Does there exist a vector space $X$ and two norms $\Vert \cdot \Vert$ and $\Vert \cdot \Vert_1$ on $X$ such that both spaces $(X, \Vert \cdot \Vert)$ and $(X, \Vert \cdot \Vert_1)$ are complete, but the two norms $\Vert \cdot \Vert$ and $\Vert \cdot \Vert_1$ are not equivalent?
A fact:
When one of these two norms is stronger than the other one, then they are equivalent. So I want to find a counter example, where the above condition does not hold.
Best Answer
Let $X$ be a separable infinite-dimensional Banach space (for instance $X=\ell_p$ for some $p\in [1,\infty]$). Every infinite-dimensional separable Banach space has Hamel basis of cardinality continuum; let us choose any such space $Y$ that is not isomorphic to $X$ (for instance, $c_0$ if $X=\ell_p$). As $X$ and $Y$ have bases of the same cardinality, there exists a bijective linear map $T\colon X\to Y$ which sends one basis onto the other. Define
$$\|x\|^\prime = \|Tx\|\quad (x\in X).$$
Then this is essentially the norm in $Y$, hence inequivalent as we have taken $Y$ to be non-isomorphic to $X$.