= Solution
Taking the <Fourier transform> gives $(1+|\xi|^2)\widehat u(\xi)=\widehat\phi(\xi)$. Thus the canonical solution is
$$
\boxed{u(x)=\frac1{(2\pi)^n}\int_{\mathbb R^n}e^{ix\cdot\xi}\frac{\widehat\phi(\xi)}{1+|\xi|^2}\,d\xi.}
$$
A smooth compactly supported function is a <Schwartz function>, as is its <Fourier transform>. The multiplier $(1+|\xi|^2)^{-1}$ and all its derivatives have polynomial bounds, so their product with $\widehat\phi$ is again a <Schwartz function>. The inverse integral can therefore be differentiated to verify $-\Delta u+u=\phi$ classically. By part (b), its Fourier-space integrand factor $\widehat\phi/(1+|\xi|^2)$ is radial. Changing variables by any <orthogonal transformation> in the inverse integral gives $u(Qx)=u(x)$, proving radiality.
This is also the unique tempered-distribution solution: the nonvanishing multiplier $1+|\xi|^2$ can be inverted on that space. Unrestricted classical solutions need not be unique, since one may add, for example, $e^{x_1}$; the Fourier construction selects the rapidly decaying solution.
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