Since $L_2(\mathbb R)$ is separable, and it is a Hilbert space, then it must has a orthonormal basis. But how to get an orthonormal basis of it? In this case $sin(nx)$ and $cos(nx)$ do not converge, then how to construct an orthonormal basis? Thanks!

  • $\begingroup$ You should have a look at Hermite Polynomials $\endgroup$
    – nicomezi
    Apr 11, 2021 at 14:25
  • $\begingroup$ @nicomezi Thank you. $\endgroup$
    – Richard
    Apr 11, 2021 at 14:27
  • 2
    $\begingroup$ $1_{x\in [k,k+1)} e^{2i\pi nx}, k,n\in \Bbb{Z}$ works too. $\endgroup$
    – reuns
    Apr 11, 2021 at 15:04

1 Answer 1


Choose any diffeomorphism $\varphi :{\mathbb R}\to (0,2\pi )$ and define $$ T:L^2(0,2\pi )\to L^2(\mathbb R) $$ by $$ T(f)|_x = |\varphi '(x)|^{1/2}f(\varphi (x)), $$ for all $f$ in $ L^2(0,2\pi )$, and $x\in {\mathbb R}$.

Show that $T$ is a unitary operator and then, for any orthonormal basis $\{e_n\}_n$ of $L^2(0, 2\pi )$, you get an orthonormal basis $\{T(e_n)\}_n$ for $L^2({\mathbb R})$.


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