Let $\| \cdot \|_1$ and $\| \cdot \|_2$ be complete norms on a vector space $X$ (which is not finite dimensional).
I would like to show that the inequality $\| x \|_1 \le C \| x \|_2$ implies the inequality $\| x \|_2 \le D \| x \|_1$ for some $C, D$.
First of all I would like to ask what is the definition of complete norms (I tried to solve that problem thinking of them as norms with limits in $X$). I haven't found any help in my sources. Secondly I would like to ask for some help in finding the solution to the problem above.
Complete norms are norms in relation to which the space is Banach. Let $X_1$ be $X$ with norm $\|\cdot\|_1$, $X_2$ be $X$ with norm $\|\cdot\|$, then $X_1$ and $X_2$ is banach spaces. By condition $\|x\|_1\leq C\|x\|_2$. Consider the operator $A:X_2\to X_1$ such that $Ax=x$. It's clear that $D(A)=X_2$, $R(A)=X_1$, $A$ -- linear, bounded and bijective. Then exists inverse operator $A^{-1}:X_1\to X_2$, moreover, $A^{-1}x=x$ and by Banach's inverse operator theorem $A^{-1}$ is bounded. Thus $\|A^{-1}x\|_2\leq\|A^{-1}\|\|x\|_1$, therefore $\|x\|_1\geq\|A^{-1}x\|_2\|A^{-1}\|^{-1}$. Finally, $\|A^{-1}x\|_2=\|x\|_2$.