Proof of Fibre Suprema along a Coordinate Map over a Translation-Closed Set and Their Semicontinuous Envelopes

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement of the lemma. The properties of the coordinate map $\Lambda$ and of the associated map $\Lambda^{\sharp}$ recorded in Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §coordinates are used throughout, as is Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §stability. Put

so that the standing hypothesis reads

Fix $C_{u}\in\mathbb{R}$ an upper bound for the set of values of $\hat{u}$ and $c_{v}\in\mathbb{R}$ a lower bound for the set of values of $\hat{v}$.

Claim 1 (clause 1). Let $\zeta\in\mathbb{R}^{m}$. The set $A$ is nonempty because it is translation-closed along $e$; fix $z_{0}\in A$ and put $z=z_{0}+\Lambda^{\sharp}\bigl(\zeta-\Lambda z_{0}\bigr)$, which lies in $A$ by that clause. By the linearity of $\Lambda$ and the identity $\Lambda\Lambda^{\sharp}\eta=\eta$, both recorded in Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §coordinates, $\Lambda z=\Lambda z_{0}+\bigl(\zeta-\Lambda z_{0}\bigr)$, and this is $\zeta$ because $\mathbb{R}^{m}$ is a real vector space. Hence $z\in A_{\zeta}$ and $A_{\zeta}$ is nonempty.

The set $\{\hat{u}(x):x\in A_{\zeta}\}$ is therefore nonempty and is bounded above by $C_{u}$, so it has a least upper bound by Least Upper Bound Property of the Real Numbers, unique by Uniqueness of the Supremum and of the Infimum; and $\{\hat{v}(y):y\in A_{\zeta}\}$ is nonempty and bounded below by $c_{v}$, so it has a greatest lower bound by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below, again unique. Thus $U$ and $\mathcal{V}$ are defined. Since $C_{u}$ is an upper bound for each of the sets whose suprema define $U$, we get $U(\zeta)\le C_{u}$ for every $\zeta$; likewise $c_{v}\le\mathcal{V}(\zeta)$ for every $\zeta$. Consequently, for every $\zeta\in\mathbb{R}^{m}$ the number $C_{u}$ belongs to the set $A_{U}(\zeta)$ of Upper and Lower Semicontinuous Envelopes of a Real-Valued Function (take $r=1$) and $c_{v}$ belongs to $B_{\mathcal{V}}(\zeta)$, so $U$ is bounded above near each point of $\mathbb{R}^{m}$ and $\mathcal{V}$ is bounded below near each point. This proves Claim 1.

Claim 2 (the fibrewise inequality). For all $\zeta,\omega\in\mathbb{R}^{m}$,

Proof. Fix $\zeta,\omega$ and write $\pi=\tfrac{\alpha}{2}\lVert\zeta-\omega\rVert^{2}$. Let $y\in A_{\omega}$. For every $x\in A_{\zeta}$ we have $\Lambda x=\zeta$ and $\Lambda y=\omega$, so the standing hypothesis gives $\hat{u}(x)-\hat{v}(y)-\pi\le M_{1}$, that is $\hat{u}(x)\le M_{1}+\hat{v}(y)+\pi$ by claim 3 of Elementary Arithmetic in an Ordered Field. Thus $M_{1}+\hat{v}(y)+\pi$ is an upper bound for $\{\hat{u}(x):x\in A_{\zeta}\}$, whence $U(\zeta)\le M_{1}+\hat{v}(y)+\pi$ by Upper Bound and Least Upper Bound, that is $U(\zeta)-M_{1}-\pi\le\hat{v}(y)$. As $y\in A_{\omega}$ was arbitrary, $U(\zeta)-M_{1}-\pi$ is a lower bound for $\{\hat{v}(y):y\in A_{\omega}\}$, so $U(\zeta)-M_{1}-\pi\le\mathcal{V}(\omega)$ by Lower Bound and Greatest Lower Bound in a Totally Ordered Set, which rearranges to the claim. This proves Claim 2.

