Can one find a stronger norm on a Banach space? 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.
 A: 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.)
A: Do you mean, can you always find a strictly stronger norm? Or does there exist a strictly stronger norm of some Banach norm?
In the first case, I do not know. But in the second case, we can take the example of measurable functions on an interval: $(L^1([0,1],||\cdot||_1)$ where here the $1$ doubles as your "first norm" and also the $L^1$ norm ($||f||_1 = \int|f|$) then take $||f||_2$ to be the essential supremum of |f| (i.e. $||f||_\infty$ adapted to measurable functions).
So note that $||f||_1 \leq ||f||_2$, so $||\cdot||_2$ is stronger. But here we see that for $f_n(x) = 1$ for $x\in [0,\frac{1}{n}]$ and $f_n(x) = 0$ for $x\in [\frac{1}{n}, 1]$, $||f||_1 = 1/n$ and $||f||_2 = 1$, so we can't find a $c$ such that $c||f_n||_2 \leq ||f_n||_1$ for every $n$. Thus $||\cdot||_1$ is strictly stronger.
I believe this works as an example.
A: I'm inclined to say no. The identity map from $V$ under $||\cdot||_2$ to $V$ under $||\cdot||_1$ is continuous, linear, and onto, but if we pick a set $A$ that's open in the $||\cdot||_2$-topology but not the $||\cdot||_1$-topology, then it seems the open mapping theorem is violated.
