[proofplan] We show $u = -\Phi * f$ is well-defined and $C^2$ by differentiating under the integral sign (which works for first derivatives but not directly for second). To compute $\Delta u$, we change variables to place the derivatives on $f$ instead of $\Phi$, yielding $\Delta u = \Phi * \Delta f$. The distributional identity $-\Delta\Phi = \delta_0$ from the [Fundamental Solution theorem](/theorems/566) then gives $-\Delta u = f$. [/proofplan] [step:Show $u$ is well-defined and compute its first derivatives] Since $f \in C_c^2(\mathbb{R}^n)$, the integrand $\Phi(x - y)\,f(y)$ is compactly supported in $y$ for each $x$. The local integrability of $\Phi$ (from the [Fundamental Solution theorem](/theorems/566)) ensures $u(x)$ is finite for every $x$. For the first derivatives: $|\nabla\Phi(x)| = C|x|^{1-n}$, and $|x|^{1-n}$ is locally integrable for $n \geq 2$. Therefore differentiation under the integral is justified by dominated convergence: \begin{align*} \partial_{x_i} u(x) = -\int_{\mathbb{R}^n} \partial_{x_i}\Phi(x - y)\,f(y) \, d\mathcal{L}^n(y). \end{align*} [/step] [step:Compute $\Delta u$ by transferring derivatives to $f$ via a change of variables] Direct second differentiation fails: $|D^2\Phi(x)| = C|x|^{-n}$ is not locally integrable. Instead, change variables $z = x - y$: \begin{align*} u(x) = -\int_{\mathbb{R}^n} \Phi(z)\,f(x - z) \, d\mathcal{L}^n(z). \end{align*} Now $f(x - z)$ is $C^2$ in $x$ for each $z$, and all derivatives fall on $f$: \begin{align*} \Delta_x u(x) = -\int_{\mathbb{R}^n} \Phi(z)\,\Delta_x f(x - z) \, d\mathcal{L}^n(z) = -(\Phi * \Delta f)(x). \end{align*} Differentiation under the integral is justified because $\Delta f \in C_c(\mathbb{R}^n)$ and $\Phi \in L^1_{\mathrm{loc}}$. [/step] [step:Apply the distributional identity $-\Delta\Phi = \delta_0$ to conclude $-\Delta u = f$] By the [Fundamental Solution theorem](/theorems/566), $-\Delta\Phi = \delta_0$ in distributions. For convolutions with compactly supported smooth functions, the distributional Laplacian commutes with convolution: \begin{align*} -\Delta u = -\Delta(-\Phi * f) = (-\Delta\Phi) * f = \delta_0 * f = f. \end{align*} The identity $\delta_0 * f = f$ holds because $(\delta_0 * f)(x) = \int_{\mathbb{R}^n} \delta_0(x - y)\,f(y) \, d\mathcal{L}^n(y) = f(x)$. [/step]