= Solution
Work with real-valued functions. The <screened sine-Gordon energy> is well defined: $0\leq1-\cos u\leq u^2/2$, and $fu$ is integrable by the <Cauchy-Schwarz inequality>. It is coercive, because
$$
E[u]\geq\tfrac12\|u\|_{H^1}^2-\|f\|_2\|u\|_2
\geq\tfrac14\|u\|_{H^1}^2-\|f\|_2^2.
$$
A minimizing sequence is bounded in the <Hilbert space> $H^1$. <Weak sequential compactness of bounded sequences in a reflexive Banach space> supplies a weakly convergent subsequence. The squared <H1 space> norm is weakly lower semicontinuous, the source pairing is weakly continuous, and the nonlinear term is covered by <local Sobolev compactness gives lower semicontinuity of a nonnegative integral>. The <direct method in the calculus of variations> therefore gives a minimizer $\phi$.
Taking its <first variation> in any $v\in H^1$ gives
$$
\int_{\mathbb R^3}\nabla\phi\cdot\nabla v+\phi v+\sin\phi\,v\,dx
=\int_{\mathbb R^3}fv\,dx.
$$
Thus the <Euler-Lagrange equation> is
$$
\boxed{-\Delta\phi+\phi+\sin\phi=f,}
$$
as a <weak solution>, equivalently in <distributions> when tested against smooth compactly supported functions. The derivative of the nonlinear term is justified by $|\sin\phi|\leq|\phi|$ and the second-order remainder bound $|\cos|\leq1$.
Now $G=f-\sin\phi\in L^2$, so the supplied <elliptic regularity> estimate puts $\phi$ in $H^2$. Applying the <Sobolev inequality> to $\phi$ and each first <weak derivative> gives $\phi\in W^{1,6}(\mathbb R^3)$. <Morrey's inequality> and <uniformly continuous integrable functions vanish at infinity> prove that its continuous representative tends to zero.
\b[The final printed supremum estimate is false in general.] The <maximum bound for a monotone reaction term> involves $h(s)=s+\sin s$, which is odd and strictly increasing: $h'=1+\cos s\geq0$, and its zeros are isolated. If $M=\|f\|_\infty<\infty$, a positive maximum $m$ of $\phi$ satisfies $h(m)\leq f\leq M$; apply the same argument to $-\phi$. The valid general estimate is
$$
\boxed{\|\phi\|_\infty\leq h^{-1}(M)\leq M+1.}
$$
The bound by $M$ is valid if $M\leq\pi$, but $h(s)$ can be smaller than $s$ for larger positive $s$.
For an explicit <failure of the source-size bound for the screened sine-Gordon equation>, take $a=3\pi/2$, $L=10$, $\phi(x)=a e^{-|x|^2/L^2}$ and $f=-\Delta\phi+\phi+\sin\phi$. These are smooth functions in the required spaces. With $q=|x|^2/L^2$,
$$
|\Delta\phi|=\frac{a}{L^2}|4q-6|e^{-q}\leq\frac{6a}{L^2},
\qquad
0\leq h(\phi)\leq h(a)=a-1.
$$
Hence $\|f\|_\infty\leq a-1+6a/L^2<a=\|\phi\|_\infty$. Moreover $V(s)=s^2/2+1-\cos s$ has $V''=1+\cos s\geq0$, so the energy is convex and this critical point is a minimizer. The example therefore satisfies even the minimizing and $C^2$ hypotheses of the printed claim.
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