The negative Laplacian of the fundamental solution is the unit Dirac distribution

First take a real-valued test function $\phi$. If its support is empty then $\phi =0$ and the desired pairing identity is immediate. Otherwise [F10] and [F13] let us choose $R>0$ with $\mathrm{supp}\phi \subset B_R(0)$. For $0<\epsilon <\min \{1,R\}$ put ${\mathrm{\Omega }}_{\epsilon }=\{z:\epsilon <\mathrm{\mid }z\mathrm{\mid }<R\}$. A point with radius strictly between $\epsilon$ and $R$ has a whole small ball in this set by [F12]; radial perturbations at either boundary sphere show $\mathrm{\partial }{\mathrm{\Omega }}_{\epsilon }=S_{\epsilon }\cup S_R$ by [F20]–[F21]. The point $((R+\epsilon )\mathrm{/}2,0,\dots ,0)$ lies in it and it is bounded by $R$. Near an outer-sphere point the inner-radius constraint is inactive and the annulus is the inside of that sphere; near an inner-sphere point the outer-radius constraint is inactive and the annulus is the outside of that sphere. The graph charts [F7], with a coordinate reflection when needed, therefore show that the annulus lies locally on one side of each $C^1$ graph. Its outward normals are $+\omega$ on $S_R$ and $-\omega$ on $S_{\epsilon }$ for $\omega =x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, since the annulus lies inside the outer sphere and outside the inner sphere. Thus [F14] makes ${\mathrm{\Omega }}_{\epsilon }$ a bounded $C^1$ domain; connectedness is not required.

Let $C_1:=\sqrt{n}{p}_1(\phi )$, where $p_1$ is the finite fixed-support seminorm from [F11]; then $C_1$ bounds $\mathrm{\mid }D\phi \mathrm{\mid }$. Since the outer radial derivative [F8] is constant on $S_{\epsilon }$, the flux identity [F28] gives $-q_n^{\mathrm{\prime }}(\epsilon ){\int }_{S_{\epsilon }}1dS=1$, so the sphere area is ${\omega }_{n-1}{\epsilon }^{n-1}$ if $n\ge 3$ and $2\pi \epsilon$ if $n=2$. By [F27], the bounded continuous inner-boundary integrand is integrable. For $n\ge 3$, its absolute integral is at most $\epsilon C_1\mathrm{/}(n-2)$; for $n=2$ it is at most $\epsilon \mathrm{\mid }\log \epsilon \mathrm{\mid }{C}_1$. Both bounds tend to zero as $\epsilon \downarrow 0$.

On $S_{\epsilon }$, [F8] gives the outward radial derivative $q_n^{\mathrm{\prime }}(\epsilon )$, and step 1.1 gives the annulus normal $-\omega$, so ${\mathrm{\partial }}_{{\nu }_{\mathrm{a}\mathrm{n}\mathrm{n}}}\mathrm{\Phi }=c_{\epsilon }:=-q_n^{\mathrm{\prime }}(\epsilon )>0$. The flux identity [F28] gives $c_{\epsilon }{\int }_{S_{\epsilon }}1dS=1$. By [F10] and [F27], the restriction of $\phi$ is surface-integrable. Therefore ${\int }_{S_{\epsilon }}\phi {\mathrm{\partial }}_{{\nu }_{\mathrm{a}\mathrm{n}\mathrm{n}}}\mathrm{\Phi }dS$ is the normalized spherical average of $\phi$. Its difference from $\phi (0)$ is at most ${\sup }_{\mathrm{\mid }z\mathrm{\mid }=\epsilon }\mathrm{\mid }\phi (z)-\phi (0)\mathrm{\mid }$, which tends to zero by continuity at the origin.

By [F15], $\mathrm{\Phi }$ is $C^2$ on a neighborhood of $\overline{{\mathrm{\Omega }}_{\epsilon }}$ and $\mathrm{\Delta }\mathrm{\Phi }=0$ there; [F10] gives the same regularity for $\phi$. Apply [F6] with $u=\mathrm{\Phi }$ and $v=\phi$. The outer-boundary terms vanish because the support lies strictly inside $B_R(0)$. On the inner sphere the normal is outward from the annulus. Hence $-{\int }_{{\mathrm{\Omega }}_{\epsilon }}\mathrm{\Phi }\mathrm{\Delta }\phi dz={\int }_{S_{\epsilon }}(\phi {\mathrm{\partial }}_{{\nu }_{\mathrm{a}\mathrm{n}\mathrm{n}}}\mathrm{\Phi }-\mathrm{\Phi }{\mathrm{\partial }}_{{\nu }_{\mathrm{a}\mathrm{n}\mathrm{n}}}\phi )dS$.

