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If $f$ is continuous, bounded and nonnegative on $[0,\infty )$, extend it to a bounded continuous function on the real line by the displayed formula. The bounded semilinear Dirichlet existence theorem gives $-\Delta u=F(u)$ with $u\in H_0^1$. Test with the negative part $u_{-}=\max \{-u,0\}$ to obtain $-\Vert \nabla u_{-}{\Vert }_2^2=\int F(u)u_{-}\ge 0$. The Poincaré inequality gives $u_{-}=0$, so $u\ge 0$ and $F(u)=f(u)$. Thus singularities or uncontrolled growth of the original formula at negative arguments need not prevent a nonnegative weak solution.

A weak solution is $u\in H_0^1(\Omega )$ such thatLet $M_f=\Vert f{\Vert }_{\infty }$ and define the Nemytskii operator $N(v)=f\circ v$ on $L^2(\Omega )$. It satisfies $\Vert N(v){\Vert }_2\le M_f|\Omega {|}^{1/2}$. It is continuous in $L^2$: from any sequence $v_j\to v$ in $L^2$, every subsequence has a further subsequence converging almost everywhere; continuity and boundedness of $f$, followed by the dominated convergence theorem, give convergence of the corresponding $N(v_j)$ in $L^2$. This subsequence criterion proves continuity of the whole sequence.

For $h\in L^2$, the Lax-Milgram theorem defines the inverse Dirichlet solution map $Sh\in H_0^1$ by $\int \nabla (Sh)\cdot \nabla \phi =\int h\phi$. Testing with $Sh$ and using the Poincaré inequality yieldsThe Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. Therefore $T=S\circ N$ is a nonlinear compact operator and its image lies in the closed $L^2$ ball of radius $R=1+C_P^2{M}_f|\Omega {|}^{1/2}$. This ball is nonempty, closed, bounded and convex. The Schauder fixed-point theorem gives $u=Tu$. Since $Tu\in H_0^1$, its defining identity is exactly the required weak formulation. Thus a weak solution exists. This is the bounded semilinear Dirichlet existence theorem.