The Dirac delta distribution and Green’s functions

1 $u_s(x)={|x|}^s$

If $s\in ℂ$ and $\mathrm{\Re }s\ge 2$, then $u_s(x)={|x|}^s$ is in $C_{\text{loc}}^2({ℝ}^n)$. $\mathrm{\Delta }:C_{\text{loc}}^2({ℝ}^n)\to C_{\text{loc}}^0({ℝ}^n)$ and, $\mathrm{\Delta }={\sum }_{i=1}^n\frac{{\partial }^2}{\partial x_i^2}$ and ${|x|}^2=x_1^2+\mathrm{\cdots }+x_n^2$,

$(\mathrm{\Delta }{u}_s)(x)$ $=$ $\mathrm{\Delta }{|x|}^s$

$=$ ${\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\partial }{\partial x_i}}{\displaystyle \frac{s}{2}}\cdot 2x_i\cdot {({|x|}^2)}^{\frac{s}{2}-1}$

$=$ ${\displaystyle \sum _{i=1}^n}\left({\displaystyle \frac{s}{2}}\cdot 2\cdot {({|x|}^2)}^{\frac{s}{2}-1}+{\displaystyle \frac{s}{2}}\left({\displaystyle \frac{s}{2}}-1\right)\cdot {(2x_i)}^2\cdot {({|x|}^2)}^{\frac{s}{2}-2}\right)$

$=$ $ns\cdot {|x|}^{s-2}+s(s-2){|x|}^{s-2}$

$=$ $s(s+n-2)\cdot {|x|}^{s-2}$

$=$ $s(s+n-2)\cdot u_{s-2}.$

We take $n>2$ in the following.

Typically we talk about functions $ℂ\to ℂ$ that are holomorphic (or meromorphic if they are defined on a subset of $ℂ$). But we can also talk about functions $ℂ\to V$ that are holomorphic/meromorphic for certain types of topological vector spaces over $ℂ$. In particular, we can talk about holomorphic/meromorphic functions that take values in the tempered distributions on ${ℝ}^n$. If $\mathrm{\Re }s>-n$, then $u_s$ is locally integrable (for any point in ${ℝ}^n$, there is a neighborhood of the point on which $u_s$ is $L^1$), and hence it is a tempered distribution for $\mathrm{\Re }s>-n$. Thus for $\mathrm{\Re }s>-n$, $\mathrm{\Delta }{u}_s$ is a tempered distribution.

For $\mathrm{\Re }s\ge 2$ we have

$$u_{s-2}=\frac{\mathrm{\Delta }{u}_s}{s(s+n-2)},$$

and hence for $\mathrm{\Re }s\ge 0$ we have

$$u_s=\frac{\mathrm{\Delta }{u}_{s+2}}{(s+2)(s+n)}.$$

As $u_0$ is a constant, $\mathrm{\Delta }{u}_0=0$, and so $s-2$ is a removable singularity of the right-hand side. It follows that $u_s$ is meromorphic and that its only possible pole is at $s=-n$. One iterates this argument and obtains that $u_s$ is meromorphic on $ℂ$, with at most simple poles at $s=-n,-n-2,-n-4,\mathrm{\dots }$.

Let $\gamma =e^{-{|x|}^2}$ and let $f$ be a Schwartz function on ${ℝ}^n$. For $\mathrm{\Re }s>-n-1$, we have $u_s\cdot (f-f(0)\gamma )\in L^1({ℝ}^n)$ (the term $f-f(0)\gamma$ is certainly integrable at infinity and will still be integrable at infinity after being multiplied by ${|x|}^s$, and while ${|x|}^s$ might not be integrable at $0$, the term $f-f(0)\gamma$ goes to $0$ like ${|x|}^2$). The tempered distribution $u_s$ maps the Schwartz function $f-f(0)\gamma$ to