Let $C_2:=n{p}_2(\phi )$, where $p_2$ is the finite fixed-support seminorm from [F11]; then $C_2$ bounds $\mathrm{\mid }\mathrm{\Delta }\phi \mathrm{\mid }$. For any sequence ${\epsilon }_j\to 0$ with ${\epsilon }_j>0$, after discarding finitely many terms we have ${\epsilon }_j<1$. The measurable functions $g_j=\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }{\mathbf{1}}_{B_{{\epsilon }_j}}$ then tend to zero off the null singleton $\{0\}$ and are dominated by $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }{\mathbf{1}}_{B_1}$, which is integrable by [F2]. Thus [F9] gives ${\int }_{B_{{\epsilon }_j}}\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }\to 0$; as this holds for every such sequence, the limit as $\epsilon \downarrow 0$ is zero. The omitted integral of $\mathrm{\Phi }\mathrm{\Delta }\phi$ is bounded by $C_2{\int }_{B_{\epsilon }}\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$, hence tends to zero and ${\int }_{{\mathrm{\Omega }}_{\epsilon }}\mathrm{\Phi }\mathrm{\Delta }\phi \to {\int }_{B_R}\mathrm{\Phi }\mathrm{\Delta }\phi$.

Taking $\epsilon \downarrow 0$ in step 2.2 and using steps 1.2, 2.1 and 2.3 yields $-{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(z)\mathrm{\Delta }\phi (z)dz=\phi (0)$; the integral over ${\mathbb{R}}^n$ equals the one over $B_R$ because $\mathrm{\Delta }\phi$ vanishes off the support. By [F2], [F3] and [F19], $T_{\mathrm{\Phi }}$ is the regular distribution; [F4] and [F25] identify the left side with $⟨-\mathrm{\Delta }{T}_{\mathrm{\Phi }},\phi ⟩$, and [F5] identifies the right side with $⟨{\delta }_0,\phi ⟩$.

For a complex-valued test, apply step 3.1 separately to its real and imaginary parts and combine by complex linearity from [F17]. The pairing is bilinear, with no conjugation, as required by the distribution convention. Hence the identity holds for every test and $-\mathrm{\Delta }{T}_{\mathrm{\Phi }}={\delta }_0$.

For any fixed $y\in {\mathbb{R}}^n$ and test $\phi$, put $h(z)=\mathrm{\Phi }(z)\phi (z+y)$. It is integrable because it is bounded by a constant times $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ on $\mathrm{supp}\phi -y$, which lies in a ball by [F12]–[F13]; use [F2] and [F11]. By [F22] and [F26], $T_{-y}(x)=x-y$ preserves Lebesgue measure; [F23] therefore gives $\int h(T_{-y}x)dx=\int h(z)dz$, that is, $\int \mathrm{\Phi }(x-y)\phi (x)dx=\int \mathrm{\Phi }(z)\phi (z+y)dz$. Thus the translate of $T_{\mathrm{\Phi }}$ is the regular distribution associated with $x\mapsto \mathrm{\Phi }(x-y)$, with the sign convention [F24]. Translate the identity in step 4.1: [F16] says constant-coefficient derivatives commute with translation and ${\delta }_0$ translates to ${\delta }_y$, so $-{\mathrm{\Delta }}_x{T}_{\mathrm{\Phi }(\cdot -y)}={\delta }_y$. The proof handles $n=2$ and $n\ge 3$ separately, excludes the distinct one-dimensional analogue by $n\ge 2$, includes the zero test and empty support, and is unchanged by the arbitrary value assigned to $\mathrm{\Phi }(0)$ because a singleton is null [F18]. The choice cost is exactly [A1], inherited through the named measure and surface Green interfaces; no full Axiom of Choice is used. The statement is not an iff.