Orthogonal basis of $L^2(\mathbb{R})$ and $L^2(\mathbb{R^+})$

Question

Let $\mathbb{R}^+ = [0, \infty)$, i.e. the positive real numbers. Let $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^+)$ be the sets of real-valued functions with domains in $\mathbb{R}$ and $\mathbb{R}^+$ respectively that are square integrate with respect to the Lebesgue measure. These sets are endowed with the inner product $$ \left( f, g \right) = \int_{-\infty}^{\infty} f g \ \mathsf{d} x, $$ and $$ \left( f, g \right) = \int_{0}^{\infty} f g \ \mathsf{d} x, $$ respectively.

I have the following questions

1. Are the sets $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^+)$ separable Hilbert Spaces?.

2. In such a case, what Schauder basis exist for these sets?. Is there an known orthogonal family of functions on these sets?

Motivation: For the case $L^2(\Omega)$ where $\Omega$ is compact, this is known to be true. The families of orthogonal functions on this set are used in a wide range of engineering applications, e.g. solving PDES in compact domains. However, I have not found similar results when $\Omega$ is unbounded. The existence of an orthogonal set of functions in $L^2(\mathbb{R})$ and $L^2(\mathbb{R}^+)$ would be useful to solve problems in unbounded domains.

Answer 1

Start with an orthonormal basis $\{ \varphi_n \}_{n=1}^{\infty}$ on $[0,1]$ with respect to ordinary Lebesgue measure. Then perform a change of variables from $[0,\infty)$ to $[0,1)$ such as $u(x)=1-e^{-x}$. Then $$ \int_0^1 f(x)\overline{g(x)}dx=\int_0^{\infty}f(u(x))\overline{g(u(x))}u'(x)dx = \int_{0}^{\infty}[f(u(x))e^{-x/2}]\overline{g(u(x))][e^{-x/2}]}dx $$ Therefore, $\Phi : L^2[0,1]\rightarrow L^2[0,\infty)$ is a unitary map, where $$ (\Phi f)(x)= f(u(x))e^{-x/2}. $$ In this way, any orthonormal basis $\{f_n\}_{n=1}^{\infty}$ of $L^2[0,1]$ is mapped to an orthonormal basis of $L^2[0,\infty)$ through $\Phi$.

Answer 2

I believe it is not difficult to prove that $L^2(\mathbb{R})$ and $L^2(\mathbb{R}_+)$ are Hilbert spaces. They are inner product spaces. Their completeness is a rather technical exercise in measure theory. Here is an example (the proof uses Fatou's lemma):

"Running through an appropriate sequence" in proof of L2 Completeness

$L^2(\mathbb{R})$ has a countable basis. It consists of Hermite functions. $L^2(\mathbb{R}_+)$ has a countable basis. It consists of Laguerre functions. Here is a link:

On the completeness of the generalized Laguerre polynomials

But separability is much easier to prove. Here is a bit more general statement.

$L^p(\mathbb{R})$ is separable ($1 \leq p < \infty$).

Proof. Let $f \in L^p(\mathbb{R})$. We know that $|f|^p$ is integrable on $\mathbb{R}$ with respect to standard Lebesgue measure only if for any positive $\varepsilon$ there exists a natural $n$ s.t. $$ \int\limits_{\mathbb{R}/[-n,n]} |f|^p \, d\mu < \varepsilon $$ i.e. on the subset $\mathbb{R}/[-n,n]$ $f$ can be approximated by zero.

Denote $L^p_0 [-n,n]$ a countable dense subset of $L^p [-n,n]$. Using the indicator of $[n,-n]$ we can embed $L^p [-n,n]$ in $L^p (\mathbb{R})$. Thus, the subset $$ L^p_0 (\mathbb{R}) = \bigcup_{n=1}^{\infty} L^p_0 [-n,n] $$ of $L^p (\mathbb{R})$ is dense. As it is obviously countable we are done.

The statement for $L^p (\mathbb{R}_+)$ is a simple corollary.