tfrac{1}{2}\,\lambda^{-1}\bigl(2\lVert q_{0}\rVert+1\bigr).$$

Since $0<\lambda^{-1}$ and $0\le\lVert q_{0}\rVert$ 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, every summand after the leading $2$ in $\Theta$ is nonnegative, so $1\le\Theta$; in particular $0<\Theta$, and $\Theta^{-1}$ exists and is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field.

Step 1 (the basic construction). Let $\varepsilon'\in\mathbb{R}$ satisfy $0<\varepsilon'\le1$. Then there are $y\in\mathbb{R}^{M}$ and a function $\psi:\mathbb{R}^{M}\to\mathbb{R}$ of class $C^{2}$ on $\mathbb{R}^{M}$ such that $v-\psi$ has a local maximum at $y$ relative to $\mathbb{R}^{M}$ and

$$d_{E}(y,z_{0})\le\Theta\varepsilon',\quad \bigl|v(y)-L\bigr|\le\Theta\varepsilon',\quad \bigl\lVert D\psi(y)-q_{0}\bigr\rVert\le\Theta\varepsilon',\quad d_{\mathcal{S}(M)}\bigl(D^{2}\psi(y),Y\bigr)\le\Theta\varepsilon'.$$

Indeed, by hypothesis the quadruple $\bigl(\eta_{0},v^{\lambda}(\eta_{0}),q_{0},Y\bigr)$ is approximable by test data from above for $v^{\lambda}$, so clause 1 of Quadruple Approximable by Test-Function Data, applied with the positive real number $\varepsilon'$, provides $\eta\in\mathbb{R}^{M}$ and a function $\varphi:\mathbb{R}^{M}\to\mathbb{R}$ of class $C^{2}$ on $\mathbb{R}^{M}$ such that $v^{\lambda}-\varphi$ has a local maximum at $\eta$ relative to $\mathbb{R}^{M}$ and, writing $q=D\varphi(\eta)\in\mathbb{R}^{M}$ and $Y'=D^{2}\varphi(\eta)\in\mathcal{S}(M)$,

$$d_{E}(\eta,\eta_{0})<\varepsilon',\quad \bigl|v^{\lambda}(\eta)-v^{\lambda}(\eta_{0})\bigr|<\varepsilon',\quad \lVert q-q_{0}\rVert<\varepsilon',\quad d_{\mathcal{S}(M)}(Y',Y)<\varepsilon'.$$

The hypotheses of Transfer of a Test Function from the Sup-Convolution to the Original Function are met, with the present $M$, $v$, $C$, $\lambda$ and $v^{\lambda}$, with $U=\mathbb{R}^{M}$, with the point $\eta$ and with the test function $\varphi$. Choose $y\in\mathbb{R}^{M}$ with

$$v^{\lambda}(\eta)=v(y)-\frac{\lambda}{2}\,\lVert y-\eta\rVert^{2},$$

one such point existing by claim 2 of The Sup-Convolution of an Upper Semicontinuous Function Attains its Supremum, put $z=y-\eta$ and let $\psi$ be the function $\varphi_{z}$ of that lemma, given by $\psi(x)=\varphi(x-z)$ on

$$U'=\{x\in\mathbb{R}^{M}:x-z\in\mathbb{R}^{M}\}=\mathbb{R}^{M}.$$

Its three claims give

$$y=\eta+\lambda^{-1}q,\qquad v(y)=v^{\lambda}(\eta)+\tfrac{1}{2}\,\lambda^{-1}\lVert q\rVert^{2},$$

and: $\psi$ is of class $C^{2}$ on $\mathbb{R}^{M}$, $D\psi(y)=q$, $D^{2}\psi(y)=Y'$, and $v-\psi$ has a local maximum at $y$ relative to $\mathbb{R}^{M}$.

