In three dimensions the <Sobolev embedding theorem> gives
$$
H^2(U)\hookrightarrow L^\infty(U),
\qquad
H^2(U)\hookrightarrow W^{1,4}(U).
$$
Consequently, for $w\in H^2(U)\cap H_0^1(U)$,
$$
\||Dw|^2w\|_{L^2}
\leq\|Dw\|_{L^4}^2\|w\|_{L^\infty}
\leq C\|w\|_{H^2}^3.
$$
Thus $f+|Dw|^2w\in L^2(U)$. The <Dirichlet Poisson regularity theorem> on a bounded $C^2$ domain says that
$$
-\Delta v=f+|Dw|^2w,qquad v|_{\partial U}=0,
$$
has a unique $v\in H^2(U)\cap H_0^1(U)$ and
$$
\|v\|_{H^2}\leq C\bigl(\|f\|_2+\||Dw|^2w\|_2\bigr).
$$
Hence $\Phi(w)=v$ is well defined.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 / 3 / i / Solution

{c}
{title2=$-\Delta:H^2(U)\cap H_0^1(U)\to L^2(U)$}

If $U\subset\mathbb R^n$ is bounded with $C^2$ boundary, then every $F\in L^2(U)$ determines a unique $v\in H^2(U)\cap H_0^1(U)$ satisfying
$$
-\Delta v=F,
\qquad
\|v\|_{H^2(U)}\leq C\|F\|_{L^2(U)}.
$$


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 Incoming links: Dirichlet Poisson regularity theorem