Are Banach space norms (up to equivalence) unique? Here is a naive question: is a "completing" norm of a vector space unique (up to equivalence) or can one find a vector space and two non-equivalent norms $\|.\|$ and $|||.|||$ that both induce a Banach space norm? That should be a classic question, but I do not find anything in text books.
Cheers, Bernhard
 A: I learned the following "construction" in the article "Equivalent complete norms and positivity." from Arendt and Nittka. On a Banach space $(X,\lVert\cdot\rVert)$, take an unbounded functional $\varphi$ and a point $y\in X$ such that $\varphi(y)=1$ and define the operator $S:X\to X$ by $Sx := x - 2\varphi(x)y$. Then you can check that $S^2=I$ and $\lvert x \rvert := \lVert Sx\rVert$ defines a complete norm on $X$ which is not equivalent to the $\lVert\cdot\rVert$-norm.
Of course there are many such norms, you might also have a look at this question.
A: Take your two favorite Banach spaces of reasonable size. They are going to have algebraic bases of the same cardinal (traditionally the cardinal of the continuum) and hence are going to be isomorphic as vector spaces. This provides an example of a Banach space with two non equivalent norms inducing Banach space topology.
One the other hand, note that if the two norms are comparable then the open mapping theorem implies that they are equivalent (the identity is a continuous surjection, hence an open map, hence its inverse is continuous)
Also if I remember correctly, the construction of discontinuous map between Banach spaces always require the axiom of choice, hence there will be no explicit counterexample.
