It's well-known that the Fourier transform plays nicely with scaling. Particularly if we define, for $\alpha >0$, $D_{\alpha}$ by $D_{\alpha}f(x) = \alpha^{-1/2} f(x/\alpha)$, then (for suitable functions),

$$\mathcal{F}D_{\alpha} = D_{\alpha^{-1}}\mathcal{F}$$

where $\mathcal{F}$ denotes the Fourier transform. The scaling property can ultimately be tied to two facts: the Fourier transform is over all of $\Bbb R$ and that the Fourier kernel is symmetric and more specifically diagonal, i.e. if $k$ represents the Fourier kernel, $k(\omega,t) = k(\omega t,1) = k(1,\omega t)$.

If we look to integral transforms in general and ask under what conditions they obey the same scaling property, it isn't hard to see that if the kernel is diagonal then the transform will obey the same relation. Furthermore, the transforms must be over all of $\Bbb R$.

My question is: is this the only case? Or are there kernels which induce integral transforms that obey the scaling property as above?


There is a pretty trivial counter-example. Take for instance $\varphi(\omega, t) = e^{-i\omega|t|}$, and define $\mathcal{T}$ by the map $f\to \int_{-\infty}^{\infty} \varphi(\omega, t)f(t)\,dt$. Then $\mathcal{T}$ satisfies the scaling property given above:

$$\mathcal{T}\mathcal{D}_{\alpha}f(\omega) = \int_{-\infty}^{\infty} e^{-i\omega |t|}\sqrt{\alpha}f(\alpha t)\,dt = \frac{1}{\sqrt{\alpha}} \int_{-\infty}^{\infty} e^{-i\omega\left|\frac{t}{\alpha}\right|}f(t)\,dt.$$

However since $\alpha > 0$, this is clearly the same as

$$\mathcal{T}\mathcal{D}_{\alpha}f(\omega) = \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty} e^{-i\frac{\omega}{\alpha}|t|}f(t)\,dt = D_{\alpha^{-1}}\mathcal{T}f(\omega).$$

Thus $\mathcal{D}_{\alpha^{-1}}\mathcal{T} = \mathcal{T}\mathcal{D}_{\alpha}$ but $\varphi$ is not a diagonal kernel since $\varphi(\omega t, 1) = \omega t\neq \omega|t| = \varphi(\omega, t)$

If however one assumes that it holds for all $\alpha \in \Bbb R\setminus\{0\}$, then it is true that $\varphi$ is a diagonal kernel and I was able to prove that. It's actually fairly straightforward.

Conjecture: If $\varphi$ is smooth and $\alpha > 0$ and the integral operator defined by $\varphi$ satisfies the dilation property, then $\varphi$ is diagonal.

The reason that I think this is true if $\varphi$ is smooth is that the only counter-examples I could conceive involved absolute values in the kernel which are not smooth. I would also be tempted to think it would work for simply differentiable $\varphi$ but I'm not sure about that. If $\varphi$ is assumed to be real-analytic, then it is very easy to prove (since the dilation structure forces the power series to be diagonal).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.