"Let $\|\cdot\|_1$ and $\|\cdot\|_2$ be two norms on the space $X$ s.t. $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$ are both complete. Assume that for any sequence $(x_n) \subseteq X$, $\|x_n\|_1 \to 0$ always implies $\|x_n\|_2 \to 0$. Prove that the two norms are equivalent, i.e. $\exists a,b>0$ s.t. $\forall x \in X$, $a\|x\|_1 \leq \|x\|_2 \leq b\|x\|_1$."
This is a problem from Erwin Kreyszig's "Introductory Functional Analysis with Applications". Sorry for the vagueness of the title; I couldn't think of an adequate title.
This problem was given as an exercise right after the section on the Open Mapping Theorem so I'm thinking that's what the author would like the reader to use. However, I don't see any connection.
If you have an idea on how OMT is related to the problem or a solution, that would be great.
Thanks in advance.
Best Answer
Consider the identity mapping $I: (X, \|\cdot\|_1) \rightarrow (X, \|\cdot\|_2)$, your given condition implies $I$ is continuous, so $\|x\|_2 \leq C\|x\|_1$
Since $I(X) = X$ is of second category in $X$, by open mapping theorem $I$ is an open mapping, which means $I^{-1} = I$ is continuous from $(X, \|\cdot\|_2)$ to $(X, \|\cdot\|_1)$, so $\|x\|_1 \leq C'\|x\|_2$