# Are topologically equivalent hilbertian norms necessarily conjugate?

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.

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

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 $$\phi \in V$$, $$\Vert S\phi \Vert_1 = \Vert \phi\Vert_2$$.