Proposition 3.1.13F8A69601BB8C

Proof. We begin with (3.1.15). Let $F\in C^2(a,b)$. The function $x\mapsto |x|$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, so $F(|x|)$ is of class $C^2$ on the annulus. For $a<|x|<b$ the chain rule gives $\partial _ jF(|x|)=F'(|x|)x_ j/|x|$ and

summing over $j$ gives the first expression in (3.1.15), and the product rule gives the second.

Write $E_ d(x)=F_ d(|x|)$ for $x\neq 0$. Differentiating the three cases of (3.1.14) and using $s_0=2$ and $s_1=2\pi $, we find

in every dimension $d\geq 1$. Hence $t^{d-1}F_ d'(t)$ is constant, and $E_ d$ is harmonic on $\mathbb R^ d\setminus \{ 0\} $ by (3.1.15). The functions $E_ d$ are continuous off the origin, and they are locally integrable. For $d\geq 2$ this follows from (3.1.2): $\int _{B(0,1)}|E_ d|\, \mathrm dx=s_{d-1}\int _0^1r^{d-1}|F_ d(r)|\, \mathrm dr$, which is a constant multiple of $\int _0^1r|\log r|\, \mathrm dr$ when $d=2$ and of $\int _0^1r\, \mathrm dr$ when $d\geq 3$, hence finite, while $E_ d$ is bounded on every compact set disjoint from $B(0,1)$. In dimension one, $E_1$ is continuous, and for every test function $\chi $ on $\mathbb R$ an integration by parts on each half-line gives

Thus $\langle E_1'',\chi \rangle =\int E_1\chi ''\, \mathrm dx=\chi (0)$, that is, $E_1''=\delta _0$.

Let $d\geq 2$, let $\chi $ be a test function, and choose $R$ with $\operatorname{Supp}\chi \subset B(0,R)$. The vector field $E_ d\nabla \chi -\chi \nabla E_ d$ is smooth on $\mathbb R^ d\setminus \{ 0\} $, which is an open neighborhood of every closed annulus $\{ \varepsilon \leq |x|\leq R\} $, and its divergence is

because $\Delta E_ d=0$ off the origin. The vector field vanishes near the sphere $|x|=R$, since $\operatorname{Supp}\chi $ is a compact subset of $B(0,R)$, and $\Delta \chi =0$ for $|x|\geq R$. Hence (3.1.4), with $c=0$ and $\nu =x/|x|$, gives

where $\partial _ r$ denotes the derivative in the direction of $x/|x|$. The first boundary term tends to zero as $\varepsilon \to 0$: its absolute value is at most $\sup |\nabla \chi |\, |F_ d(\varepsilon )|\, s_{d-1}\varepsilon ^{d-1}$, which is $\sup |\nabla \chi |$ times $\varepsilon |\log \varepsilon |$ if $d=2$, and times $\varepsilon /(d-2)$ if $d\geq 3$. On the sphere $|x|=\varepsilon $ we have $\partial _ rE_ d=F_ d'(\varepsilon )=(s_{d-1}\varepsilon ^{d-1})^{-1}$, so the second term equals $s_{d-1}^{-1}\int _{S^{d-1}}\chi (\varepsilon \omega )\, \mathrm dS(\omega )=M_\chi (0,\varepsilon )$, which tends to $\chi (0)$ because $\chi $ is continuous at $0$. On the left, $E_ d\Delta \chi $ is integrable, so dominated convergence gives the limit $\int E_ d\Delta \chi \, \mathrm dx$. Hence $\langle \Delta E_ d,\chi \rangle =\langle E_ d,\Delta \chi \rangle =\chi (0)$, that is, $\Delta E_ d=\delta _0$.

Finally, let $f\in C_ c^\infty (\mathbb R^ d)$ and write $(E_ d*f)(x)=\int E_ d(y)f(x-y)\, \mathrm dy$. Fix a compact set $K\subset \mathbb R^ d$, and let $L$ be the compact set of points $x'-z$ with $\operatorname {dist}(x',K)\leq 1$ and $z\in \operatorname{Supp}f$. If $g\in C_ c(\mathbb R^ d)$ is supported in $\operatorname{Supp}f$, then for $x\in K$ the function $y\mapsto E_ d(y)g(x-y)$ vanishes outside $L$ and is bounded by $\sup |g|\, |E_ d|\mathbf1_ L$, which is integrable; by dominated convergence, $E_ d*g$ is continuous on $K$. Moreover, for $x\in K$, a coordinate vector $e_ k$ and $0<|t|\leq 1$, the mean value theorem gives

so dominated convergence shows that $\partial _ k(E_ d*f)=E_ d*\partial _ kf$ on $K$. Since $K$ is arbitrary and $\partial _ kf$ is again a test function supported in $\operatorname{Supp}f$, induction on $|\alpha |$ shows that $E_ d*f$ is smooth and $\partial ^\alpha (E_ d*f)=E_ d*\partial ^\alpha f$. In particular, since $\Delta _ y[f(x-y)]=(\Delta f)(x-y)$,