Equivalence of norms implies isomorphism between Hilbert spaces If I have 2 Hilbert spaces with 2 norms, and a map between the Hilbert spaces, and I know that the norms are equivalent, does this mean that the spaces are isomorphic? 
 A: I assume that the mentioned map between the Hilbert spaces (say, $\mathcal H_1$ and  $\mathcal H_2$) is a linear bijection, say $\varphi:\mathcal H_1\to\mathcal H_2$, such that the norms are equivalent w.r.t. this $\varphi$, i.e.
$$ \exists c,C>0: \ \forall x\in\mathcal H_1: \  
c||x||_1 \le ||\varphi(x)||_2 \le C||x||_1 $$
I think we also need to assume separability, at least, it would need more work for the general case.
Let us fix an orthonormal basis $(b_i)$ for $\mathcal H_1$, and consider a mapping
$\psi:\mathcal H_1\to\mathcal H_2$ obtained via the Gram-Schmidt orthogonalization project from $\varphi(b_1),\varphi(b_2),\dots$, yielding an orthonormal basis $\psi(b_1)$ starting as
$$\psi(b_1):=\frac{\varphi(b_1)}{||\varphi(b_1)||},\ 
\psi_0(b_2):=\varphi(b_2)-\langle \varphi(b_2),\psi(b_1)\rangle\cdot \psi(b_1) \\
\psi(b_2):=\frac{\psi_0(b_2)}{||\psi_0(b_2)||},\ \ldots
$$
Using that $\varphi$ is bounded. This way, $\psi$ preserves the norm and orthogonality of the basis vectors, hence, is an isomorphism.
A: Any two Hilbert spaces for which the orthonormal bases have the same cardinality, are linearly isometric. If the bases do not have the same cardinality, then they are not even isomorphic.
