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If I have an isomorphism of two separable Hilbert spaces that preserves norms, does the isomorphism map orthnormal basis to orthonormal basis? I can't show it.

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You need to show that the isomorphism preserves scalar products. Try expressing a scalar product as a function of the norm by means of a polarization identity. – Giuseppe Negro Jan 16 '13 at 12:14
up vote 2 down vote accepted

I'm not sure where separability and completeness with respect to norm come in, it seems to hold for any linear isometry between inner product spaces:

Let $T: H \to H'$ be a linear isometry. Let $e_i$ be an orthonormal basis of $H$. We want to show that $\langle Te_i , Te_j \rangle = \langle e_i , e_j \rangle$.

$T$ is an isometry, that is, $\|Tx\| = \|x\|$, and the norm is given by $\|x\|^2 = \langle x,x \rangle$. You don't need the "full" polarisation identity: note that $\langle x-y , x-y\rangle = \|x-y\|^2 = \|x\|^2 - 2 \langle x,y \rangle + \|y\|^2$ and hence $2 \langle x,y \rangle = \|x\|^2 - \|x-y\|^2 + \|y\|^2$.


$\begin{align} 2 \langle Te_i , Te_j \rangle &= \|Te_i\|^2 - \|Te_i - Te_j\|^2 + \|Te_j\|^2\\ &= \|e_i\|^2 - \|e_i-e_j\|^2 + \|e_j\|^2 \\ &= 2 \langle e_i , e_j \rangle \end{align} $

which proves the claim.

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