Fundamental solution for the positive operator minus Laplacian

By [F1] and [F2], ${\omega }_{n-1}=n\mathrm{\mid }{B}_1\mathrm{\mid }$ is finite and strictly positive. In dimension two, [F3] gives $\mathrm{\mid }{B}_1\mathrm{\mid }=\pi$, so ${\omega }_1=2\pi$ and the logarithmic coefficient in the Statement is exactly $1\mathrm{/}{\omega }_1$. The operator sign is $L=-\mathrm{\Delta }$ using the definition [F9]. This uses the Countable Choice assumption [A1] through the measure convention.

In dimension one, ${\int }_{-R}^R\mathrm{\mid }{\mathrm{\Phi }}_1(x)\mathrm{\mid }dx=R^2\mathrm{/}2$, so ${\mathrm{\Phi }}_1$ is locally integrable. For a test $\phi$, [F7] gives $⟨-{\mathrm{\Phi }}_1^{\mathrm{\prime }\mathrm{\prime }},\phi ⟩=-{\int }_{\mathbb{R}}{\mathrm{\Phi }}_1(x){\phi }^{\mathrm{\prime }\mathrm{\prime }}(x)dx$. The slopes of ${\mathrm{\Phi }}_1$ are $+1\mathrm{/}2$ to the left of zero and $-1\mathrm{/}2$ to the right; integrating by parts separately on the two half-lines, the compact-support endpoint terms vanish and ${\int }_{\mathbb{R}}{\mathrm{\Phi }}_1{\phi }^{\mathrm{\prime }\mathrm{\prime }}=-\phi (0)$, so the negative integral equals $\phi (0)={\delta }_0(\phi )$ by [F8]. Thus $-{\mathrm{\Phi }}_1^{\mathrm{\prime }\mathrm{\prime }}={\delta }_0$ and [F5] identifies it as the one-dimensional fundamental-solution analogue for $L=-d^2\mathrm{/}d{x}^2$.

Let $0<R\le 1$. For $n\ge 3$, [F4] gives ${\int }_{B_R}\mathrm{\mid }{\mathrm{\Phi }}_n(x)\mathrm{\mid }dx=\frac{1}{n-2}{\int }_0^{R}rdr=\frac{R^2}{2(n-2)}<\mathrm{\infty }$. For $n=2$, using ${\omega }_1=2\pi$ from step 1.1, it gives ${\int }_{B_R}\mathrm{\mid }{\mathrm{\Phi }}_2(x)\mathrm{\mid }dx={\int }_0^{R}r(-\log r)dr=-\frac{R^2}{2}\log R+\frac{R^2}{4}<\mathrm{\infty }$. The polar integration use [F4] also assumes [A1].

The formulas are continuous away from zero, so they are integrable on every compact set avoiding the pole. Step 2.1 proves integrability in a neighborhood of the pole; hence ${\mathrm{\Phi }}_n\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. By [F10], changing its value at the single point zero does not change its almost-everywhere class, and [F6] defines the corresponding regular functional.

The zero-dimensional case is outside the stated $n\ge 2$ kernel and has no unit-sphere convention used here. The polar endpoint $r=0$ is included as an improper integral in step 2.1; away from zero the kernel is smooth. Countable Choice is used only through [F1], [F2], and [F4]; no full Axiom of Choice is used.