Equivalence of norms given continuous identity It is known that $\parallel \; \parallel_{1}$ & $\parallel \; \parallel_{2}$ are equivalent norms over $X$ if there are $A,B>0$ such that $A\parallel x \parallel_{1} \leq  \parallel x \parallel_{2} \leq B\parallel x \parallel_{1}$ for all $x \in X$
Now, consider $i:(X, \parallel \; \parallel_{1}) \longrightarrow (X, \parallel \; \parallel_{2})$ the identity function and suppose that it is a continuous function.
Are $\parallel \; \parallel_{1}$ & $\parallel \; \parallel_{2}$ equivalent norms?
I know that it is valid if $i$ is a diffeomorphism, but I don't know how to prove that its inverse is continuous too. 
Is it necessary another further condition over $X$ aside of being a Normed Vector Space?
Thanks for any hint given

If $X$ is complete in both norms, then the open mapping theorem guarantees that the inverse is bounded, implying that the norms are equivalent. 
Otherwise, let $X=C[0,1]$, with $\|f\|_1=\max\{|f(t):\ t\in[0,1]\} $, $\|f\|_2=\int_0^1|f(t)|\, dt$ (the choice of the subscripts is unfortunate,  but I'm respecting the original notation). As $\|f\|_2\leq\|f\|_1$, the map $i$ is bounded; but its inverse is not,  as the norms are not equivalent. 
