Let $L$ be a linear differential operator with constant coefficients, and $E$ a fundamental solution: $LE = \delta$.

Claim: If $f \in \mathcal{D}(\mathbb{R}^n)$ (smooth, compactly supported), then $u = E * f$ satisfies $Lu = f$.

Step 1: Convolution is Well-Defined

Since $f$ has compact support and $E$ is a distribution, the convolution $E * f$ is well-defined as a distribution: $\langle E * f, \phi \rangle = \langle E, \tilde{f} * \phi \rangle$ where $\tilde{f}(x) = f(-x)$.

Moreover, since $f \in \mathcal{D}$, the result $E * f$ is actually a $C^\infty$ function (regularity of convolutions with smooth functions).

Step 2: Apply the Operator

We compute $L(E * f)$ using the fact that differentiation commutes with convolution for distributions: $L(E * f) = (LE) * f$

This identity holds because: $\langle L(E * f), \phi \rangle = \langle E * f, L^*\phi \rangle = \langle E, \tilde{f} * (L^*\phi) \rangle$ $= \langle E, L^*(\tilde{f} * \phi) \rangle = \langle LE, \tilde{f} * \phi \rangle = \langle (LE) * f, \phi \rangle$

Step 3: Use $LE = \delta$

Since $LE = \delta$, we have: $L(E * f) = (LE) * f = \delta * f = f$

The last equality uses the fundamental property of delta: $\delta * f = f$ for any function or distribution.

Step 4: Verify Explicitly

To see $\delta * f = f$ explicitly: $(\delta * f)(x) = \langle \delta_y, f(x - y) \rangle = f(x - 0) = f(x)$

Therefore, $u = E * f$ satisfies $Lu = f$.

Step 5: Extension to General $f$

For $f \in L^1_{\text{loc}}$ or more general distributions, the convolution $E * f$ may require additional care (ensuring $E * f$ is defined) but the formal computation remains valid in appropriate function spaces.