Does it hold $L^2([0,1]) \simeq L^2([0,\infty)$?

Question

Does it hold that $L^2([0,1])$ is isomorphic (as a normed vector space) to $L^2([0,\infty))$? Intuitively, the $L^2$-space of a space with a finite measure is something different than with an infinite measure.

Answer 1

In addition to the general argument by GEdgar:

All separable, infinite-dimensional Hilbert spaces are isomorphic. A good proof would involve choosing an orthonormal basis, and showing both are isomorphic to $l^2$.

I give a concrete one, based on change of variable. Pick a nice bijection $\phi: [0,\infty) \to [0,1)$, for example $\phi(x)=x/(x+1)$. By the change of variables, $$ \int_0^1 |f(t)|^2\,dt = \int_0^\infty |f(\phi(x))|^2\,\phi'(x)\,dx = \int_0^\infty \left|\frac{f(\phi(x))}{x+1}\right|^2 \,dx $$ Thus, if we let $Tf(x)= f(\phi(x))/(x+1)$, the map $T$ is an isometry from $L^2[0,1]$ to $L^2[0,\infty)$. It is a bijection, with the inverse given by $Tg(t) = f(\phi^{-1}(t))/(1-t)$.

As a bonus, this argument applies to $L^p$ with $p\ne 2$ as well.