The Dirac delta distribution and Green’s functions

Jordan Bell
February 7, 2015

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

If $s\in\mathbb{C}$ and $\Re s\geq 2$, then $u_{s}(x)=|x|^{s}$ is in $C^{2}_{\textrm{loc}}(\mathbb{R}^{n})$.11 1 This is all an expansion and gloss on Paul Garrett’s note Meromorphic continuations of distributions, which is on his homepage. $\Delta:C^{2}_{\textrm{loc}}(\mathbb{R}^{n})\to C^{0}_{\textrm{loc}}(\mathbb{R}% ^{n})$ and, $\Delta=\sum_{i=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}$ and $|x|^{2}=x_{1}^{2}+\cdots+x_{n}^{2}$,

 $\displaystyle(\Delta u_{s})(x)$ $\displaystyle=$ $\displaystyle\Delta|x|^{s}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\frac{\partial}{\partial x_{i}}\frac{s}{2}\cdot 2x_% {i}\cdot(|x|^{2})^{\frac{s}{2}-1}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{n}\left(\frac{s}{2}\cdot 2\cdot(|x|^{2})^{\frac{s}{2}% -1}+\frac{s}{2}\left(\frac{s}{2}-1\right)\cdot(2x_{i})^{2}\cdot(|x|^{2})^{% \frac{s}{2}-2}\right)$ $\displaystyle=$ $\displaystyle ns\cdot|x|^{s-2}+s(s-2)|x|^{s-2}$ $\displaystyle=$ $\displaystyle s(s+n-2)\cdot|x|^{s-2}$ $\displaystyle=$ $\displaystyle s(s+n-2)\cdot u_{s-2}.$

We take $n>2$ in the following.

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

For $\Re s\geq 2$ we have

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

and hence for $\Re s\geq 0$ we have

 $u_{s}=\frac{\Delta u_{s+2}}{(s+2)(s+n)}.$

As $u_{0}$ is a constant, $\Delta u_{0}=0$, 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 $\mathbb{C}$, with at most simple poles at $s=-n,-n-2,-n-4,\ldots$.

Let $\gamma=e^{-|x|^{2}}$ and let $f$ be a Schwartz function on $\mathbb{R}^{n}$. For $\Re s>-n-1$, we have $u_{s}\cdot(f-f(0)\gamma)\in L^{1}(\mathbb{R}^{n})$ (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

 $u_{s}(f-f(0)\gamma)=\int_{\mathbb{R}^{n}}|x|^{s}\cdot(f(x)-f(0)\gamma(x))dx.$

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

 $\textrm{Res}_{s=-n}u_{s}(f-f(0)\gamma)=0.$

Thus

 $\displaystyle\textrm{Res}_{s=-n}u_{s}(f)$ $\displaystyle=$ $\displaystyle\textrm{Res}_{s=-n}u_{s}(f-f(0)\gamma)+\textrm{Res}_{s=-n}u_{s}(f% (0)\gamma)$ $\displaystyle=$ $\displaystyle\textrm{Res}_{s=-n}u_{s}(f(0)\gamma)$ $\displaystyle=$ $\displaystyle f(0)\textrm{Res}_{s=-n}u_{s}(\gamma)$ $\displaystyle=$ $\displaystyle\delta(f)\textrm{Res}_{s=-n}u_{s}(\gamma).$

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

 $\displaystyle\textrm{Res}_{s=-n}u_{s}(\gamma)$ $\displaystyle=$ $\displaystyle\textrm{Res}_{s=-n}\int_{\mathbb{R}^{n}}|x|^{s}e^{-|x|^{2}}dx$ $\displaystyle=$ $\displaystyle\textrm{Res}_{s=-n}\int_{0}^{\infty}\int_{S^{n-1}}|rx^{\prime}|^{% s}e^{-|rx^{\prime}|^{2}}r^{n-1}d\sigma(x^{\prime})dr$ $\displaystyle=$ $\displaystyle\sigma(S^{n-1})\textrm{Res}_{s=-n}\int_{0}^{\infty}r^{s+n-1}e^{-r% ^{2}}dr$ $\displaystyle=$ $\displaystyle\frac{\sigma(S^{n-1})}{2}\textrm{Res}_{s=-n}\int_{0}^{\infty}t^{% \frac{s+n-2}{2}}e^{-t}dt$ $\displaystyle=$ $\displaystyle\frac{\sigma(S^{n-1})}{2}\textrm{Res}_{s=-n}\Gamma(\frac{s+n}{2}).$

As $\Gamma(z+1)=z\Gamma(z)$,

 $\displaystyle\textrm{Res}_{s=-n}u_{s}(\gamma)$ $\displaystyle=$ $\displaystyle\frac{\sigma(S^{n-1})}{2}\textrm{Res}_{s=-n}\frac{2}{s+n}$ $\displaystyle=$ $\displaystyle\frac{\sigma(S^{n-1})}{2}\cdot 2$ $\displaystyle=$ $\displaystyle\sigma(S^{n-1}).$

Therefore for any Schwartz function $f$, we have

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

hence

 $\textrm{Res}_{s=-n}u_{s}=\sigma(S^{n-1})\cdot\delta.$

We know that $u_{s}$ has poles at most at $s=-n,-n-2,-n-4,\ldots$, 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 $\textrm{Res}_{s=-n}u_{s}$. But

 $u_{s}=\frac{\Delta u_{s+2}}{(s+2)(s+n)},$

so

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

i.e.

