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From the Wikipedia page on integral transforms, it states that: integral transform is any transform $T$ of the following form: $$ (Tf)(u)=\int^{t_2}_{t_1}K(t,u)f(t)dt $$ ...There are numerous useful integral transforms. Each is specified by a choice of the function $K$ of two variables, the kernel function or nucleus of the transform.

I was wondering if the kernel function has to have any special properties in order for the integral transform to work. On the bottom of the Wikipedia page, it shows some of the common kernel functions such as $\frac{e^{-iut}}{\sqrt{2\pi}}$ and $e^{-ut}$ for the Fourier Transform and the Laplace Transform. However, as an example, with the kernel function $$ K(t,u)=\ln|t+u|,\space t_1=0,\space t_2=\infty $$ the integral transform would diverge for almost all $f(t)$. Or, for another example, the kernel function $$ K(t,u)=ut,\space t_1=-\infty,\space t_2=0 $$ also diverges for almost all $f(t)$. My thinking is that there has to be some certain property of the Fourier Transform and the Laplace transform that make it so they don't diverge for almost all $f(t)$. So, my basic question is whether a certain property has to hold in order for the integral transform to not diverge to infinity given any function $f(t)$.

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Depends on what kind of $f$ you plan on applying it to. – Qiaochu Yuan Aug 24 '12 at 3:53
Elementary functions I guess? The Laplace transform works for all elementary functions (from what I've seen, correct me if I'm wrong), so something similar to that. – Envious Page Aug 24 '12 at 4:11
No, it doesn't. For example, $e^{t^2}$ doesn't have a Laplace transform. The problem is that if you want to take the integral over all of $\mathbb{R}$ then you need to specify some growth condition on $K$, and exactly what growth condition you need is dictated by the growth conditions you specify on the $f$ you want. – Qiaochu Yuan Aug 24 '12 at 4:12
So $K$ can't be or isn't easy to find given a group of functions for $f$ unless they all have the same growth conditions? Or am I misinterpreting what you said? – Envious Page Aug 24 '12 at 4:20
What I'm saying is that if you want some conditions on $K$ for the integral operator to be well-defined you have to specify a class of functions $f$ that you want to integrate $K$ against. Without this data, you can't say anything. If the functions $f$ you want to integrate against grow very quickly then $K$ needs to decay very quickly. – Qiaochu Yuan Aug 24 '12 at 4:26

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