Is there a transform based on Mellin convolution analogous to the Hilbert transform which is based on Fourier convolution

convolutionfourier transformmellin-transformtransformation

Fourier convolution is defined as follows.

(1) $\quad\left(f(x)\ *_\mathcal{F}\ g(x)\right)(y)=\int\limits_{-\infty}^\infty f(x)\ g(y-x)\ dx$

The Hilbert transform of $f(x)$ defined in (2) below is based on Fourier convolution where $g(x)=\frac{1}{x}$. Note the inverse Hilbert transform is defined in (3) below is the negative of the Hilbert transform defined in (2) below $\left(\mathcal{H}_y^{-1}[f(y)](x)=-\mathcal{H}_y[f(y)](x)\right)$.

(2) $\quad\hat{f}(y)=\mathcal{H}_x[f(x)](y)=\int\limits_{-\infty}^\infty f(x)\ \frac{1}{y-x}\ dx$

(3) $\quad f(x)=\mathcal{H}_y^{-1}[f(y)](x)=-\int\limits_{-\infty}^\infty f(y)\ \frac{1}{x-y}\ dy$

Mellin convolution can be defined in a couple of ways as follows.

(4) $\quad\left(f(x)\ *_{\mathcal{M}_1}\ g(x)\right)(y)=\int\limits_0^\infty f(x)\ g(\frac{y}{x})\ \frac{dx}{x}$

(5) $\quad\left(f(x)\ *_{\mathcal{M}_2}\ g(x)\right)(y)=\int\limits_0^\infty f(x)\ g(y\ x)\ dx$

Question (1): Are there any notable transforms based on either of the Mellin convolutions defined above?

Question (2): Is there a function $g(x)$ for which either of the Mellin convolutions defined above is its own inverse or the negative of its own inverse as illustrated in (6) below analogous to the inverse Hilbert transform defined in (3) above?

(6) $\quad \hat{f}(y)=(f(x)\ *_\mathcal{M}\ g(x))(y)\implies f(x)=\pm\left(\hat{f}(y)\ *_\mathcal{M}\ g(y)\right)(x)$

Question (3): Is question (2) above misguided? Should I really be looking for a function $g(x)$ that satisfies something like the following as the Mellin convolutions analogous to the Fourier convolutions illustrated formulas (2) and (3) above?

(7) $\quad \hat{f}(y)=\left(f(x)\ *_\mathcal{M}\ g(x)\right)(y)\implies f(x)=\left(\hat{f}(y)\ *_\mathcal{M}\ g(\frac{1}{y})\right)(x)$

Best Answer

  • Mellin and Fourier convolution are the same up to a change of variable $y= e^x$. Same for the Fourier/Laplace and Mellin transform.

  • The Hilbert transform is defined for all $C^1$ function with a $O(1/x)$ decay as $$H[f](x)=\lim_{h\to 0}\int_{-\infty}^{-h}+\int_h^\infty f(x-t)\frac1\pi\frac1tdt$$ (ie. principal value $pv$)

  • The change of variable of $pv$ is the $pv$ of the change of variable ie.

$$H_{Mellin}[F](y)=\lim_{h\to 0}\int_0^{1-h}+\int_{1+h}^\infty F(y/u)\frac1\pi\frac1{\log u}\frac{du}{u}$$

$f(x)=F(e^x)$, $H[f](x)=H_{Mellin}[F](e^x)$.

  • Hilbert transform is well defined for many other functions/distributions but they need a special argument to show it is.

  • The only way to show that the Fourier transform of $pv(1/(\pi x))$ is $i\ sign(\xi)$ is with the distribution theory, which doesn't have anything difficult itself. Instead of considering $pv(1/(\pi x))$ as a function we consider it as a convolution operator, once convoluted with a Schwartz function its Fourier transform is well-defined.

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