📂Fourier Analysis

Mellin Transform Convolution

Definition

The convolution of the Mellin transform is as follows.

$$(f \times g) (y) = \int _{0}^{\infty} f(x)g \left(\frac{y}{x} \right)\frac{dx}{x}$$

Explanation

It is also called multiplicative convolution¹.

Proof

$$\mathcal{M}(f \times g)=(\mathcal{M}f)(\mathcal{M}g)$$

It suffices to show that the above equation holds.

$$\begin{align*} \mathcal{M}(f\times g)(s) &= \int _{0} ^{\infty} x^{s-1} (f\times g)(x)dx \\ &= \int _{0} ^{\infty} x^{s-1} (f\times g)(x)dx \\ &= \int _{0} ^{\infty} x^{s-1} \left( \int _{0}^{\infty}f(y)g \left( \frac{x}{y} \right)\frac{dy}{y} \right)dx \\ &= \int _{0} ^{\infty} \int _{0}^{\infty}x^{s-1}f(y)g \left( \frac{x}{y} \right)\frac{dy}{y} dx \\ &= \int _{0} ^{\infty} \int _{0}^{\infty}y^{s-1}z^{s-1}f(y)g (z)dydz \\ &= \int _{0} ^{\infty} y^{s-1}f(y)dy \int_{0}^{\infty} z^{s-1}g (z)dz \\ &= \mathcal{M}f(s) \mathcal{M}g(s) \end{align*}$$

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