Let $(X,\|\cdot\|_{1})$ and $(X,\|\cdot\|_{2})$ be complete normed vector spaces and $\|x\|_{1}\le\|x\|_{2}$ $\forall x\in X$. I want to prove that $\exists M>0$ such that $\|x\|_{2}\le M\|x\|_{1}$.
We know that any Cauchy sequence $(x_{n})_{n\in\mathbb{N}}\subset X$ converges. And of course, $\frac{1}{M}\|x\|_{1}\le\|x\|_{1}\le M\|x\|_{1}$.
Does anyone have any hints on how I can proceed? I assume that I have to use this completeness property somehow.
The identity mapping $I : (X,\|\cdot\|_2) \to (X,\|\cdot\|_1)$ is a bijection and by assumption is continuous. Since the spaces are complete you can appeal to the open mapping theorem. The inverse is also continuous.