Hilbert transform of Dirac delta function

Question

I just read the definition of the Hilbert transform and I was wondering if you could apply it to the Dirac mass at the origin. What does this look like rigorously? I guess you should get $1/x$, but where does this function live, and what happens at $x = 0$?

Answer 1

This is an interesting and suggestive question. On one hand, for any distribution $u$, with a suitable definition of convolution, $\delta*u=u*\delta=u$, so we know this in advance. Yes, there are potential issues about associativity of convolution of distributions, e.g., the classic $(1*\delta')*H=0*H=0$, while $1*(\delta'*H)=1*\delta=1$, where $H$ is the Heaviside function.

A slightly subtle point is about "$1/x$". Hopefully-unsurprisingly, since the function $1/x$ is (just barely) not locally integrable, we cannot identify that function with any distribution. Yet, perhaps-innocently, the corresponding "principal value integral" is readily proven to be a distribution. (In fact, tempered.) But "principal value of integral" is most definitely not a literal integral, even if "just barely".

In particular, whether or not one manages to compute some pointwise values of the convolution of $\delta$ with "$1/x$", this has limited bearing on the more-authentic question of direct computation of convolution of the distributions. (Not so far away, sure, but not elementarily identical.)

A more robust way to think about the Hilbert transform $H$ (for a moment suspending judgement on where we can define it) is as the composition of Fourier transform, multiplication by $\mathrm{sgn}\, x$, and Fourier transform again (up to normalization constants). Thus, we can imagine some sense of $H$ on anything admitting a Fourier transform, e.g., tempered distributions. Then the question is on what class of tempered distributions $u$ multiplication by $\mathrm{sgn}\,x$ is well-defined and/or continuous. Certainly it suffices that $u$ be locally integrable near $0$. This is the case for $\widehat{\delta}=1$, and we get (up to constants) the sign function.

Another interesting point is that the Fourier transform of the sign function is (up to constants) the principal-value integral against $1/x$. That is, not literally "$1/x$", but the principal-value integral-against.

So, yes, for not-so-fragile reasons, up to constants, the sign function and the principal-value integral against $1/x$ are mutual Fourier transforms... which gives a direct computation of the convolution, incidentally.

Answer 2

Some non-rigorous arguments:

When $t\neq 0$, the support of $\delta(\tau)$ (this is $\{0\}$) and the singular support of $\operatorname{p.v.}\frac{1}{t-\tau}$ (this is $\{t\}$) are disjoint so the integral is defined and we get $H(\delta)(t) = \frac{1}{\pi} \operatorname{p.v.}\frac{1}{t}$.

When $t=0$, we have an even "function" ($\delta$) multiplied with an odd function $\frac{1}{-\tau}$, i.e. the product is an odd expression, integrated over a symmetric interval, so the integral will be $0$. Thus $H(\delta)(0) = 0$.

Answer 3

The Hilbert transform of the Dirac delta function is a standard problem. It's covered by page 11 in Hilbert Transform in Signal Processing by Stefan L. Hahn. The result is

$$H[\delta(t)]=\frac{1}{\pi t}*\delta(t)=\frac{1}{\pi t}$$

where the $*$ denotes the convolution.