$$u_s(f-f(0)\gamma )={\int }_{{ℝ}^n}{|x|}^s\cdot (f(x)-f(0)\gamma (x))𝑑x.$$

In the above equation (for fixed $f$), the right-hand side is holomorphic for $\mathrm{\Re }s>-n-1$, thus so is the left. Hence the residue of the left side at $s=-n$ is $0$:

$${\text{Res}}_{s=-n}{u}_s(f-f(0)\gamma )=0.$$

Thus

${\text{Res}}_{s=-n}{u}_s(f)$ $=$ ${\text{Res}}_{s=-n}{u}_s(f-f(0)\gamma )+{\text{Res}}_{s=-n}{u}_s(f(0)\gamma )$

$=$ ${\text{Res}}_{s=-n}{u}_s(f(0)\gamma )$

$=$ $f(0){\text{Res}}_{s=-n}{u}_s(\gamma )$

$=$ $\delta (f){\text{Res}}_{s=-n}{u}_s(\gamma ).$

Using polar coordinates, with $\sigma (S^{n-1})=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }(n/2)}$,

${\text{Res}}_{s=-n}{u}_s(\gamma )$ $=$ ${\text{Res}}_{s=-n}{\displaystyle {\int }_{{ℝ}^n}}{|x|}^s{e}^{-{|x|}^2}𝑑x$

$=$ ${\text{Res}}_{s=-n}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{S^{n-1}}}{|r{x}^{\prime }|}^s{e}^{-{|r{x}^{\prime }|}^2}{r}^{n-1}𝑑\sigma (x^{\prime })𝑑r$

$=$ $\sigma (S^{n-1}){\text{Res}}_{s=-n}{\displaystyle {\int }_0^{\mathrm{\infty }}}{r}^{s+n-1}{e}^{-r^2}𝑑r$

$=$ ${\displaystyle \frac{\sigma (S^{n-1})}{2}}{\text{Res}}_{s=-n}{\displaystyle {\int }_0^{\mathrm{\infty }}}{t}^{\frac{s+n-2}{2}}{e}^{-t}𝑑t$

$=$ ${\displaystyle \frac{\sigma (S^{n-1})}{2}}{\text{Res}}_{s=-n}\mathrm{\Gamma }({\displaystyle \frac{s+n}{2}}).$

As $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$,

${\text{Res}}_{s=-n}{u}_s(\gamma )$ $=$ ${\displaystyle \frac{\sigma (S^{n-1})}{2}}{\text{Res}}_{s=-n}{\displaystyle \frac{2}{s+n}}$

$=$ ${\displaystyle \frac{\sigma (S^{n-1})}{2}}\cdot 2$

$=$ $\sigma (S^{n-1}).$

Therefore for any Schwartz function $f$, we have

$${\text{Res}}_{s=-n}{u}_s(f)=\sigma (S^{n-1})\cdot \delta (f),$$

hence

$${\text{Res}}_{s=-n}{u}_s=\sigma (S^{n-1})\cdot \delta .$$

We know that $u_s$ has poles at most at $s=-n,-n-2,-n-4,\mathrm{\dots }$, and we have just explicitly found its residue at $s=-n$.

This fact has an important consequence. As $u_s$ has a simple pole at $s=-n$, the value of $(s+n)u_s$ at $s=-n$ is ${\text{Res}}_{s=-n}{u}_s$. But

$$u_s=\frac{\mathrm{\Delta }{u}_{s+2}}{(s+2)(s+n)},$$

so

$$\mathrm{\Delta }{u}_{-n+2}=(-n+2)\cdot \sigma (S^{n-1})\cdot \delta ,$$

i.e.

$$\mathrm{\Delta }\frac{1}{{|x|}^{n-2}}=(-n+2)\cdot \sigma (S^{n-1})\cdot \delta ,$$

with $\sigma (S^{n-1})=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }(n/2)}$. Recall that we have assumed $n>2$. In other words, we have just determined the Green’s function of the Laplace operator on ${ℝ}^n$, $n>2$.

