Let $(X,||.||)$ be a Banach space . From Can every Banach space be given an inner product homeomorphically? I know that there is an inner product on $X$ which generates the same topology as that of $(X,||.||)$ . My question is : does there always exist an inner product on $X$ giving it a norm $||.||_1$ such that $(X,||.||)$ and $(X,||.||_1)$ are linearly isomorphic i.e. there is a continuous, bijective , linear map between them with a continuous inverse ? If this is not always true , then what if we start with a separable Banach space $(X,||.||)$ ?

  • $\begingroup$ Have you looked at math.stackexchange.com/questions/692522/… $\endgroup$ – Artur Araujo Mar 23 '18 at 15:27
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    $\begingroup$ As remarked in the other answers to the question you have linked, no this is not true. For example, $\ell^p(\Bbb N)$ and $\ell^q(\Bbb N)$ are not isomorphic if $p\neq q$, but any separable complete inner product space is isomorphic to $\ell^2(\Bbb N)$. $\endgroup$ – s.harp Mar 23 '18 at 15:28
  • $\begingroup$ @ArturAraujo : that only says that $||.||$ need not necessarily come from an inner product .... I am talking about a much weaker thing $\endgroup$ – user Mar 23 '18 at 15:29
  • $\begingroup$ How is that weaker? Given an isomorphism as in your question, you can pull back the inner product to make the first Banach space a Hilbert one on the nose. $\endgroup$ – Artur Araujo Mar 23 '18 at 15:33
  • $\begingroup$ @ArturAraujo: I don't think so ... $\mathbb R^n$ with the Euclidean norm and the taxicab norm are linearly isomorphic , yet one of them comes from an inner product while the other doesn't. $\endgroup$ – user Mar 23 '18 at 15:38

s.harp, in his/her comment, provides a correct answer. A more detailed explanation/reference for why $\ell^r$ is not linearly homeomorphic to $\ell^2$ for $2 < r$ can be found here Proof of Pitt's theorem.

A different approach would be to exhibit a non-reflexive separable Banach space. A Banach space is called reflexive if the natural embedding of $X$ in $X^{**}$ is surjective. This is true for Hilbert spaces and for $L^p$ spaces, for $1 < p < \infty$, and is a property preserved under linear homeomorphism. $c_0$ and $\ell_1$ are separable and non-reflexive, https://en.wikipedia.org/wiki/Reflexive_space#Examples

Edit: In response to a question, here is how to show that reflexivity is preserved under linear homeomorphism. Let $T: X \to Y$ be a linear homeomorphism of Banach spaces. Then $T^* : Y^* \to X^*$ and $T^{**} : Y^{**} \to X^{**}$ are linear homeomorphisms. Let $\iota_X$ and $\iota_Y$ be the natural embeddings of $X$ and $Y$ into their double dual spaces. Then check that $\iota_Y\circ T = T^{**} \circ \iota_X$; hence, $\iota_Y = T^{**} \circ \iota_X\circ T^{-1}$. Thus if $\iota_X$ is surjective, so is $\iota_Y$.

  • $\begingroup$ How would one show that reflexivity is preserved under linear homeomorphisms? $\endgroup$ – mechanodroid Mar 23 '18 at 21:07
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    $\begingroup$ See edit for the justification, @mechandroid. $\endgroup$ – fredgoodman Mar 23 '18 at 22:01

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