4
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Mar 6, 2016 at 8:33
  • $\begingroup$ 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 $\endgroup$
    – user99914
    Mar 6, 2016 at 8:36
  • 1
    $\begingroup$ We may assume that $x_n$ converges to something using the closed graph theorem. $\endgroup$ Mar 6, 2016 at 8:40
  • $\begingroup$ @SebastianBechtel You are right, you may write an answer. $\endgroup$
    – user99914
    Mar 6, 2016 at 8:42

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Isn't the same argument holds for general $X$? Where did you use that $X$ is separable? $\endgroup$
    – user99914
    Mar 6, 2016 at 8:52
  • $\begingroup$ 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. $\endgroup$ Mar 6, 2016 at 8:56
  • 1
    $\begingroup$ Sorry, you are right! I have worked to much with locally convex vector spaces in recent time and there you have to take care on that point. $\endgroup$ Mar 6, 2016 at 9:16
  • 1
    $\begingroup$ I could not understand how using closed graph theorem one can say that $(x_n)$ converges somewhere in $(X,\|.\|_2)$? $\endgroup$
    – Anupam
    Mar 6, 2016 at 14:45
  • 1
    $\begingroup$ 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. $\endgroup$
    – Svetoslav
    Mar 6, 2016 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .