Equivalent norm in homeomorphic spaces

Question

Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on a real vector space $X$ such that the resulting spaces with induced topologies are homeomorphic, i.e., there exists a bicontinuous bijection $X\to X$.

Is it true that these two norms have to be necessarily equivalent? [I think the answer is positive as in the classical case of different norms in $\mathbb{R}^n$]

Answer 1

$\newcommand{\nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{\norm}{\nrm\bullet}$ No. Pick a $\norm_1$-discontinuous linear isomorphism $T:X\to X$ and call $\nrm{x}_2:=\nrm{Tx}_1$

Since $T$ is discontinuous as a map of normed spaces $(X,\norm_1)\to(X,\norm_1)$, by definition $\nrm{x}_2\le C\nrm{x}_1$ cannot hold for all $x$ and a fixed constant $C$.

However, $T:(X,\norm_2)\to(X,\norm_1)$ is a bijective isometry.

Answer 2

Yes!

By definition, two metrics, $\rho$ and $\sigma$ on a metric space, $X$, is said to be equivalent if the identity mapping of $\langle X,\rho\rangle$ onto $\langle X, \sigma \rangle$ is homomorphism.

Now if two metrics are (different) norms, the inverse image of every open set is open in the other space, hence (by definition) the identity map is continuous (in both directions), hence it's homomorphism, Hence they are equivalent.