Proof of Uniqueness of a Bounded Continuous Viscosity Solution on a Hilbert Triple

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

By Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on a Hilbert Triple (iii) each of $u_{1}$ and $u_{2}$ is both a viscosity subsolution and a viscosity supersolution of $F$ on $H$. By claim 6 of Properties of the Absolute Value in an Ordered Field, the hypotheses $|u_{1}(x)|\le C$ and $|u_{2}(x)|\le C$ give

$$u_{1}(x)\le C,\quad -C\le u_{1}(x),\quad u_{2}(x)\le C,\quad -C\le u_{2}(x)\qquad\text{for every }x\in H .$$

Apply A Comparison Principle on a Hilbert Triple under the First-Order Structure Condition (ii) with $u_{1}$ as the subsolution, $u_{2}$ as the supersolution and the constant $C$: it gives $u_{1}(x)\le u_{2}(x)$ for every $x\in V$. Applying it again with the roles of $u_{1}$ and $u_{2}$ exchanged gives $u_{2}(x)\le u_{1}(x)$ for every $x\in V$. Hence $u_{1}(x)=u_{2}(x)$ for every $x\in V$.

The subspace $V$ is nonempty and dense in $H$ by Hilbert Triples: Standing Notation and Background (i), and $u_{1}$ and $u_{2}$ are continuous on $H$, so Extension of a Uniformly Continuous Real Function from a Dense Subset (ii), applied in the metric space $(H,d_{H})$ with the dense subset $V$, gives $u_{1}(x)=u_{2}(x)$ for every $x\in H$.