# Tempered distribution by convolution with singular kernel

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

• Why should $$\langle f,\phi\rangle=\int_\mathbb{R}(\phi(x)-\phi(0)1_{|x|<1})f(x)dx$$ define a tempered distribution? It is not even a function, since after integrating over $x$, there is no variable left. Feb 1, 2023 at 15:37
• my notation may be bad (which is why i asked for references in the question), but i found the original post. @Nuke_Gunray please see reuns' answer in this post: math.stackexchange.com/questions/3336437/…
– APIs
Feb 1, 2023 at 15:42
• @Nuke_Gunray This is indeed the good way to define a tempered distribution. A tempered distribution is a linear form on smooth and decaying functions. The tempered distribution is $f$, and its action on test functions is $\langle f,\phi\rangle$ Feb 1, 2023 at 20:34
• @LL3.14 Yeah I know, but it was formulated in a way as if $\langle f,\phi\rangle$ was a distribution for a choice of $f,\phi$, as opposed to the actual map $\phi\mapsto \langle f,\cdot\rangle$. Feb 2, 2023 at 14:28

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 $$g(x) = \mathrm{sgn}(x) \ln(1-e^{-|x|})$$.

• nice answer... so i forgot the translation for the indicator function and $\varphi(y)$.
– APIs
Feb 1, 2023 at 21:39
• I found Gelfand/Shilov "Generalized functions vol 1" where they cover this in section 1.7. There is a proposition there that if $f(x)r^m$ is locally summable ($m\in\mathbb{Z}_{>0}$, and r is distance to singularity?) then these functions can be regularized. later they state that other singular functions can be regularized with other classes of test functions. is $C^\infty_c$ then the correct class to regularize $f(x)=\frac{1}{e^{|x|}-1}$?
– APIs
Mar 8, 2023 at 22:16
• one more followup (i didn't think this warranted a separate question): to take the derivative of the function in your remark, we would just: $\int_{\mathbb{R}}\frac{\varphi'(x)-\varphi'(0)1_{|x|<1}}{|x|}dx$? Thank you again!
– APIs
Mar 8, 2023 at 22:18
• You are talking of the derivative of $u'$, so $u''$ ? Then yes, but with a minus sign, since by definition of distributional derivative $\langle u'',\varphi\rangle = -\langle u',\varphi'\rangle$. For the previous comment, in your case $x f(x)$ is indeed locally summable. There is no problem to use $C^\infty_c$ as a space of test functions. Mar 9, 2023 at 9:23