2 $w_s(x)={|x|}^s\cdot \log |x|$

If $\mathrm{\Re }s>2$, then $w_s(x)={|x|}^s\cdot \log |x|\in C_{\text{loc}}^2({ℝ}^2)$. Let $u_s(x)={|x|}^s$. We have

$(\mathrm{\Delta }{w}_s)(x)$ $=$ ${\displaystyle \frac{1}{2}}{\displaystyle \sum _{i=1}^2}{\displaystyle \frac{{\partial }^2}{\partial x_i^2}}\left({(x_1^2+x_2^2)}^{\frac{s}{2}}\log (x_1^2+x_2^2)\right)$

$=$ ${\displaystyle \frac{1}{2}}{\displaystyle \sum _{i=1}^2}{\displaystyle \frac{\partial }{\partial x_i}}({\displaystyle \frac{s}{2}}{(x_1^2+x_2^2)}^{\frac{s}{2}-1}\cdot 2x_i\cdot \log (x_1^2+x_2^2)$

$+{(x_1^2+x_2^2)}^{\frac{s}{2}}\cdot {\displaystyle \frac{2x_i}{x_1^2+x_2^2}})$

$=$ ${\displaystyle \sum _{i=1}^2}{\displaystyle \frac{\partial }{\partial x_i}}\left({\displaystyle \frac{s}{2}}{(x_1^2+x_2^2)}^{\frac{s}{2}-1}\cdot x_i\cdot \log (x_1^2+x_2^2)+{(x_1^2+x_2^2)}^{\frac{s}{2}-1}\cdot x_i\right)$

$=$ ${\displaystyle \sum _{i=1}^2}{\displaystyle \frac{s}{2}}\left({\displaystyle \frac{s}{2}}-1\right){(x_1^2+x_2^2)}^{\frac{s}{2}-2}\cdot 2x_i^2\cdot \log (x_1^2+x_2^2)$

$+{\displaystyle \frac{s}{2}}{(x_1^2+x_2^2)}^{\frac{s}{2}-1}\cdot \log (x_1^2+x_2^2)+{\displaystyle \frac{s}{2}}{(x_1^2+x_2^2)}^{\frac{s}{2}-1}{\displaystyle \frac{2x_i^2}{x_1^2+x_2^2}}$

$+\left({\displaystyle \frac{s}{2}}-1\right){(x_1^2+x_2^2)}^{\frac{s}{2}-2}\cdot 2x_i^2+{(x_1^2+x_2^2)}^{\frac{s}{2}-1}$

$=$ ${\displaystyle \sum _{i=1}^2}s(s-2){|x|}^{s-4}\cdot x_i^2\cdot \log |x|+s{|x|}^{s-2}\cdot \log |x|+s{|x|}^{s-4}\cdot x_i^2$

$+(s-2){|x|}^{s-4}\cdot x_i^2+{|x|}^{s-2}$

$=$ $s(s-2){|x|}^{s-2}\log |x|+2s{|x|}^{s-2}\log |x|+s{|x|}^{s-2}+(s-2){|x|}^{s-2}+2{|x|}^{s-2}$

$=$ $s^2{w}_{s-2}(x)+2s{u}_{s-2}(x).$

Hence

$$\mathrm{\Delta }{w}_s=s^2{w}_{s-2}+2s{u}_{s-2},$$

and so

$${(s+2)}^2{w}_s=-2(s+2)u_s+\mathrm{\Delta }{w}_{s+2}.$$

We calculate

${\displaystyle {\int }_{{ℝ}^2}}{|x|}^s\log |x|e^{-{|x|}^2}dx$ $=$ ${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{S^1}}{|r{x}^{\prime }|}^s\log |r{x}^{\prime }|e^{-{|r{x}^{\prime }|}^2}rd\sigma (x^{\prime })𝑑r$