Since $1\le\Theta$ and $0<\varepsilon'$ we have $\varepsilon'\le\Theta\varepsilon'$, so the third and the fourth of the asserted bounds follow at once from $\lVert q-q_{0}\rVert<\varepsilon'$ and $d_{\mathcal{S}(M)}(Y',Y)<\varepsilon'$.

For the first, the vector space identities of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space give

$$y-z_{0}=\bigl(\eta+\lambda^{-1}q\bigr)-\bigl(\eta_{0}+\lambda^{-1}q_{0}\bigr)=(\eta-\eta_{0})+\lambda^{-1}(q-q_{0}),$$

so by claim 6 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (the triangle inequality) and claim 5 there (homogeneity), with $|\lambda^{-1}|=\lambda^{-1}$ because $\lambda^{-1}$ is positive,

$$d_{E}(y,z_{0})=\lVert y-z_{0}\rVert\le\lVert\eta-\eta_{0}\rVert+\lambda^{-1}\lVert q-q_{0}\rVert\le\varepsilon'+\lambda^{-1}\varepsilon'=\bigl(1+\lambda^{-1}\bigr)\varepsilon'\le\Theta\varepsilon'.$$

For the second, subtracting the two displayed value identities gives

$$v(y)-L=\bigl(v^{\lambda}(\eta)-v^{\lambda}(\eta_{0})\bigr)+\tfrac{1}{2}\,\lambda^{-1}\bigl(\lVert q\rVert^{2}-\lVert q_{0}\rVert^{2}\bigr).$$

Applying Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vii) with the matrix $I_{M}$, and using Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi) in the case $a=1$, which gives $x\cdot(I_{M}x)=\lVert x\rVert^{2}$ for every $x$ and $\lVert I_{M}\rVert=1$, we obtain

$$\bigl|\lVert q\rVert^{2}-\lVert q_{0}\rVert^{2}\bigr|\le\lVert q-q_{0}\rVert\,\lVert q+q_{0}\rVert .$$

By the triangle inequality $\lVert q\rVert\le\lVert q-q_{0}\rVert+\lVert q_{0}\rVert<\varepsilon'+\lVert q_{0}\rVert$, whence

$$\lVert q+q_{0}\rVert\le\lVert q\rVert+\lVert q_{0}\rVert<2\lVert q_{0}\rVert+\varepsilon'\le 2\lVert q_{0}\rVert+1 .$$

Combining, and using claims 5 and 4 of Properties of the Absolute Value in an Ordered Field (the triangle inequality and multiplicativity of the absolute value),

$$\bigl|v(y)-L\bigr|\le\varepsilon'+\tfrac{1}{2}\,\lambda^{-1}\,\varepsilon'\bigl(2\lVert q_{0}\rVert+1\bigr)=\Bigl(1+\tfrac{1}{2}\,\lambda^{-1}\bigl(2\lVert q_{0}\rVert+1\bigr)\Bigr)\varepsilon'\le\Theta\varepsilon'.$$

This proves Step 1.

Step 2 (proof of claim 1). We first show $v(z_{0})\le L$. By Sup-Convolution of a Function on $\mathbb{R}^M$ the number $v^{\lambda}(\eta_{0})$ is an upper bound for the set $S_{\lambda,v}(\eta_{0})$, so taking the point $z_{0}$ in that set,

$$v(z_{0})-\frac{\lambda}{2}\,\lVert z_{0}-\eta_{0}\rVert^{2}\le v^{\lambda}(\eta_{0}).$$

Here $z_{0}-\eta_{0}=\lambda^{-1}q_{0}$, so $\lVert z_{0}-\eta_{0}\rVert=\lambda^{-1}\lVert q_{0}\rVert$ by 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, hence $\lVert z_{0}-\eta_{0}\rVert^{2}=\lambda^{-1}\lambda^{-1}\lVert q_{0}\rVert^{2}$ and, since $\lambda\lambda^{-1}=1$,

$$\frac{\lambda}{2}\,\lVert z_{0}-\eta_{0}\rVert^{2}=\tfrac{1}{2}\,\lambda^{-1}\lVert q_{0}\rVert^{2}.$$

Therefore $v(z_{0})\le L$.

