Norm equivalence on $l^1$. Suppose that $\|\cdot\|$ is a norm on $l^1$ such that: 
a) $(l^1, \|\cdot\|)$ is a Banach space, 
b) for all $x \in l^1$ $\|x\|_{\infty} \leq \|x\|$. 
Prove that the norms $\|.\|$ and $\|.\|_1$ are equivalent. 
($\|\cdot\|_{\infty}$ - supremum norm and $\|\cdot\|_1$ - standard $l^1$) 
Becouse both $(l^1, \|\cdot\|)$ and $(l^1, \|\cdot\|_1)$ are Banach so it's enough to prove the existence some $M > 0$ such that $M\|x\|_1 \leq \|x\|$ or  $\|x\|_1 \geq M\|x\|$ but I don't know which one may be easier to prove.

At first I thought one or the other inequality must be obvious, but I don't see it after a little thought. Big Hint: It's trivial from the Closed Graph Theorem.
