Equivalence between norms when in one $X$ is a Banach space I'm trying to prove next propisition:
Let $X$ be a Banach space in either of the norms $||\cdot||_{1}$ or $||\cdot||_{2}.$ Suppose that $||\cdot||_{1}\leq C||\cdot||_{2}$ for some $C>0.$ Prove that there is a $D$ with $||\cdot||_{2}\leq D||\cdot||_{1}.$
I was trying to prove that $X$ is Banach space respect both norm to use inverse mapping theorem (a consequence of Open Map Theorem) applied to identity map and get the desire inequality, until I realized that $X$ could be complete respect $||\cdot||_{2}$ but not necessarily with respect to the other.
I'm stuck proving this. Any kind of help is thanked in advanced.
 A: So there are two (or three) ways of viewing this problem:  Assuming that $X$ is complete in both the $\|\cdot\|_1$ and $\|\cdot\|_2$ norms, or with one of the norms.  But only one of these is correct.
If we consider the first case, that is where $X$ is complete with both $\|\cdot\|_1$ and $\|\cdot\|_2$, and $\|\cdot\|_1\leq C\|\cdot\|_2$ for some $C>0$, then the two norms are equivalent.  Indeed, this means the identity map $I:(X,\|\cdot\|_2)\to(X,\|\cdot\|_1)$ is bounded and bijective, hence the result follows from the open mapping theorem.
The second case is incorrect.  Indeed, consider the case $X=\ell^1$, with $\|x\|_1=\sum_k|x(k)|$ (with which $X$ is Banach) and $\|x\|_2=\left(\sum_k|x(k)|^2\right)^{1/2}$ (with which $X$ is not Banach).  Then we have $\|\cdot\|_2\leq\|\cdot\|_1$, but if we consider the sequence $(x_n)$ in $\ell_1$ defined by 
$$x_n(k)=\sum_{j=1}^n\frac{1}{j}e_j(k),$$
(where $e_j(k)=\delta_{jk}$ are the coordinate functions)  then we have 
$$\|x_n\|_2=\left(\sum_{k=1}^n\frac{1}{k^2}\right)^{1/2}\leq\frac{\pi}{\sqrt{6}}$$
for all $n$, while 
$$\|x_n\|_1=\sum_{k=1}^n\frac{1}{k}\to\infty$$
ass $n\to\infty$.  This means precisely that there is no $D>0$ such that $\|\cdot\|_1\leq D\|\cdot\|_2$.  
A: So what you have by the. Condition is that the identity map is continuous. Now since with respect to both norm, the space is Banach , the identity map is open so, the inverse identity map, which is again the identity map from $(X,||.||_{2})$ to $(X,||.||_{1})$  is continous. Now the result follows because continuity and an operator being bounded is the same thing. This is sometimes known as two-norm lemma.
