1
$\begingroup$

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 $(X,||.||)$ ?

$\endgroup$
  • $\begingroup$ Have you looked at math.stackexchange.com/questions/692522/… $\endgroup$ – Artur Araujo Mar 23 '18 at 15:27
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $\iota_Y$.

$\endgroup$
  • $\begingroup$ How would one show that reflexivity is preserved under linear homeomorphisms? $\endgroup$ – mechanodroid Mar 23 '18 at 21:07
  • 1
    $\begingroup$ See edit for the justification, @mechandroid. $\endgroup$ – fredgoodman Mar 23 '18 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.