Claim 3 (the envelope inequality). $U^{*}$ is upper semicontinuous on $\mathbb{R}^{m}$, $\mathcal{V}_{*}$ is lower semicontinuous on $\mathbb{R}^{m}$, and for all $\zeta,\omega\in\mathbb{R}^{m}$,

Proof. The semicontinuity assertions are claim 2 of Properties of the Upper Semicontinuous Envelope and claim 3 of Properties of the Lower Semicontinuous Envelope, by Duality, whose hypotheses are supplied by Claim 1.

Fix $\zeta,\omega\in\mathbb{R}^{m}$ and write $D=\lVert\zeta-\omega\rVert$. Let $\varepsilon\in\mathbb{R}$ be positive. Choose a positive $\delta\le1$ with

which is possible: the number $2+2|\alpha|(D+1)$ is positive, so the quotient $\tfrac{\varepsilon}{2+2|\alpha|(D+1)}$ exists and is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field, and taking for $\delta$ the lesser of it and $1$ gives $2\delta+2|\alpha|\delta(D+1)=\delta\bigl(2+2|\alpha|(D+1)\bigr)\le\varepsilon$.

By claim 5 of Properties of the Upper Semicontinuous Envelope, applied to $U$ at $\zeta$ with $\delta$, there is $\zeta'\in\mathbb{R}^{m}$ with $d_{E}(\zeta',\zeta)\le\delta$ and $|U(\zeta')-U^{*}(\zeta)|<\delta$. By claim 1 of Properties of the Lower Semicontinuous Envelope, by Duality the function $-\mathcal{V}$, whose value at $\xi$ is the additive inverse of $\mathcal{V}(\xi)$, is bounded above near each point of $\mathbb{R}^{m}$ and satisfies $(-\mathcal{V})^{*}=-\mathcal{V}_{*}$; so claim 5 of Properties of the Upper Semicontinuous Envelope, applied to $-\mathcal{V}$ at $\omega$ with $\delta$, gives $\omega'\in\mathbb{R}^{m}$ with $d_{E}(\omega',\omega)\le\delta$ and $|\mathcal{V}(\omega')-\mathcal{V}_{*}(\omega)|<\delta$. Since $d_{E}(\xi,\xi')=\lVert\xi-\xi'\rVert$ by Euclidean Distance is a Metric on $\mathbb{R}^n$, we have $\lVert\zeta-\zeta'\rVert\le\delta$ and $\lVert\omega-\omega'\rVert\le\delta$.

By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions §stability, $\lVert\zeta'-\omega'\rVert\le D+2\delta\le D+2$ and

using $\delta\le1$, claim 4 of Properties of the Absolute Value in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field. Combining, and using claim 6 of Properties of the Absolute Value in an Ordered Field on the three absolute values,

the last step by Claim 2 applied at $(\zeta',\omega')$ and by the choice of $\delta$. As $\varepsilon$ was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack gives $U^{*}(\zeta)-\mathcal{V}_{*}(\omega)-\tfrac{\alpha}{2}D^{2}\le M_{1}$. This proves Claim 3.

Claim 4 (clause 3). Since $\bar{x}\in A_{\bar{\zeta}}$ and $\bar{y}\in A_{\bar{\omega}}$, we have $\hat{u}(\bar{x})\le U(\bar{\zeta})$ and $\mathcal{V}(\bar{\omega})\le\hat{v}(\bar{y})$ by Upper Bound and Least Upper Bound and Lower Bound and Greatest Lower Bound in a Totally Ordered Set; and $U(\bar{\zeta})\le U^{*}(\bar{\zeta})$ and $\mathcal{V}_{*}(\bar{\omega})\le\mathcal{V}(\bar{\omega})$ by claim 1 of Properties of the Upper Semicontinuous Envelope and claim 2 of Properties of the Lower Semicontinuous Envelope, by Duality. Writing $\pi=\tfrac{\alpha}{2}\lVert\bar{\zeta}-\bar{\omega}\rVert^{2}$, these inequalities give

