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We know that

$$ f(s)= \int_{-\infty}^{\infty}f(x)\delta (x-s) d x$$

however, is there a similar delta function so for the Mellin transform

$$ f(s)=\int_{0}^{\infty}f(x)m(xs) d x$$ ?

That is a function $ m(x) $ or distribution with similar properties of the delta function but for the Mellin transform.

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1 Answer 1

up vote 4 down vote accepted

The expression

$$ f(s) = \int_0^\infty f(x) m(xs) \mathrm{d}x $$

cannot be satisfied for all $f$. Let $g(x) = f(\lambda x)$ for $\lambda > 0$. Then

$$ \int_0^\infty g(x) m(xs) \mathrm{d}x = \int_0^\infty f(\lambda x) m(\lambda x \cdot \frac{s}{\lambda}) \frac{1}{\lambda}\mathrm{d}(\lambda x) = \frac{1}\lambda\int_0^\infty f(y) m(y s/\lambda) \mathrm{d}y = \frac{1}{\lambda} f(s/\lambda) \neq g(s) = f(s\lambda)$$

in general.

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Very nice. Could you please explain why the same phenomenon does not occur with the Dirac delta? Would it be possible to give an explanation in terms of (formal) dimensional analysis, in the spirit of the blog post by T.Tao you recommended here? –  Giuseppe Negro Mar 1 '13 at 17:53
1  
@GiuseppeNegro: actually, a similar problem (using additive translations instead of multiplicative scalings) would occur with $\delta$ if instead of the convolution one ponder the possibility of the expression $$ f(s) = \int f(x) \psi(x + s) \mathrm{d}x$$ Note that had the OP written $m(x/s)$ instead of $m(xs)$, and had $m$ be identified not as a function but some sort of density with the right automatic scaling properties, the argument above could be circumvented. –  Willie Wong Mar 1 '13 at 18:01
    
This is a very nice observation. Thank you very much. I had never realized that the minus sign in the convolution had its importance. I have always thought that it was just a convention. –  Giuseppe Negro Mar 1 '13 at 18:22

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