$=$ $2\pi {\displaystyle {\int }_0^{\mathrm{\infty }}}{r}^{s+1}\cdot \log r\cdot e^{-r^2}dr$

$=$ $2\pi \cdot {\displaystyle \frac{1}{4}}\mathrm{\Gamma }\left(1+{\displaystyle \frac{s}{2}}\right)\psi \left(1+{\displaystyle \frac{s}{2}}\right),$

where $\psi (z)=\frac{{\mathrm{\Gamma }}^{\prime }(z)}{\mathrm{\Gamma }(z)}$, namely the digamma function. Using $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$ and $\psi (z+1)=\psi (z)+\frac{1}{z}$, with $\gamma (x)=e^{-x^2}$,

${\text{Res}}_{s=-2}(s+2)w_s(\gamma )$ $=$ ${\displaystyle \frac{\pi }{2}}\cdot {\text{Res}}_{s=-2}(s+2){\displaystyle \frac{1}{1+\frac{s}{2}}}\left(\psi \left(1+{\displaystyle \frac{s}{2}}+1\right)-{\displaystyle \frac{1}{1+\frac{s}{2}}}\right)$

$=$ $\pi \cdot {\text{Res}}_{s=-2}\left(-C-{\displaystyle \frac{2}{s+2}}\right)$

$=$ $-2\pi ,$

where $C$ is Euler’s constant; it is a fact that $\psi (1)=-C$. Thus like in the previous section, if $f$ is a Schwartz function then

$${\text{Res}}_{s=-2}(s+2)w_s(f)=\delta (f){\text{Res}}_{s=-2}(s+2)w_s(\gamma )=-2\pi \cdot \delta (f).$$

Because

$${(s+2)}^2{w}_s=-2(s+2)u_s+\mathrm{\Delta }{w}_{s+2},$$

the value of ${(s+2)}^2{w}_s$ at $s=-2$ is $\mathrm{\Delta }{w}_0-2\cdot {\text{Res}}_{s=-2}{u}_s$. On the other hand, the value of ${(s+2)}^2{w}_s$ at $s=-2$ is ${\text{Res}}_{s=-2}(s+2)w_s=-2\pi \cdot \delta$, hence

$$\mathrm{\Delta }{w}_0=-2\pi \cdot \delta +2\cdot {\text{Res}}_{s=-2}{u}_s.$$

We can calculate ${\text{Res}}_{s=-2}{u}_s$ just like in the previous section. If $f$ is a Schwartz function and $\gamma =e^{-{|x|}^2}$, then

${\text{Res}}_{s=-2}{u}_s(f)$ $=$ $\delta (f){\text{Res}}_{s=-2}{u}_s(\gamma )$

$=$ $\delta (f){\text{Res}}_{s=-2}{\displaystyle \frac{1}{2}}\mathrm{\Gamma }\left(1+{\displaystyle \frac{s}{2}}\right)$

$=$ $2\pi \cdot \delta (f).$

Therefore

$$\mathrm{\Delta }{w}_0=2\pi \cdot \delta ,$$

i.e.,

$$\mathrm{\Delta }\log |x|=2\pi \cdot \delta .$$

Recall that here $n=2$. In other words, we have just determined the Green’s function of the Laplace operator on ${ℝ}^2$.

3 Dirac comb

Let $𝕋=ℝ/\mathbb{Z}$. On ${ℝ}^n$, tempered distributions integrate against a larger class of functions than do distributions, so it’s stronger to be a tempered distribution. But on $𝕋$, any Schwartz function has compact support, and moreover, any $C^{\mathrm{\infty }}$ function on $𝕋$ is a Schwartz function. Thus distributions on $𝕋$ integrate smooth functions on $𝕋$. For $\mathrm{\Re }s>2$, define the following distribution on $𝕋$:

Why is this in fact a distribution? If $f\in C^{\mathrm{\infty }}(𝕋)$ then $f$ is certainly bounded (indeed, $u_s$ can take any continuous function on $𝕋$ as an argument, not just smooth functions). Let $|f(t)|\le K$ for all $t\in 𝕋$. Then,

Since $\mathrm{\Re }s>2$, this series converges.

