0
$\begingroup$

Let $V$ be a $\mathbb{C}$-vector space. Let $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ be two norms that each come from an inner product, each of which gives to $V$ the structure of a separable Hilbert space.

We assume that there exists a positive constant $c$ such that for every $\phi \in V$, $c^{-1}\Vert \phi \Vert_1 \leq \Vert \phi \Vert_2 \leq c \Vert \phi \Vert_1$.

Question:

Is it true that there exists a linear $S : V \rightarrow V$ which is bounded (for any choice - between $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ - of norms) and such that for every $\phi \in V$, $\Vert S\phi\Vert_1 = \Vert \phi \Vert_2$?

My attempt: Let $(e_n)_n$ be an orthonormal basis of $V$ for $\Vert \cdot \Vert_1$. Let $S$ be the map defined by linear extension on $span\{ e_n \ \vert \ n \in \mathbb{N}\}$ by $\forall n \in \mathbb{N},\ Se_n := \Vert e_n\Vert_2 e_n$.

Then, for every $n \in \mathbb{N}$, $\Vert Se_n \Vert_1 = \Vert e_n\Vert_2$.

However, for linear combinations of $e_i$'s, it may not work: $\Vert S e_1+e_2 \Vert_1 = \Vert \Vert e_1\Vert_2 e_1 + \Vert e_2 \Vert_2 e_2 \Vert_1 = \Vert e_1\Vert^2_2 + \Vert e_2\Vert^2_2$; however, $e_1$ and $e_2$ may not be orthogonal with respect to $\Vert \cdot \Vert_2$, so $\Vert e_1 + e_2\Vert^2_2 = \Vert e_1\Vert^2_2 + \Vert e_2\Vert^2_2$ isn't necessarily true.

EDIT: The question is trivial. I was aware that all separable Hilbert spaces are isomorphic; I just was somehow troubled by the fact that the two Hilbert spaces $(V,\Vert \cdot \Vert_1)$ and $(V,\Vert \cdot \Vert_2)$ share the same underlying vector space...

EDIT2: Now I don't understand why my question is trivial. Maybe I should take a break.

$\endgroup$
2
  • $\begingroup$ Do you know that all separable Hilbert spaces are isometrically isomorphic? $\endgroup$
    – gerw
    Commented Jan 24, 2022 at 13:29
  • $\begingroup$ Yes, yes, see my edit. I am totally confused and cannot write something correct at this moment. $\endgroup$
    – Plop
    Commented Jan 24, 2022 at 13:31

1 Answer 1

1
$\begingroup$

Ok, I got it. Sorry for the trivial question and answer, I got messed up with everything.

Let $(e_n)_n$ be an orthonormal basis for $\Vert \cdot \Vert_1$ and $(f_n)_n$ be an orthonormal basis for $\Vert \cdot \Vert_2$. Let, for all $n$, $Sf_n := e_n$ and extend $S$ to a bounded linear operator $V \rightarrow V$. Then for every $n$, $\Vert S f_n \Vert_1 = \Vert e_1\Vert_1 = \Vert f_1\Vert_2$. Moreover, for every square-summable sequence $(\lambda_n)_n$, $\Vert S \sum_n \lambda_n f_n \Vert^2_1 = \Vert \sum_n \lambda_n e_n \Vert^2_1 = \sum_n \vert \lambda_n\vert^2 = \Vert \sum_n \lambda_n f_n \Vert^2_2$, so for every $\phi \in V$, $\Vert S\phi \Vert_1 = \Vert \phi\Vert_2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .