# 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?

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How are norms on different spaces equivalent? – Hagen von Eitzen Feb 4 '13 at 13:06
It would help if you define precisely what you mean by "the norms are equivalent" and "the spaces are isomorphic". There are a few different possible interpretations. – Nate Eldredge Feb 4 '13 at 14:00
@NateEldredge I mean by equivalence, if $T:X \to Y$, then $K|Tx| \leq |x| \leq C|Tx|.$ By isomorphic I want a 1-1 mapping bijective surjective mapping. – combo Feb 5 '13 at 8:49

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.

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I'm not sure it is good -- where is equivalence of norms used? – Berci Feb 4 '13 at 16:41
If $\varphi$ is a bounded linear bijection, then the equivalence condition is automatic. What you showed is that the two Hilbert spaces are isometric, that is $||\psi(x)||=||x||$. – Theo Feb 4 '13 at 18:05

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.

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