Proof of The Viscosity Property Depends Only on the Values on the Trace of $V$

Each result cited is universally quantified over the data in its own statement.

Claim 1. Let $\delta\in\mathbb{R}$ be positive and suppose that $u$ and $\tilde{u}$ are bounded above near each point of $U$. By The $\delta$-Envelopes $u^-_\delta$ and $u^+_\delta$ of a Function on an Open Subset of a Hilbert Triple (i) the envelope $u^{-}_{\delta}$ is the upper semicontinuous envelope of the function $V\cap U\to\mathbb{R}$ whose value at $x$ is $u(x)-\delta h(x)$, and $\tilde{u}^{-}_{\delta}$ is that of the function whose value at $x$ is $\tilde{u}(x)-\delta h(x)$. These two functions on $V\cap U$ are equal, because $u(x)=\tilde{u}(x)$ for every $x\in V\cap U$; hence their upper semicontinuous envelopes are equal, that is, $u^{-}_{\delta}(x)=\tilde{u}^{-}_{\delta}(x)$ for every $x\in V\cap U$. The same argument with The $\delta$-Envelopes $u^-_\delta$ and $u^+_\delta$ of a Function on an Open Subset of a Hilbert Triple (ii) and the lower semicontinuous envelopes gives the second assertion.

Claim 2. Assume $u$ and $\tilde{u}$ are bounded above near each point of $U$ and that $u$ is a viscosity subsolution of $F$ on $U$. The condition of Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on a Hilbert Triple (i) is then to be verified for $\tilde{u}$, which is bounded above near each point of $U$ by hypothesis. So let $\delta\in\mathbb{R}$ be positive, let $\varphi\in C^{2}(U)$, let $\hat{x}\in V\cap U$ be a point at which the function $V\cap U\to\mathbb{R}$ with value $\tilde{u}^{-}_{\delta}(x)-\varphi(x)$ at $x$ has a local maximum relative to $V\cap U$, and let $\varepsilon\in\mathbb{R}$ be positive.

By claim 1 the function just named is the function $V\cap U\to\mathbb{R}$ with value $u^{-}_{\delta}(x)-\varphi(x)$ at $x$, so the latter has a local maximum relative to $V\cap U$ at $\hat{x}$. Since $u$ is a viscosity subsolution of $F$ on $U$, there are $y\in W$, $s\in\mathbb{R}$, $q\in H$ and $Y\in\mathrm{Sym}(H)$ with

$$|y-\hat{x}|_{H}<\varepsilon,\quad |u^{-}_{\delta}(y)-u^{-}_{\delta}(\hat{x})|<\varepsilon,\quad |s-u^{-}_{\delta}(\hat{x})|<\varepsilon,\quad |q-D\varphi(\hat{x})|_{H}<\varepsilon,\quad \lVert Y-D^{2}\varphi(\hat{x})\rVert<\varepsilon$$

and $F^{-}_{\delta}(y,s,q,Y)\le\varepsilon$. By claim 1 again, $u^{-}_{\delta}(y)=\tilde{u}^{-}_{\delta}(y)$ and $u^{-}_{\delta}(\hat{x})=\tilde{u}^{-}_{\delta}(\hat{x})$ (both $y$ and $\hat{x}$ lie in $V\cap U$, since $W\subseteq V\cap U$), so the same $y$, $s$, $q$ and $Y$ satisfy the displayed conditions with $\tilde{u}^{-}_{\delta}$ in place of $u^{-}_{\delta}$, together with $F^{-}_{\delta}(y,s,q,Y)\le\varepsilon$. As $\delta$, $\varphi$, $\hat{x}$ and $\varepsilon$ were arbitrary, $\tilde{u}$ is a viscosity subsolution of $F$ on $U$.

Claim 3. The argument is the same, with Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on a Hilbert Triple (ii), the envelopes $u^{+}_{\delta}$ and $\tilde{u}^{+}_{\delta}$, which agree by claim 1, local minima in place of local maxima, and the final inequality $-\varepsilon\le F^{+}_{\delta}(y,s,q,Y)$.

Claim 4. Assume $u$ and $\tilde{u}$ are bounded above near each point of $U$ and bounded below near each point of $U$, and that $u$ is a viscosity solution of $F$ on $U$. By Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on a Hilbert Triple (iii) the function $u$ is both a viscosity subsolution and a viscosity supersolution of $F$ on $U$, so by claims 2 and 3 so is $\tilde{u}$; by the same clause, $\tilde{u}$ is a viscosity solution of $F$ on $U$.