The Dirac delta distribution and Green’s functions

Jordan Bell
February 7, 2015

1 us(x)=|x|s

If s and s2, then us(x)=|x|s is in Cloc2(n).11 1 This is all an expansion and gloss on Paul Garrett’s note Meromorphic continuations of distributions, which is on his homepage. Δ:Cloc2(n)Cloc0(n) and, Δ=i=1n2xi2 and |x|2=x12++xn2,

(Δus)(x) = Δ|x|s
= i=1nxis22xi(|x|2)s2-1
= i=1n(s22(|x|2)s2-1+s2(s2-1)(2xi)2(|x|2)s2-2)
= ns|x|s-2+s(s-2)|x|s-2
= s(s+n-2)|x|s-2
= s(s+n-2)us-2.

We take n>2 in the following.

Typically we talk about functions that are holomorphic (or meromorphic if they are defined on a subset of ). But we can also talk about functions V that are holomorphic/meromorphic for certain types of topological vector spaces over . In particular, we can talk about holomorphic/meromorphic functions that take values in the tempered distributions on n. If s>-n, then us is locally integrable (for any point in n, there is a neighborhood of the point on which us is L1), and hence it is a tempered distribution for s>-n. Thus for s>-n, Δus is a tempered distribution.

For s2 we have


and hence for s0 we have


As u0 is a constant, Δu0=0, and so s-2 is a removable singularity of the right-hand side. It follows that us is meromorphic and that its only possible pole is at s=-n. One iterates this argument and obtains that us is meromorphic on , with at most simple poles at s=-n,-n-2,-n-4,.

Let γ=e-|x|2 and let f be a Schwartz function on n. For s>-n-1, we have us(f-f(0)γ)L1(n) (the term f-f(0)γ 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)γ goes to 0 like |x|2). The tempered distribution us maps the Schwartz function f-f(0)γ to


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



Ress=-nus(f) = Ress=-nus(f-f(0)γ)+Ress=-nus(f(0)γ)
= Ress=-nus(f(0)γ)
= f(0)Ress=-nus(γ)
= δ(f)Ress=-nus(γ).

Using polar coordinates, with σ(Sn-1)=2πn/2Γ(n/2),

Ress=-nus(γ) = Ress=-nn|x|se-|x|2𝑑x
= Ress=-n0Sn-1|rx|se-|rx|2rn-1𝑑σ(x)𝑑r
= σ(Sn-1)Ress=-n0rs+n-1e-r2𝑑r
= σ(Sn-1)2Ress=-n0ts+n-22e-t𝑑t
= σ(Sn-1)2Ress=-nΓ(s+n2).

As Γ(z+1)=zΓ(z),

Ress=-nus(γ) = σ(Sn-1)2Ress=-n2s+n
= σ(Sn-1)22
= σ(Sn-1).

Therefore for any Schwartz function f, we have




We know that us has poles at most at s=-n,-n-2,-n-4,, and we have just explicitly found its residue at s=-n.

This fact has an important consequence. As us has a simple pole at s=-n, the value of (s+n)us at s=-n is Ress=-nus. But






with σ(Sn-1)=2πn/2Γ(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 n, n>2.

2 ws(x)=|x|slog|x|

If s>2, then ws(x)=|x|slog|x|Cloc2(2). Let us(x)=|x|s. We have

(Δws)(x) = 12i=122xi2((x12+x22)s2log(x12+x22))
= 12i=12xi(s2(x12+x22)s2-12xilog(x12+x22)
= i=12xi(s2(x12+x22)s2-1xilog(x12+x22)+(x12+x22)s2-1xi)
= i=12s2(s2-1)(x12+x22)s2-22xi2log(x12+x22)
= i=12s(s-2)|x|s-4xi2log|x|+s|x|s-2log|x|+s|x|s-4xi2
= s(s-2)|x|s-2log|x|+2s|x|s-2log|x|+s|x|s-2+(s-2)|x|s-2+2|x|s-2
= s2ws-2(x)+2sus-2(x).



and so


We calculate

2|x|slog|x|e-|x|2dx = 0S1|rx|slog|rx|e-|rx|2rdσ(x)𝑑r
= 2π0rs+1logre-r2dr
= 2π14Γ(1+s2)ψ(1+s2),

where ψ(z)=Γ(z)Γ(z), namely the digamma function. Using Γ(z+1)=zΓ(z) and ψ(z+1)=ψ(z)+1z, with γ(x)=e-x2,

Ress=-2(s+2)ws(γ) = π2Ress=-2(s+2)11+s2(ψ(1+s2+1)-11+s2)
= πRess=-2(-C-2s+2)
= -2π,

where C is Euler’s constant; it is a fact that ψ(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 γ. Thus like in the previous section, if f is a Schwartz function then




the value of (s+2)2ws at s=-2 is Δw0-2Ress=-2us. On the other hand, the value of (s+2)2ws at s=-2 is Ress=-2(s+2)ws=-2πδ, hence


We can calculate Ress=-2us just like in the previous section. If f is a Schwartz function and γ=e-|x|2, then

Ress=-2us(f) = δ(f)Ress=-2us(γ)
= δ(f)Ress=-212Γ(1+s2)
= 2πδ(f).





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

3 Dirac comb

Let 𝕋=/. On 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 function on 𝕋 is a Schwartz function. Thus distributions on 𝕋 integrate smooth functions on 𝕋. For s>2, define the following distribution on 𝕋:


Why is this in fact a distribution? If fC(𝕋) then f is certainly bounded (indeed, us can take any continuous function on 𝕋 as an argument, not just smooth functions). Let |f(t)|K for all t𝕋. Then,


Since 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)

ζ(s)us = d11dsq11qsgcd(p,q)=10<pqδp/q
= d1q11(qd)sgcd(pd,qd)=d0<pdqdδdpdq
= q=11qsd1d|qgcd(p,q)=d0<pqδp/q
= q=11qs0<pqδp/q
= vs.

(The last equality is a definition.) To summarize: ζ(s)us=vs.

Supposing we are interested in us, using the above formula we can instead investigate vs, which for some purposes is more analytically tractable. We shall determine the Fourier series of vs. For s>2 and for n (recalling that vs is a distribution, i.e. it integrates functions)

v^s(n) = vs(e-2πinx)
= q=11qs0<pqδp/q(e-2πinx)
= q=11qs0<pqe-2πinp/q.

pe-2πinp/q, /q, is a character, and, unless it is the trivial character, the sum over /q is equal to 0. So if qn then the inner sum is 0, and if q|n then the inner sum is equal to q. (If the language of characters of /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


For n=0, we get


Otherwise, the above can be written using a standard arithmetic function, the sum of powers of positive divisors. Let σα(n) denote the sum of the αth powers of the positive divisors of n. Thus for n0 we have


Using ζ(s)us=vs 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 s2 we shall define us to be this distribution. In summary: us is originally defined as a distribution for s>2, and now we have defined it to be a distribution for s2 and ζ(s)0. Thus us is a meromorphic distribution valued functions on with poles at s=2 and at the zeros of the Riemann zeta function.

Since ζ(1)=, if n0 then u^1(n)=0. The only pole of the Riemann zeta function is at s=1, hence ζ(0)/ζ(1)=0. Thus u^1(0)=0, and it follows that as a distribution on 𝕋,


(although the distribution u1 is 0, this doesn’t mean that we can put s=1 into the original definition of us and assert that this is 0, as the original definition of us was only for s>2, and we have analytically continued us as a meromorphic distribution valued function on . Likewise, although ζ(0)=-12, it is incorrect to conclude that 1+1+1+=-12, although for certain formal arguments this may be a correct interpretation.).