TITLE: Can one find a stronger norm on a Banach space?
QUESTION [5 upvotes]: Given a Banach space $V$ of infinite dimension with norm $\|\cdot\|_1$, is that possible to find a norm $\|\cdot\|_2$ on $V$ such that the topology induced by $\|\cdot\|_2$ is strictly stronger than that of $\|\cdot\|_1$? 
I guess the answer is no, but I can only prove that $\|\cdot\|_2$ cannot be complete (simply  by open mapping theorem). Any idea will be appreciated.

REPLY [3 votes]: Yes, assuming the axiom of choice, one can find a stronger norm.
Use the axiom of choice to construct a linear operator $T : (V, \|\cdot\|_1) \to (V, \|\cdot\|_1)$ which is discontinuous.  Define $\|x\|_2 = \|x\|_1 + \|Tx\|_1$.  It's easy to check this is a norm, and since $\|x\|_2 \ge \|x\|_1$, it follows that the topology induced by $\|\cdot\|_2$ is at least as strong as the topology induced by $\|\cdot\|_1$.  
If the topologies are the same, then the identity map, considered as a map from $(V, \|\cdot\|_1)$ to $(V, \|\cdot\|_2)$ is a continuous linear map and hence bounded.  This means there is a constant $C$ such that for all $x \in V$ we have $\|x\|_2 \le C \|x\|_1$.  But this implies $\|Tx\|_1 \le (C-1)\|x\|_1$, contradicting the assumption that $T$ was discontinuous.
Hence $\|\cdot\|_2$ induces a strictly stronger topology than $\|\cdot\|_1$.  
If you reject the axiom of choice, then it is consistent that no stronger norm exists.  (If it did, then the identity map would be a discontinuous linear map from a Banach space to a normed space.  In a model like Shelah's or Solovay's, where every subset of every Polish space has the property of Baire, no such maps exist.)