Define $\Psi(x) = v(\lvert x \rvert)$. Find all the functions $v:(0, \infty) \to \mathbb{R}$ such that for $x \neq 0$, $\Delta \Psi(x) = 0$.
Determine all those solutions, if they exist, satisfying that for all $f \in \mathscr{C}^\infty(\mathbb{R}^d)$: $$ \int_{\mathbb{R}^d} \Psi(x) \Delta f(x) dx = f(0). $$
For the first part, I note that $\Psi$ is a radially symmetric function, and since $v$ is defined on only positive $x$, $\Psi(x) = v(x)$, and so $v$ must be a harmonic function, i.e. $v_{xx} = 0$. Integrating twice wrt $x$ then gives us that $v(x) = c_1 x + c_2$.
For the second part, I don't really see how to proceed, other than noting it's similarity in form to a solution to Poisson's equation $$ - \Delta u = f$$ $$ u(x) = \int_{\mathbb{R}^d} \Phi(x-y) f(y) dy $$
where $ \Phi $ here is the fundamental solution of Laplace's equation (which is also radially symmetric).
A second attempt:
To find v such that $\Delta \Psi(x) = 0$, we consider the Laplacian in hyperspherical coordinates: $$ \Delta \Psi(x) = \Psi_{rr} + \frac{n-1}{r} \Psi_r + \frac{\Delta_s \Psi}{r^2} $$ where $\Delta_s$ depends only on the angles. Since $\Psi$ is radially symmetric, $\Delta_s \Phi = 0$, and $r = \lvert x \rvert$, we thus have $$ \Delta \Psi (x) = \Psi_{xx}(x) + \frac{n-1}{\lvert x \rvert}\Psi_x(x) =0$$
$\Phi(x) = v(\lvert x \rvert)$ and $v:(0, \infty) \to \mathbb{R}$ leads us to solve the ODE $$ \frac{d^2}{dx^2}v + \frac{n-1}{\lvert x \rvert} \frac{d}{dx}v =0$$
Which is seperable, giving: $\frac{d}{dx}v = x^{1-n}+C$ and so $$ v = \frac{1}{2-n} x^{2-n} + Cx +d$$ (Which isn't singular as @qbert suggests?).
For the second part, using Green's identity $$\int_{\mathbb{R}^d} \Psi(x) \Delta f(x) dx = \int_{\delta \mathbb{R}^d} \Psi \frac{df}{d\nu} - f \frac{d\Psi}{d\nu} dS + \int_\mathbb{R}^d f \Delta \Psi dx $$
Since $\delta \mathbb{R}^d = \emptyset$, and $\Delta \Psi =0$ the right hand side is 0, so the $f$ are those that have $f(0)=0$?