Let $H$ be a Hilbert space with a norm $\| \cdot \|_1$. Let $\| \cdot \|_2$ be another norm on $H$ which is equivalent with $\| \cdot \|_1$. It is easy to see that $(H, \| \cdot \|_2)$ is a Banach space since the norms are equivalent. Is it also true that $(H, \| \cdot \|_2)$ is a Hilbert space? I think that the answer is no, but I cannot find a counterexample.
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Consider $\ell_1^n$ and $\ell_2^n$. – Tom Cooney May 31 '13 at 15:33
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5This is not true even in the finite dimensional case. Take $\mathbb{C}^n$: There is a unique LCTVS topology. In particular, all $l^p$ norms, $1 \leq p \leq \infty$, are equivalent. But only one, $p=2$, is a Hilbert space norm. – Michael May 31 '13 at 15:33
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Sure, I was silly. Thank you. – jpinder May 31 '13 at 15:53
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@Michael That's an answer, not a comment... – ˈjuː.zɚ79365 Jun 01 '13 at 04:40