the last inequality by Claim 3. Hence all four quantities are equal. Put $a=U^{*}(\bar{\zeta})-\hat{u}(\bar{x})$ and $b=\hat{v}(\bar{y})-\mathcal{V}_{*}(\bar{\omega})$; both are nonnegative by the inequalities above and claim 3 of Elementary Arithmetic in an Ordered Field, and $a+b=0$ because the first and third quantities in the display are equal. Then $a\le a+b=0$ and $b\le a+b=0$ by the compatibility of the order with addition (an axiom of Ordered Field), so $a=0$ and $b=0$, that is $U^{*}(\bar{\zeta})=\hat{u}(\bar{x})$ and $\mathcal{V}_{*}(\bar{\omega})=\hat{v}(\bar{y})$. The chain of inequalities $\hat{u}(\bar{x})\le U(\bar{\zeta})\le U^{*}(\bar{\zeta})=\hat{u}(\bar{x})$ then gives $U(\bar{\zeta})=\hat{u}(\bar{x})$, and likewise $\mathcal{V}(\bar{\omega})=\hat{v}(\bar{y})$. This proves Claim 4, which is clause 3.

Clause 2. By Claim 4 the right-hand side of the display in clause 2 equals $M_{1}$, so the display is exactly the inequality of Claim 3. Together with the semicontinuity assertions of Claim 3 this proves clause 2.

Claim 5 (clause 4). Let $\zeta_{1}\in\mathbb{R}^{m}$ and let $r,\varepsilon\in\mathbb{R}$ be positive; let $\sigma$ be the lesser of $r$ and $\tfrac{\varepsilon}{2}$, positive because it is one of them (claim 9 of Elementary Order Arithmetic in an Ordered Field and claim 8 of the same lemma for $\tfrac{\varepsilon}{2}$).

By claim 5 of Properties of the Upper Semicontinuous Envelope applied to $U$ at $\zeta_{1}$ with $\sigma$, there is $\zeta'\in\mathbb{R}^{m}$ with $d_{E}(\zeta',\zeta_{1})\le\sigma$ and $|U(\zeta')-U^{*}(\zeta_{1})|<\sigma$; hence $U^{*}(\zeta_{1})-\tfrac{\varepsilon}{2}\le U^{*}(\zeta_{1})-\sigma<U(\zeta')$ by claim 6 of Properties of the Absolute Value in an Ordered Field. Since $U(\zeta')$ is the least upper bound of $\{\hat{u}(x):x\in A_{\zeta'}\}$ and $U(\zeta')-\tfrac{\varepsilon}{2}<U(\zeta')$, the number $U(\zeta')-\tfrac{\varepsilon}{2}$ is not an upper bound of that set, so there is $x\in A_{\zeta'}$ with $U(\zeta')-\tfrac{\varepsilon}{2}<\hat{u}(x)$. Then $x\in A$, $\lVert\Lambda x-\zeta_{1}\rVert=\lVert\zeta'-\zeta_{1}\rVert=d_{E}(\zeta',\zeta_{1})\le\sigma\le r$, and

For $y$, apply the same argument to $-\mathcal{V}$, using $(-\mathcal{V})^{*}=-\mathcal{V}_{*}$ from claim 1 of Properties of the Lower Semicontinuous Envelope, by Duality: there is $\omega'\in\mathbb{R}^{m}$ with $d_{E}(\omega',\zeta_{1})\le\sigma$ and $-\mathcal{V}_{*}(\zeta_{1})-\sigma<-\mathcal{V}(\omega')$, and then $y\in A_{\omega'}$ with $-\mathcal{V}(\omega')-\tfrac{\varepsilon}{2}<-\hat{v}(y)$, because $\mathcal{V}(\omega')$ is the greatest lower bound of $\{\hat{v}(y):y\in A_{\omega'}\}$ and $\mathcal{V}(\omega')+\tfrac{\varepsilon}{2}$ is therefore not a lower bound of it. Combining as above and negating (claim 4 of Elementary Order Arithmetic in an Ordered Field) gives $\lVert\Lambda y-\zeta_{1}\rVert\le r$ and $\hat{v}(y)<\mathcal{V}_{*}(\zeta_{1})+\varepsilon$. This proves Claim 5, which is clause 4, and completes the proof of the lemma.