= Solution
The <Sobolev embedding theorem> $H^2(\Omega)\hookrightarrow L^\infty(\Omega)$ in dimension three, combined with the preceding estimate, gives
$$
\boxed{\|u\|_\infty\le C\|f\|_2.}
$$
One direct justification of this embedding is a bounded <Sobolev extension operator> into $H^2(\mathbb R^3)$, followed by Fourier inversion and the <Cauchy-Schwarz inequality>: $\int_{\mathbb R^3}(1+|\xi|^2)^{-2}d\xi<\infty$ bounds the inverse Fourier integral by the $H^2$ <norm>.
For radial forcing, the preceding part makes $u=U(r)$ and $f=F(r)$. Set $a=1/2$, $b=2$ and $H(r)=\int_a^rs^2F(s)\,ds$. The radial equation gives $r^2U'(r)=C_0+H(r)$. Both zero boundary values imply
$$
C_0=-\frac{\int_a^br^{-2}H(r)\,dr}{\int_a^br^{-2}\,dr},\qquad |C_0|\le\|H\|_\infty.
$$
Moreover $\|H\|_\infty\le(\int_a^bs^2ds)^{1/2}(\int_a^bs^2|F|^2ds)^{1/2}\le C\|f\|_2$. Since $r\ge a$, this proves the <radial Poisson gradient estimate>:
$$
\boxed{\|\nabla u\|_\infty=\|U'\|_\infty\le C\|f\|_2\quad\text{for radial }f.}
$$
For a nonradial counterexample choose $x_0$ in the shell and a nonconstant $\varphi\in C_c^\infty(B(0,1))$. For sufficiently small $\varepsilon$, set $u_\varepsilon(x)=\varepsilon^{1/2}\varphi((x-x_0)/\varepsilon)$ and $f_\varepsilon=\Delta u_\varepsilon$. These are smooth with support away from the boundary. Scaling gives
$$
\boxed{\|f_\varepsilon\|_2=\|\Delta\varphi\|_2,\qquad\|\nabla u_\varepsilon\|_\infty=\varepsilon^{-1/2}\|\nabla\varphi\|_\infty\longrightarrow\infty.}
$$
Thus the <failure of L2-to-Linfinity Poisson gradient bounds in three dimensions> is explicit, despite the valid $H^2$ and value-supremum estimates.
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