Is it true that every Hilbert space has a closed subspace isometrically isomorphic to $L^2[0,1]$? Can someone sketch a proof of this, or at least point me in the right direction to understanding it? If this is something I should be able to prove on my own, then hints would be nice rather than a full solution.


1 Answer 1


Things to prove

You need the other space $H$ to be infinite dimensional.

To do this just get an orthonormal basis of $L^2[0,1]$. Why there is one, or how to construct one?

Since $L^2[0,1]$ is separable Why?, then it has a countable orthonormal basis $x_1,x_2,..$. Why?

Then take an orthonormal basis $\{y_i\}_{i\in I}$ of $H$. Since $H$ is infinite dimensional, its basis is going to be infinite (perhaps not countable).

Take countably many elements of that basis $y_1,y_2,...$

The linear map that sends $x_i\mapsto y_i$ is the isometry we need from $L^2[0,1]$ to $\overline{\text{span}\{y_1,y_2,...\}}\subset H$. Check that it is.

  • $\begingroup$ Does it follow from this construction that all separable Hilbert spaces are isometrically isomorphic? That seems a likely conclusion. $\endgroup$
    – Mnifldz
    Commented Jan 28, 2015 at 7:23
  • $\begingroup$ @Mnifldz Yes, that's true. $\endgroup$
    – Pp..
    Commented Jan 28, 2015 at 7:24

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