Proof of Local Bounds, Semicontinuous Envelopes and Local Extrema Are Unchanged on the Trace of an Open Set

Throughout, $A_{u}(x)$ and $B_{u}(x)$ denote the sets attached to a function and a point of its domain in Upper and Lower Semicontinuous Envelopes of a Real-Valued Function, the ambient metric space being $(M,d)$. Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named at the point of use.

Claim 1. Suppose $u$ is bounded above near each point of $S$ and let $x\in S'$. Then $A_{u}(x)$ is nonempty by Upper and Lower Semicontinuous Envelopes of a Real-Valued Function §near-bounds: there are $c\in\mathbb{R}$ and a positive $r\in\mathbb{R}$ with $u(y)\le c$ for every $y\in S$ with $d(y,x)\le r$. Since $S'\subseteq S$, every $y\in S'$ with $d(y,x)\le r$ lies in $S$ and satisfies $u|_{S'}(y)=u(y)\le c$. Hence $c\in A_{u|_{S'}}(x)$, so this set is nonempty; as $x\in S'$ was arbitrary, $u|_{S'}$ is bounded above near each point of $S'$. The statement for lower bounds is obtained in the same way from the sets $B_{u}(x)$ and $B_{u|_{S'}}(x)$, with the inequalities $c\le u(y)$ in place of $u(y)\le c$.

A radius adapted to $O$. Let $x\in S'$. Then $x\in O$, and $O$ is open in $(M,d)$, so by Open Subset of a Metric Space there is a positive $\rho\in\mathbb{R}$ with $B_{d}(x,\rho)\subseteq O$. Put $r=\tfrac{\rho}{2}$, so that $0<r$ and $r<\rho$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, and put

If $y\in S_{r}$, then $d(x,y)=d(y,x)\le r<\rho$ by the symmetry axiom, so $d(x,y)<\rho$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, whence $y\in B_{d}(x,\rho)$ by Open Ball in a Metric Space and therefore $y\in O$ and $y\in S\cap O=S'$. Thus $S_{r}\subseteq S'$. Consequently $S'_{r}=S_{r}$: the inclusion $S'_{r}\subseteq S_{r}$ holds because $S'\subseteq S$, and the reverse because every $y\in S_{r}$ lies in $S'$ and satisfies $d(y,x)\le r$. In particular $u|_{S_{r}}$ and $(u|_{S'})|_{S'_{r}}$ have the same domain and the same value $u(z)$ at each point $z$ of it, so they are the same function. Finally $x\in S_{r}$, since $d(x,x)=0\le r$ by the metric axioms.

Claim 2. Suppose $u$ is bounded above near each point of $S$, let $x\in S'$, and let $\rho$, $r$, $S_{r}$ and $S'_{r}$ be as above. We apply Semicontinuity and the Semicontinuous Envelopes are Local Notions twice, in both cases with the metric space $(M,d)$, the point $x$ and the radius $r$.

First, with the nonempty set $S$ and the function $u$: the set called $S_{r}$ there is the set $S_{r}$ above, and claim 4 of that lemma gives $\bigl(u|_{S_{r}}\bigr)^{*}(y)=u^{*}(y)$ for every $y\in S$ with $d(y,x)<r$. Taking $y=x$, which is legitimate because $d(x,x)=0<r$, we obtain

Secondly, with the set $S'$, which is nonempty and contains $x$, and the function $u|_{S'}$, which is bounded above near each point of $S'$ by claim 1: the set called $S_{r}$ there is now $S'_{r}$, and claim 4 of that lemma gives

Since $(u|_{S'})|_{S'_{r}}$ and $u|_{S_{r}}$ are the same function, the left-hand sides of the two displays are equal, and therefore $\bigl(u|_{S'}\bigr)^{*}(x)=u^{*}(x)$.

Claim 3. Suppose $u$ is bounded below near each point of $S$ and let $x\in S'$. The argument of claim 2 applies verbatim, using the second half of claim 1 of the present lemma for the hypothesis on $u|_{S'}$ and claim 5 of Semicontinuity and the Semicontinuous Envelopes are Local Notions in place of claim 4; it yields $\bigl(u|_{S'}\bigr)_{*}(x)=u_{*}(x)$.

Claim 4. Suppose first that $u$ has a local maximum at $\bar{x}$ relative to $S$. By Local Maximum of a Function Relative to a Subset of a Metric Space there is a positive $\delta\in\mathbb{R}$ such that every $y\in S$ with $d(\bar{x},y)<\delta$ satisfies $u(y)\le u(\bar{x})$. Since $S'\subseteq S$ and $\bar{x}\in S'$, every $y\in S'$ with $d(\bar{x},y)<\delta$ satisfies $u|_{S'}(y)=u(y)\le u(\bar{x})=u|_{S'}(\bar{x})$, so the same $\delta$ witnesses that $u|_{S'}$ has a local maximum at $\bar{x}$ relative to $S'$.

Conversely, suppose $u|_{S'}$ has a local maximum at $\bar{x}$ relative to $S'$, and let $\delta$ be a positive real number as in Local Maximum of a Function Relative to a Subset of a Metric Space for this local maximum. Since $\bar{x}\in S'\subseteq O$ and $O$ is open in $(M,d)$, there is a positive $\rho\in\mathbb{R}$ with $B_{d}(\bar{x},\rho)\subseteq O$ by Open Subset of a Metric Space. Put $\delta'=\min\{\delta,\rho\}$, their minimum; by claim 2 of Elementary Properties of the Minimum of Two Elements, $\delta'$ is $\delta$ or $\rho$, hence positive, and by claim 1 of that lemma $\delta'\le\delta$ and $\delta'\le\rho$. Let $y\in S$ satisfy $d(\bar{x},y)<\delta'$. Then $d(\bar{x},y)<\rho$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so $y\in B_{d}(\bar{x},\rho)\subseteq O$ by Open Ball in a Metric Space and hence $y\in S\cap O=S'$; and $d(\bar{x},y)<\delta$ by the same claim, so $u(y)=u|_{S'}(y)\le u|_{S'}(\bar{x})=u(\bar{x})$. Thus $\delta'$ witnesses that $u$ has a local maximum at $\bar{x}$ relative to $S$.

Claim 5. The argument of claim 4 applies verbatim with Local Minimum of a Function Relative to a Subset of a Metric Space in place of Local Maximum of a Function Relative to a Subset of a Metric Space and the inequalities $u(\bar{x})\le u(y)$ in place of $u(y)\le u(\bar{x})$.