Editing Mellin transform (section)

== As an isometry on ''L''<sup>2</sup> spaces ==
In the study of [[Hilbert space]]s, the Mellin transform is often posed in a slightly different way.  For functions in <math>L^2(0,\infty)</math> (see [[Lp space]]) the fundamental strip always includes <math>\tfrac{1}{2}+i\mathbb{R}</math>, so we may define a [[linear operator]] <math>\tilde{\mathcal{M}}</math> as
<math display="block">\tilde{\mathcal{M}}\colon L^2(0,\infty)\to L^2(-\infty,\infty),</math>
<math display="block">\{\tilde{\mathcal{M}}f\}(s) := \frac{1}{\sqrt{2\pi}}\int_0^{\infty} x^{-\frac{1}{2} + is} f(x)\,dx.</math>
In other words, we have set
<math display="block">\{\tilde{\mathcal{M}}f\}(s):=\tfrac{1}{\sqrt{2\pi}}\{\mathcal{M}f\}(\tfrac{1}{2} + is).</math>
This operator is usually denoted by just plain <math>\mathcal{M}</math> and called the "Mellin transform", but <math>\tilde{\mathcal{M}}</math> is used here to distinguish from the definition used elsewhere in this article.  The [[Mellin inversion theorem]] then shows that <math>\tilde{\mathcal{M}}</math> is invertible with inverse
<math display="block">\tilde{\mathcal{M}}^{-1}\colon L^2(-\infty,\infty) \to L^2(0,\infty),</math>
<math display="block">\{\tilde{\mathcal{M}}^{-1}\varphi\}(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} x^{-\frac{1}{2}-is} \varphi(s)\,ds.</math>
Furthermore, this operator is an [[isometry]], that is to say <math>\|\tilde{\mathcal{M}} f\|_{L^2(-\infty,\infty)}=\|f\|_{L^2(0,\infty)}</math> for all <math>f\in L^2(0,\infty)</math> (this explains why the factor of <math>1/\sqrt{2\pi}</math> was used).