How to prove $L_{2} ([0,1])$ $\cong$ $L_{2}(\mathbb{R})$

Question

I want to show that $L_{2} ([0,1])$ $\cong$ $L_{2}(\mathbb{R})$.

So that intuitively, Any Hilbert space is isomorphic with $L_{2} ([0,1])$ (please correct me if I am wrong here).

I saw that people try to prove any two Hilbert spaces of the same dimension are isomorphic by stating they are isomorphic to the $L_{2}$(convergent series) space. But I can't fully appreciate the idea of $L_{2}$ space. Is there any way to prove this directly?

If so, can someone please give me a sketch of the idea of the proof?

Any help is appreciated!

Answer 1

Hint:

You may try first to consider Legendre polynomials on $L_2[-1,1]$, and the Hermite polynomials on $L_2(\mathbb{R},\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx)$. From there, it should be straightforward.

A more general way of doing this is through a result due to Kolmogorov:

Theorem: Suppose $\mu$ is a finite positive measure on $\big(\mathbb{R},\mathscr{B}(\mathbb{R})\big)$. If $\int e^{-\delta_0|x|}\mu(x)<\infty$ for some $\delta_0>0$, then $\operatorname{span}\big(\{p_n(x)=x^n: n\in\mathbb{Z}_+\}\big)$ is dense in $L_2(\mathbb{R},\mu)$.

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Setting $\mu(dx)=\mathbb{1}_{[0,1]}\,dx$ and $\nu(dx)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx$, you obtained two orthogonal and complete systems $\{p_n:n\in\mathbb{Z}_+\}$ for $L_2[0,1]$ and $\{q_n(x)e^{-x^2/4}:n\in\mathbb{Z}_+\}$ for $L_2(\mathbb{R})$ (use the Gran-Schmidt orthogonalization procedure). Mapping $p_n\mapsto q_n\cdot e^{-x^2/4}$ gives you the desired isometric isomorphism.

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Here is a short proof of Kolmogorov's theorem:

By hypothesis, $\int |x|^n\,\mu(dx)<\infty$ for all nonnegative integers $n$. Assume the statement is false. The Hahn--Banach theorem and the Riesz representation theorem for Hilbert spaces imply that for some $h\in L_2$ not identically zero, $\langle p_n,h\rangle=\int x^n h(x)\,\mu(dx)=0$ for all integers $n\geq0$. By hypothesis, for any $0<\delta<\frac12\delta_0$ the map $x\mapsto e^{\delta|x|}h(x)$ is in $L_1(\mu)$. Hence, setting $\mu_h:=h\cdot d\mu$, we have that \begin{align} \widehat{\mu_h}(t)=\int e^{i tx}h(x)\,\mu(dx) \end{align} can be extended analytically to the strip $H=\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\frac12\delta\}$. Our assumption implies that $\widehat{\mu_h}^{(n)}(0)=(i)^n\langle p_n,h\rangle =0$ and so, $\widehat{\mu_h}(z)\equiv0$. This means that $h=0$ ,$\mu$--a.s. which is a contradiction.

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Answer 2

Let $f : \mathbb{R} \to (0,\infty)$ be a strictly positive measurable function satisfying $\int_{\mathbb{R}} f(x)\,dx = 1$ (e.g. $f$ could be the Gaussian density). Set $F(x) = \int_{-\infty}^x f(y)\,dy$ so that $F : \mathbb{R} \to (0,1)$ is continuous and strictly increasing, with $F'=f$.

For $\phi \in L^2([0,1])$ let $$(T\phi)(x) = \phi(F(x)) \sqrt{f(x)}.$$

Then verify that $T : L^2([0,1]) \to L^2(\mathbb{R})$ is an isometric isomorphism. I leave this as an exercise. (Hint: change of variables.)