Consider the Mellin transform defined initially for continuous function $f$ of compact support in $R^+=${$t\in R:t>0$} and $x\in R$ by
$Mf(x)=\int_0^\infty f(t)t^{ix-1}dt$
Prove that ($2\pi$)$^{-1/2}M$ extends to a unitary operator from $L^2(R^+,dt/t)$ to $L^2(R)$
I guess it follows the similar way of Theorem 1.1 in chapter 5 (Extension of Fourier transform)
So my plane is 1) $S=\{f\mid f\mbox{ is continuous of compact support in }R^+\}$ is dense in $R^+$
2) By Lemma 1.3 , $M:S\to L^2$ can be extended to $M':L^2(R^+,dt/t)\to L^2$ (dense $\to$ extend possible under some condition , in fact 3) )
3) For any $f \in S$ , $||f||_{L^2(R^+,dt/t)}=\lVert Mf\rVert_{L^2(R)}$
But it's hard to me.... how can I draw the next step?
Further question : 1. What's the mean of $dt/t$? Does it mean that if $f\in L^2(R^+,dt/t)$ then $\int_0^{\infty} \frac {|f(t)|^2}t <\infty$?
- Why does the term ($2\pi$)$^{-1/2}$ appear in the extension form?
For the further questions: 1. You are right about the meaning of $dt/t$.
In order to show that for each $f\in S$, $$\lVert f\rVert_{L^2(\mathbb R^+,dt/t)}=(2\pi)^{-1/2} \lVert Mf\rVert_{L^2(\mathbb R)},$$ we can do the substitution $t=e^u$ in the integral defining the Mellin transform. A Fourier transform will appear and we can use Plancherel's formula.