# Equivalence of complete norms

Let $\|.\|_1$ and $\|.\|_2$ be two complete norms on a linear space $X$ such that if a sequence $(x_n)$ converges to $x$ in $(X,\|.\|_1)$ and to $y$ in $(X,\|.\|_2)$, then $x=y$. We have to prove that $\|.\|_1$ and $\|.\|_2$ are equivalent norms.

I know that if there exists $K>0$ such that $\|x\|_1\leq K\|x\|_2$ for all $x\in X$, then by the consequence of open mapping theorem, two norms are equivalent. But how to get this from the available information? Please suggest!

• Is $X$ separable or does this furthermore hold for nets? Then the assumption just asserts that the Identity is continuous in $0$ and hence bounded. – Sebastian Bechtel Mar 6 '16 at 8:33
• It seems that some more works are needed? If $x_n \to 0$ in $\|\cdot\|_1$-norm, from the assumption it is not sure if $x_n$ converges in $\|\cdot\|_2$-norm to anything at all. @SebastianBechtel – user99914 Mar 6 '16 at 8:36
• We may assume that $x_n$ converges to something using the closed graph theorem. – Sebastian Bechtel Mar 6 '16 at 8:40
• @SebastianBechtel You are right, you may write an answer. – user99914 Mar 6 '16 at 8:42

We show that the identity is continuous in $0$ and hence bounded which is just the claim.
Let $(x_n)$ converge to $0$ in $\|\cdot\|_1$. By closed graph theorem $(x_i)$ converge to something in $\|\cdot\|_2$. By assumption this something is $0$ as well, hence $\text{id}$ is continuous in $0$.
For the sake of completeness: We have that $\text{id}: (X,\|\cdot\|_1) \to (X,\|\cdot\|_2)$ is a continuous linear map between Banach spaces. By continuous inverse theorem its inverse map $\text{id}: (X,\|\cdot\|_2) \to (X,\|\cdot\|_1)$ is continuous as well which yields the other direction of norm equivalence.
• Isn't the same argument holds for general $X$? Where did you use that $X$ is separable? – user99914 Mar 6 '16 at 8:52
• In general spaces it is not sufficient just to consider sequences. If $X$ is separable, then further metrizable and hence sequences suffice. If we consider nets then we may deal with arbitrary topological spaces. – Sebastian Bechtel Mar 6 '16 at 8:56
• I could not understand how using closed graph theorem one can say that $(x_n)$ converges somewhere in $(X,\|.\|_2)$? – Anupam Mar 6 '16 at 14:45
• I upvoted your answer, but I think it can be written much more clearly: First show that the identity operator $I:X_1\to X_2$ is a closed graph operator (here $X_i$ is $X$ normed with $\|.\|_i$). Then, because $I$ is defined on the whole space $X_1$ which is Banach, use the closed graph theorem to conclude that $I$ is bdd. – Svetoslav Mar 6 '16 at 15:58