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

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

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

If $\Re s>2$, then $w_{s}(x)=|x|^{s}\cdot\log|x|\in C^{2}_{\textrm{loc}}(\mathbb{R}^{2})$. Let $u_{s}(x)=|x|^{s}$. We have

 $\displaystyle(\Delta w_{s})(x)$ $\displaystyle=$ $\displaystyle\frac{1}{2}\sum_{i=1}^{2}\frac{\partial^{2}}{\partial x_{i}^{2}}% \Big{(}(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}}\log(x_{1}^{2}+x_{2}^{2})\Big{)}$ $\displaystyle=$ $\displaystyle\frac{1}{2}\sum_{i=1}^{2}\frac{\partial}{\partial x_{i}}\Big{(}% \frac{s}{2}(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}\cdot 2x_{i}\cdot\log(x_{1}^{2% }+x_{2}^{2})$ $\displaystyle+(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}}\cdot\frac{2x_{i}}{x_{1}^{2}+% x_{2}^{2}}\Big{)}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{2}\frac{\partial}{\partial x_{i}}\Big{(}\frac{s}{2}(x% _{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}\cdot x_{i}\cdot\log(x_{1}^{2}+x_{2}^{2})+(% x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}\cdot x_{i}\Big{)}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{2}\frac{s}{2}\left(\frac{s}{2}-1\right)(x_{1}^{2}+x_{% 2}^{2})^{\frac{s}{2}-2}\cdot 2x_{i}^{2}\cdot\log(x_{1}^{2}+x_{2}^{2})$ $\displaystyle+\frac{s}{2}(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}\cdot\log(x_{1}^% {2}+x_{2}^{2})+\frac{s}{2}(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}\frac{2x_{i}^{2% }}{x_{1}^{2}+x_{2}^{2}}$ $\displaystyle+\left(\frac{s}{2}-1\right)(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-2}% \cdot 2x_{i}^{2}+(x_{1}^{2}+x_{2}^{2})^{\frac{s}{2}-1}$ $\displaystyle=$ $\displaystyle\sum_{i=1}^{2}s(s-2)|x|^{s-4}\cdot x_{i}^{2}\cdot\log|x|+s|x|^{s-% 2}\cdot\log|x|+s|x|^{s-4}\cdot x_{i}^{2}$ $\displaystyle+(s-2)|x|^{s-4}\cdot x_{i}^{2}+|x|^{s-2}$ $\displaystyle=$ $\displaystyle s(s-2)|x|^{s-2}\log|x|+2s|x|^{s-2}\log|x|+s|x|^{s-2}+(s-2)|x|^{s% -2}+2|x|^{s-2}$ $\displaystyle=$ $\displaystyle s^{2}w_{s-2}(x)+2su_{s-2}(x).$

Hence

 $\Delta w_{s}=s^{2}w_{s-2}+2su_{s-2},$

and so

 $(s+2)^{2}w_{s}=-2(s+2)u_{s}+\Delta w_{s+2}.$

We calculate

 $\displaystyle\int_{\mathbb{R}^{2}}|x|^{s}\log|x|e^{-|x|^{2}}dx$ $\displaystyle=$ $\displaystyle\int_{0}^{\infty}\int_{S^{1}}|rx^{\prime}|^{s}\log|rx^{\prime}|e^% {-|rx^{\prime}|^{2}}rd\sigma(x^{\prime})dr$ $\displaystyle=$ $\displaystyle 2\pi\int_{0}^{\infty}r^{s+1}\cdot\log r\cdot e^{-r^{2}}dr$ $\displaystyle=$ $\displaystyle 2\pi\cdot\frac{1}{4}\Gamma\left(1+\frac{s}{2}\right)\psi\left(1+% \frac{s}{2}\right),$

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

 $\displaystyle\textrm{Res}_{s=-2}(s+2)w_{s}(\gamma)$ $\displaystyle=$ $\displaystyle\frac{\pi}{2}\cdot\textrm{Res}_{s=-2}(s+2)\frac{1}{1+\frac{s}{2}}% \left(\psi\left(1+\frac{s}{2}+1\right)-\frac{1}{1+\frac{s}{2}}\right)$ $\displaystyle=$ $\displaystyle\pi\cdot\textrm{Res}_{s=-2}\left(-C-\frac{2}{s+2}\right)$ $\displaystyle=$ $\displaystyle-2\pi,$

where $C$ is Euler’s constant; it is a fact that $\psi(1)=-C$.22 2 Historical note: In the papers of Euler’s that I’ve seen where he mentions the Euler constant, the notation he uses is either $C$ or $O$, not once the modern $\gamma$. Thus like in the previous section, if $f$ is a Schwartz function then

 $\textrm{Res}_{s=-2}(s+2)w_{s}(f)=\delta(f)\textrm{Res}_{s=-2}(s+2)w_{s}(\gamma% )=-2\pi\cdot\delta(f).$

