[proofplan] We use the constant-coefficient hypothesis to commute the differential operator through convolution. The equation $LE=\delta_0$ then reduces the result to the identity property of the Dirac delta under convolution with a [test function](/page/Test%20Function). [/proofplan] [step: Commute the operator through convolution] Because $L$ is a constant-coefficient linear differential operator on $\mathbb R^n$, there is an integer $m\ge 0$ and constants $a_\alpha\in\mathbb R$ or $\mathbb C$, indexed by multi-indices $\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb N^n$ with $|\alpha|=\alpha_1+\cdots+\alpha_n\le m$, such that \begin{align*} L=\sum_{|\alpha|\le m}a_\alpha\partial^\alpha \end{align*} where $\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}$. Since $f\in C_c^\infty(\mathbb R^n)$ and $E$ is a distribution, the convolution $E*f$ is a distribution and distributional [derivatives commute with convolution](/theorems/6159) by $f$: \begin{align*} \partial^\alpha(E*f)=(\partial^\alpha E)*f. \end{align*} Therefore \begin{align*} L(E*f)=(LE)*f. \end{align*} The constancy of the coefficients is exactly what allows the same coefficients $a_\alpha$ to pass through this computation. [/step] [step: Use the defining property of the fundamental solution] By hypothesis $LE=\delta_0$ in $\mathcal D'(\mathbb R^n)$. Hence \begin{align*} L(E*f)=(LE)*f=\delta_0*f. \end{align*} The Dirac delta is the identity element for convolution with a test function, so $\delta_0*f=f$ as a distribution. Thus $Lu=f$ in $\mathcal D'(\mathbb R^n)$ for $u=E*f$. [guided] Since $L$ has constant coefficients, write $L=\sum_{|\alpha|\le m}a_\alpha\partial^\alpha$ for some $m\ge0$, constants $a_\alpha$, and multi-indices $\alpha\in\mathbb N^n$. For a distribution $E\in\mathcal D'(\mathbb R^n)$ and a test function $f\in C_c^\infty(\mathbb R^n)$, convolution is defined distributionally and satisfies $\partial^\alpha(E*f)=(\partial^\alpha E)*f$. Hence \begin{align*} L(E*f)=\sum_{|\alpha|\le m}a_\alpha\partial^\alpha(E*f)=\sum_{|\alpha|\le m}a_\alpha(\partial^\alpha E)*f=(LE)*f. \end{align*} The hypothesis says $LE=\delta_0$, so this becomes \begin{align*} L(E*f)=(LE)*f=\delta_0*f. \end{align*} Finally $\delta_0*f=f$ in $\mathcal D'(\mathbb R^n)$ because the Dirac delta acts as the identity for convolution with a test function. Thus, for $u=E*f$, one obtains $Lu=f$ in the distributional sense. [/guided] [/step]