We now show $L\le v(z_{0})$. Suppose, for a contradiction, that $v(z_{0})<L$, and put

$$\varepsilon=\bigl(L-v(z_{0})\bigr)\cdot 2^{-1}\cdot 2^{-1},$$

which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field applied twice. Since $v$ is upper semicontinuous at $z_{0}$ relative to $\mathbb{R}^{M}$, there is a positive $\delta\in\mathbb{R}$ such that every $x\in\mathbb{R}^{M}$ with $d_{E}(z_{0},x)<\delta$ satisfies $v(x)<v(z_{0})+\varepsilon$. Put

$$\varepsilon'=\min\bigl\{\,1,\ \delta\,\Theta^{-1}2^{-1},\ \varepsilon\,\Theta^{-1}2^{-1}\,\bigr\},$$

which exists by claim 9 of Elementary Order Arithmetic in an Ordered Field applied twice and is positive by claims 5, 7 and 8 there. Let $y$ and $\psi$ be as furnished by Step 1 for this $\varepsilon'$. From $\varepsilon'\le\delta\,\Theta^{-1}2^{-1}$ and $0<\Theta$ we get $\Theta\varepsilon'\le\delta\cdot2^{-1}<\delta$, so

$$d_{E}(z_{0},y)=d_{E}(y,z_{0})\le\Theta\varepsilon'<\delta$$

by the symmetry of the metric $d_{E}$, and therefore $v(y)<v(z_{0})+\varepsilon$. Likewise $\bigl|v(y)-L\bigr|\le\Theta\varepsilon'\le\varepsilon\cdot2^{-1}<\varepsilon$, and by claims 2 and 3 of Properties of the Absolute Value in an Ordered Field we have $L-v(y)\le\bigl|v(y)-L\bigr|$, so $L<v(y)+\varepsilon$. Combining the two estimates and using claim 8 of Elementary Order Arithmetic in an Ordered Field twice,

$$L<v(y)+\varepsilon<v(z_{0})+\varepsilon+\varepsilon=v(z_{0})+\bigl(L-v(z_{0})\bigr)2^{-1}<v(z_{0})+\bigl(L-v(z_{0})\bigr)=L,$$

a contradiction. Hence $L\le v(z_{0})$, and with the previous paragraph and the antisymmetry of $\le$ we conclude $v(z_{0})=L$, which is claim 1.

Step 3 (proof of claim 2). Let $\varepsilon\in\mathbb{R}$ be positive and put $\varepsilon'=\min\bigl\{1,\ \varepsilon\,\Theta^{-1}2^{-1}\bigr\}$, positive as in Step 2, so that $\Theta\varepsilon'\le\varepsilon\cdot2^{-1}<\varepsilon$. Let $y$ and $\psi$ be as furnished by Step 1 for this $\varepsilon'$. By claim 1 we have $v(z_{0})=L$, so the four bounds of Step 1 read

$$d_{E}(y,z_{0})<\varepsilon,\quad \bigl|v(y)-v(z_{0})\bigr|<\varepsilon,\quad \bigl\lVert D\psi(y)-q_{0}\bigr\rVert<\varepsilon,\quad d_{\mathcal{S}(M)}\bigl(D^{2}\psi(y),Y\bigr)<\varepsilon,$$

while $\psi$ is of class $C^{2}$ on $\mathbb{R}^{M}$ and $v-\psi$ has a local maximum at $y$ relative to $\mathbb{R}^{M}$. Since $\varepsilon$ was an arbitrary positive real number, clause 1 of Quadruple Approximable by Test-Function Data shows that the quadruple $\bigl(z_{0},v(z_{0}),q_{0},Y\bigr)$ is approximable by test data from above for $v$, which is claim 2.