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Let $H$ be any Hilbert space. Must there exist a measure space $(X,\scr{M},\mu)$ such that we have a Hilbert space isomorphism: $$H\cong L^2(\mu)$$

Thank you

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Yes. If $\kappa>0$ is the (Hilbert) dimension of $H$, then you can take for $\mu$ the counting measure on $\kappa$ and then it's easy to see that $H\cong L^2(\mu)=\ell^2(\kappa)$.

If $\kappa=0$, then take any $\mu$ such that $\mu(X)=\infty$ for any nonempty set $X$. Then $L^2(\mu)=\{0\}$.

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  • $\begingroup$ Where is the origin of this theorem? $\endgroup$ – user51514 Jul 27 '15 at 14:34
  • $\begingroup$ @user51514: What theorem? $\endgroup$ – tomasz Jul 27 '15 at 15:27
  • $\begingroup$ $H\cong L^2(\mu)\cong \ell^2(k)$? $\endgroup$ – user51514 Jul 27 '15 at 15:33
  • $\begingroup$ The last one is equality, not isomorphism. The congruence follows from the fact that two Hilbert spaces of the same dimension are isomorphic (which is immediate from the definition). I believe it is entirely elementary, calling it a theorem is a gross overstatement. Basically the only things this argument uses are the definitions and the fact that every Hilbert space has a (Hilbert) basis. $\endgroup$ – tomasz Jul 27 '15 at 15:36
  • $\begingroup$ It is true for separable Hilbert space. Is it true that every Hilbert space $H$ is the same as $L^2(\mu)$? $\endgroup$ – user51514 Jul 27 '15 at 16:02

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