Doing some series manipulations we get (probably the hardest step to see is that summing over the products of $d$ and $q$ is the same as summing over $q$ and then over those $d$ that divide it)

$\zeta (s)\cdot u_s$ $=$

$=$

$=$

$=$

$=$ $v_s.$

(The last equality is a definition.) To summarize: $\zeta (s)\cdot u_s=v_s$.

Supposing we are interested in $u_s$, using the above formula we can instead investigate $v_s$, which for some purposes is more analytically tractable. We shall determine the Fourier series of $v_s$. For $\mathrm{\Re }s>2$ and for $n\in \mathbb{Z}$ (recalling that $v_s$ is a distribution, i.e. it integrates functions)

${\widehat{v}}_s(n)$ $=$ $v_s(e^{-2\pi inx})$

$=$

$=$

$p\mapsto e^{-2\pi inp/q}$, $\mathbb{Z}/q\to ℂ$, is a character, and, unless it is the trivial character, the sum over $\mathbb{Z}/q$ is equal to $0$. So if $q|̸n$ then the inner sum is 0, and if $q|n$ then the inner sum is equal to $q$. (If the language of characters of $\mathbb{Z}/q$ isn’t familiar, you can check this fact directly; to show the inner sum is 0, you show that the inner sum is equal to itself times something that is nonzero.) Thus

$${\widehat{v}}_s(n)=\sum _{\stackrel{q|n}{q\ge 1}}\frac{1}{q^{s-1}}.$$

For $n=0$, we get

$${\widehat{v}}_s(0)=\zeta (s-1).$$

Otherwise, the above can be written using a standard arithmetic function, the sum of powers of positive divisors. Let ${\sigma }_{\alpha }(n)$ denote the sum of the $\alpha$th powers of the positive divisors of $n$. Thus for $n\ne 0$ we have

$${\widehat{v}}_s(n)={\sigma }_{1-s}(n).$$

Using $\zeta (s)\cdot u_s=v_s$ we get

The expression on the right-hand side has poles at $s=2$ and at the zeros of the Riemann zeta function. Otherwise, for a fixed $s$, the right-hand side has at most polynomial growth in $n$, and therefore it is the Fourier series of a distribution on $𝕋$ (see Katznelson, p. 48, Chapter 1, Exercise 7.5), and for $\mathrm{\Re }s\le 2$ we shall define $u_s$ to be this distribution. In summary: $u_s$ is originally defined as a distribution for $\mathrm{\Re }s>2$, and now we have defined it to be a distribution for $s\ne 2$ and $\zeta (s)\ne 0$. Thus $u_s$ is a meromorphic distribution valued functions on $ℂ$ with poles at $s=2$ and at the zeros of the Riemann zeta function.

Since $\zeta (1)=\mathrm{\infty }$, if $n\ne 0$ then ${\widehat{u}}_1(n)=0$. The only pole of the Riemann zeta function is at $s=1$, hence $\zeta (0)/\zeta (1)=0$. Thus ${\widehat{u}}_1(0)=0$, and it follows that as a distribution on $𝕋$,

$$u_1=0$$

(although the distribution $u_1$ is $0$, this doesn’t mean that we can put $s=1$ into the original definition of $u_s$ and assert that this is $0$, as the original definition of $u_s$ was only for $\mathrm{\Re }s>2$, and we have analytically continued $u_s$ as a meromorphic distribution valued function on $ℂ$. Likewise, although $\zeta (0)=-\frac{1}{2}$, it is incorrect to conclude that $1+1+1+\mathrm{\cdots }=-\frac{1}{2}$, although for certain formal arguments this may be a correct interpretation.).