I want to ask two questions that are related.

  1. Suppose the Banach space $X$ isn't a Hilbert space, is it possible that it can be homeomorphic to a Hilbert space, or we say we can have two equivalent norms on the same linear space with one satisfies Parallelogram law while another not.

  2. Suppose we have two equivalent norms on the same linear space such that one is induced by the inner product, is it true that another is also induced by inner product

  • 2
    $\begingroup$ It's been shown that any two separable infinite dimensional Banach spaces are topologically homeomorphic. See this. $\endgroup$ – David Mitra Aug 18 '17 at 8:29

The following is a counterexample, in finite dimension, to both (1) and (2).

It is well-known that all norms on $\mathbb{R}^n$ are equivalent, in particular all the $p$-norms $$||x||_p := \left( \sum_{i=1}^n |x_i|^p \right)^{1/p}$$ are equivalent. However, such a norm satisfies the parallelogram law if and only if $p=2$, i.e., when it is the usual euclidean norm. In particular, $p=2$ is the only case where the $p$-norm is induced by a scalar product.

  • $\begingroup$ I can't see why that's a counterexample to (1). $\mathbb{R}^n$ with the $2$-norm (induced from an inner product) is indeed homeomorphic to $\mathbb{R}^n$ with any $p$-norm that is not induced by an inner product, and that's because the norms are equivalent. $\endgroup$ – Scientifica Aug 18 '17 at 8:47
  • $\begingroup$ Actually (1) asks two things. This is an example of two norms in the same linear space, one satisfying the parallelogram law and the other not. $\endgroup$ – Francesco Polizzi Aug 18 '17 at 8:48
  • $\begingroup$ For $p\neq 2$ we do not have a Hilbert space, but in any case the topology is the same as in the $p=2$ case. So this also answers in a positive way the first part of question 1, right? $\endgroup$ – Francesco Polizzi Aug 18 '17 at 8:52
  • $\begingroup$ Ah I see. Yes it gives indeed an example where (1) is satisfied. But (1) asks for an arbitrary Banach space. $\endgroup$ – Scientifica Aug 18 '17 at 9:20
  • $\begingroup$ Well, I interpret the expression "is it possible that it can be..." as "is there an example where..." $\endgroup$ – Francesco Polizzi Aug 18 '17 at 9:29

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