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It is a known fact that any two classical Hilbert spaces are isomorphic, in particular, any classical Hilbert space is isomorphic to $L^2$.

Consider now weighted Hilbert spaces, that is, Hilbert spaces with a weighted inner product

$ \langle f, g \rangle = \int f \ g \ w \ dx $

where $w(x)$ is the weight function. This situation appears frequently in Sturm-Liouville eigenvalue problems. Let's call $H_w$ to such weighted Hilbert space.

My question is: are all weighted Hilbert spaces also isomorphic?. In other words, is any weighted Hilbert space $H_w$ isomorphic to $L^2_w$?

Thank you very much.

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  • $\begingroup$ Consider $w(t) = 0$. The answer depends on the class where $w(t)$ lies. $\endgroup$ – Pavel Ievlev Dec 3 '17 at 9:51
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    $\begingroup$ So, the right question here would be "what are the conditions on $w(t)$ for $L_{2, \, w}$ and $L_{2}$ to be isomorphic?". This question could be generalized: "what are the conditions on measures $\mu$ and $\nu$ for $L_{2, \, \mu}$ and $L_{2, \, \nu}$ to be isomorphic?" $\endgroup$ – Pavel Ievlev Dec 3 '17 at 9:57
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    $\begingroup$ It is not exactly that all Hilbert spaces are isomorphic. It depends on the "dimension" $\endgroup$ – user99914 Dec 3 '17 at 9:58
  • $\begingroup$ Hilbert spaces can have a variety of inner products and also not be composed of functions. How, fundamentally, is a weighted Hilbert space differ from the a Hilbert space as defined by the axioms? $\endgroup$ – Theo Bendit Dec 3 '17 at 10:09
  • $\begingroup$ Sorry for not being specific. I am currently working on a Sturm-Liouville problem and my question refers to infinite dimensional functional spaces with an inner product defined by the integral shown in my question. Of course, the question could be extended to other cases, too. $\endgroup$ – fjgg1549 Dec 3 '17 at 10:42
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All Hilbert spaces of the same dimension are isomorphic, no matter how they're defined. Whatever "classical" means, that has nothing to do with it.

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