I am trying to learn more about convolutions with singularities and was hoping that someone could point me to a reference and perhaps check the following (i have it in my notes that this is from a MSE post, but i can't find the original question):
Suppose we have an even function with a singularity at the origin:
$$f(x)=\frac{1}{e^{|x|}-1} $$
We can construct a tempered distribution by taking $\phi\in C^\infty_c$ and integrating around the origin:
$$\langle f,\phi\rangle=\int_\mathbb{R}(\phi(x)-\phi(0)1_{|x|<1})f(x)dx $$
I believe we can then define a convolution (also tempered):
$$f*\phi(y)=\int_\mathbb{R}(\phi(y-x)-\phi(y)1_{|x|<1})f(y)dx$$
Question: is the convolution also equal to:
$$f*\phi(y)=\int_\mathbb{R}(\phi(x)-\phi(0)1_{|x|<1})f(y-x)dx$$
Also, i would greatly appreciate any references that people could point me to that handle this (and perhaps other methods for convolution with singular kernels).
thanks
Best Answer
So all this is about the theory of distributions. In this theory, indeed, a generalized function $f$ is a linear form on a space of test functions. It is thus defined by its action on tests functions $\varphi\in C^\infty_c$, that one can indeed write $\langle f,\varphi\rangle$. The fact that its generalizes functions comes from the fact that for each locally integrable function, one associate the distribution defined by $$ \langle f,\varphi\rangle = \int_{\Bbb R} f(x)\,\varphi(x)\,\mathrm d x. $$ But in general, $\langle f,\varphi\rangle$ might not be possible to write as the integral of a product of functions. For example, the Dirac delta $\delta_0$ is defined by $\langle \delta_0,\varphi\rangle = \varphi(0)$.
Now, one can extend the usual operations by getting looking at what happens when $f$ is a nice function. For example the derivative of a distribution is defined by $\langle f',\varphi\rangle = -\langle f,\varphi'\rangle$, because for nice functions $f$, this relation is true. Similarly, to define the convolution of a distribution with a test function, one can look at the case of nice functions $f$ for which $$ f*\varphi(y) = \int_{\Bbb R} f(x)\,\varphi(y-x)\,\mathrm d x = \langle f,\varphi(y-\cdot)\rangle $$ and so one can define more generally $f*\varphi(y) := \langle f,\varphi(y-\cdot)\rangle$ even if $f$ is not a locally integrable function. In your case, this gives indeed $$ f*\varphi(y) = \int_{\Bbb R} f(x)\left(\varphi(y-x)-\varphi(y)\mathbf{1}_{|x|<1}\right)\mathrm d x $$ Now, this is a classical integral. You can do the change of variable $x\mapsto y-x$, but it will give you $$ f*\varphi(y) = \int_{\Bbb R} f(y-x)\left(\varphi(x)-\varphi(y)\mathbf{1}_{|x-y|<1}\right)\mathrm d x, $$ which is a bit different from your last expression. And indeed, your last expression cannot be right, since the difference $\varphi(x)-\varphi(0)$ is initially here to compensate the singularity of $f$ at $0$. But in your last expression, the singularity of $f$ occurs at $x=y$, but $\varphi(y)-\varphi(0)$ is in general not small. At the contrary, in my last expression $\varphi(x)-\varphi(y)$ is indeed small when $x\to y$.
Remarks:
If the test functions are in $C^\infty_c$ then you are dealing with distributions. Tempered distributions are the particular case of distributions where the test functions can be chosen in $C^\infty$ with fast decay at infinity. It is indeed the case for your particular $f$.
This regularization does not come from nowhere. Indeed, you get these kind of distributions by taking derivatives of functions in the sense of distributions (that is using the rule $\langle f',\varphi\rangle = -\langle f,\varphi'\rangle$). You can look for example at this post where I proved (take $d=1$) that the derivative of $u(x) = \mathrm{sgn}(x) \ln(|x|)$ is the distribution defined by $$ \langle u',\varphi\rangle = \int_{\Bbb R} \frac{\varphi(x)-\varphi(0)\mathbf{1}_{|x|<1}}{|x|}\,\mathrm{d}x. $$ I suppose you get something close to your distribution by the same proof and looking at the derivative of the function $g(x) = \mathrm{sgn}(x) \ln(1-e^{-|x|})$.