Because

 $(s+2)^{2}w_{s}=-2(s+2)u_{s}+\Delta w_{s+2},$

the value of $(s+2)^{2}w_{s}$ at $s=-2$ is $\Delta w_{0}-2\cdot\textrm{Res}_{s=-2}u_{s}$. On the other hand, the value of $(s+2)^{2}w_{s}$ at $s=-2$ is $\textrm{Res}_{s=-2}(s+2)w_{s}=-2\pi\cdot\delta$, hence

 $\Delta w_{0}=-2\pi\cdot\delta+2\cdot\textrm{Res}_{s=-2}u_{s}.$

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

 $\displaystyle\textrm{Res}_{s=-2}u_{s}(f)$ $\displaystyle=$ $\displaystyle\delta(f)\textrm{Res}_{s=-2}u_{s}(\gamma)$ $\displaystyle=$ $\displaystyle\delta(f)\textrm{Res}_{s=-2}\frac{1}{2}\Gamma\left(1+\frac{s}{2}\right)$ $\displaystyle=$ $\displaystyle 2\pi\cdot\delta(f).$

Therefore

 $\Delta w_{0}=2\pi\cdot\delta,$

i.e.,

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

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

3 Dirac comb

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

 $u_{s}=\sum_{\stackrel{{\scriptstyle 0<\frac{p}{q}\leq 1}}{{\gcd(p,q)=1}}}\frac% {1}{q^{s}}\cdot\delta_{p/q}.$

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

 $|u_{s}(f)|\leq K\sum_{\stackrel{{\scriptstyle 0<\frac{p}{q}\leq 1}}{{\gcd(p,q)% =1}}}\frac{1}{q^{\Re s}}

Since $\Re s>2$, 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)

 $\displaystyle\zeta(s)\cdot u_{s}$ $\displaystyle=$ $\displaystyle\sum_{d\geq 1}\frac{1}{d^{s}}\sum_{q\geq 1}\frac{1}{q^{s}}\sum_{% \stackrel{{\scriptstyle 0 $\displaystyle=$ $\displaystyle\sum_{d\geq 1}\sum_{q\geq 1}\frac{1}{(qd)^{s}}\sum_{\stackrel{{% \scriptstyle 0 $\displaystyle=$ $\displaystyle\sum_{q=1}^{\infty}\frac{1}{q^{s}}\sum_{\stackrel{{\scriptstyle d% |q}}{{d\geq 1}}}\sum_{\stackrel{{\scriptstyle 0 $\displaystyle=$ $\displaystyle\sum_{q=1}^{\infty}\frac{1}{q^{s}}\sum_{0 $\displaystyle=$ $\displaystyle v_{s}.$

(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 $\Re s>2$ and for $n\in\mathbb{Z}$ (recalling that $v_{s}$ is a distribution, i.e. it integrates functions)

 $\displaystyle\widehat{v}_{s}(n)$ $\displaystyle=$ $\displaystyle v_{s}(e^{-2\pi inx})$ $\displaystyle=$ $\displaystyle\sum_{q=1}^{\infty}\frac{1}{q^{s}}\sum_{0 $\displaystyle=$ $\displaystyle\sum_{q=1}^{\infty}\frac{1}{q^{s}}\sum_{0

$p\mapsto e^{-2\pi inp/q}$, $\mathbb{Z}/q\to\mathbb{C}$, is a character, and, unless it is the trivial character, the sum over $\mathbb{Z}/q$ is equal to $0$. So if $q\not|n$ 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

 $\widehat{v}_{s}(n)=\sum_{\stackrel{{\scriptstyle q|n}}{{q\geq 1}}}\frac{1}{q^{% s-1}}.$

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 $n\neq 0$ we have

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

Using $\zeta(s)\cdot u_{s}=v_{s}$ we get

 $\widehat{u}_{s}(n)=\begin{cases}\frac{\zeta(s-1)}{\zeta(s)}&n=0,\\ \frac{\sigma_{1-s}(n)}{\zeta(s)}&n\neq 0.\end{cases}$

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 $\mathbb{T}$ (see Katznelson, p. 48, Chapter 1, Exercise 7.5), and for $\Re s\leq 2$ we shall define $u_{s}$ to be this distribution. In summary: $u_{s}$ is originally defined as a distribution for $\Re s>2$, and now we have defined it to be a distribution for $s\neq 2$ and $\zeta(s)\neq 0$. Thus $u_{s}$ is a meromorphic distribution valued functions on $\mathbb{C}$ with poles at $s=2$ and at the zeros of the Riemann zeta function.

Since $\zeta(1)=\infty$, if $n\neq 0$ 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 $\mathbb{T}$,

 $u_{1}=0$

(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 $\Re s>2$, and we have analytically continued $u_{s}$ as a meromorphic distribution valued function on $\mathbb{C}$. Likewise, although $\zeta(0)=-\frac{1}{2}$, it is incorrect to conclude that $1+1+1+\cdots=-\frac{1}{2}$, although for certain formal arguments this may be a correct interpretation.).