The Mellin transform is known to be an isomorphism see wikipedia between $M:L^2(0, \infty) \rightarrow L^2(-\infty, \infty)$
where $$M(f):= \frac{1}{\sqrt{2\pi}}\int_0^{\infty} x^{-\frac{1}{2} + is} f(x)\,dx.$$
It is pretty straightforward to see that $L^2((0,\infty),r^n \ dr )$ gets mapped onto $L^2((-\infty,\infty)-ni).$ This just follows from the substitution $$M(r^nf)(s):= \frac{1}{\sqrt{2\pi}}\int_0^{\infty} x^{-\frac{1}{2} + i(s-in)} f(x)\,dx = M(f)(s-in).$$
Yet, what seems to be difficult to me is to understand onto what the weights $r+1$ and $1/(1+r)$ get mapped.
In other words, what are the ranges of $L^2((0,\infty),(1+r) \ dr)$ and $L^2((0,\infty), 1/(1+r) \ dr)$ under Mellin transform?
This question is in fact very important as the Fourier transform of Sobolev spaces gives precisely those weighted spaces, i.e. $ \mathcal F H^1 \simeq L^2((0,\infty),(1+r) \ dr)$ and $\mathcal F H^{-1} \simeq L^2((0,\infty), 1/(1+r))$ (after switching to spherical coordinates)
I am grateful for any remark, comment or question this. Thank you!
