Proof of Basic Properties of the Sublevel Sets of a Penalty

Each result cited is universally quantified over the data in its own statement. Since $P$ is of class $C^{2}$ on $D$ by Penalty on an Open Subset of Euclidean Space §regularity, claim 3 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous (with $m=1$, $k=2$) shows that $P$ is continuous relative to $D$ at every point of $D$, as a map into $(\mathbb{R},d_{\mathbb{R}})$. For $t\in\mathbb{R}$ put $K_{t}=\{x\in D:P(x)\le t\}$. By Penalty on an Open Subset of Euclidean Space §sublevel each $K_{t}$ is compact, hence closed by Compact Subset of $\mathbb{R}^n$ is Closed and bounded by Compact Subset of $\mathbb{R}^n$ is Bounded.

Claim 1. If $D$ is empty, $m=0$ satisfies the claim vacuously. Otherwise fix $x_{0}\in D$ and put $t_{0}=P(x_{0})$. Then $x_{0}\in K_{t_{0}}$, so $K_{t_{0}}$ is nonempty and compact, and the restriction of $P$ to $K_{t_{0}}$ is continuous on $K_{t_{0}}$ by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map. By Extreme Value Theorem on a Compact Subset of a Metric Space there is $x_{m}\in K_{t_{0}}$ with $P(x_{m})\le P(x)$ for every $x\in K_{t_{0}}$; put $m=P(x_{m})$. Let $x\in D$. If $P(x)\le t_{0}$, then $x\in K_{t_{0}}$ and $m\le P(x)$. Otherwise $t_{0}<P(x)$, and $m\le P(x_{0})=t_{0}$ because $x_{0}\in K_{t_{0}}$, so $m\le P(x)$ by transitivity.

Claim 2. Fix $R\in\mathbb{R}$.

$D_{R}$ is open. Let $x\in D_{R}$ and put $\varepsilon=R-P(x)$, positive by claim 1 of Elementary Order Arithmetic in an Ordered Field, adding $-P(x)$ to both sides of $P(x)<R$. Since $D$ is open in $(\mathbb{R}^{n},d_{E})$ by Second-Order Equations on Euclidean Open Sets §space, there is a positive $\rho$ such that every $y\in\mathbb{R}^{n}$ with $d_{E}(y,x)<\rho$ lies in $D$; by continuity of $P$ at $x$ there is a positive $\eta$ such that every $y\in D$ with $d_{E}(y,x)<\eta$ satisfies $|P(y)-P(x)|<\varepsilon$, hence $P(y)<P(x)+\varepsilon=R$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Let $\sigma$ be the least of $\rho$ and $\eta$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), which is positive. Every $y\in\mathbb{R}^{n}$ with $d_{E}(y,x)<\sigma$ lies in $D$ and satisfies $P(y)<R$, that is, lies in $D_{R}$. So $D_{R}$ is open in $(\mathbb{R}^{n},d_{E})$, hence open by Second-Order Equations on Euclidean Open Sets §space.

$D_{R}$ is bounded. $D_{R}\subseteq K_{R}$, since $P(x)<R$ implies $P(x)\le R$, and $K_{R}$ is bounded; a subset of a bounded set is bounded, since any ball containing $K_{R}$ contains $D_{R}$.

The closure. $K_{R}$ is closed and contains $D_{R}$, so $\overline{D_{R}}\subseteq K_{R}$ by claim 3 of The Closure is the Smallest Closed Superset; that is, $\overline{D_{R}}\subseteq\{x\in D:P(x)\le R\}$, and in particular $\overline{D_{R}}\subseteq D$. Moreover $\overline{D_{R}}$ is closed by claim 2 of The Closure is the Smallest Closed Superset and bounded as a subset of the bounded set $K_{R}$, so it is compact by Heine-Borel Theorem in $\mathbb{R}^n$.

Claim 3. Fix $R\in\mathbb{R}$ and $x\in\partial D_{R}$. Since $D_{R}$ is open by claim 2, claim 4 of The Interior is the Largest Open Subset gives $\operatorname{int}_{\mathbb{R}^{n}}(D_{R})=D_{R}$, so by the definition of the boundary $\partial D_{R}=\overline{D_{R}}\setminus D_{R}$. Thus $x\in\overline{D_{R}}\subseteq K_{R}$ by claim 2, so $x\in D$ and $P(x)\le R$; and $x\notin D_{R}$ with $x\in D$ means that $P(x)<R$ fails. Hence $P(x)=R$.