Here is the question I have:

Let $X,Y$ be Banach spaces and $T:X\to Y$ be linear. Suppose that whenever $x_n\to 0$ and $Tx_n\to y$, then $y=0$. Show that $T$ is continuous.

So this is what I have:

Let $x_n\in X$ such that $x_n\to 0$. Also, let $T:X\to Y$ be linear. Finally, let $Tx_n\to y$. Then, as $Tx_n\to y$ we know that for $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\|Tx_n-y\|\lt\epsilon, \forall n\geq N$. As $x_n\to 0$, $lim_{n\to\infty}\|Tx_n-y\|=\|T(0)-y\|\lt\epsilon$

Where I am stuck is how show that $y=0$, as after that I can use the closed graph theorem to prove that $T$ is continuous. Suggestions?

  • $\begingroup$ I don't understand the stuff after "Then, ...": you seem to be showing that $y=0$ but that is one of the assumptions. $\endgroup$ – Rudy the Reindeer Oct 20 '14 at 5:27
  • $\begingroup$ The title is misleading. $\endgroup$ – Did Oct 20 '14 at 6:40
  • $\begingroup$ I tried to summarize what the question I am trying to prove was - sorry if it seems vague. $\endgroup$ – user3784030 Oct 20 '14 at 6:43

Your statement $y = \|Tx_n - T(0)\| = \ldots$ is nonsense. $y$ is a member of the Banach space $Y$, $\|T x_n - T(0)\|$ is a real number; they can't be equal.

In the hypotheses of the closed graph theorem, $x_n \to x$ and $T x_n \to y$ implies $y = Tx$. You only have this conclusion in the case $x = 0$. In order to be able to apply the closed graph theorem, you need it it for all $x$. So you have something still to prove.

  • $\begingroup$ Thanks for the response, and sorry about that... I know that by the linearity of $T$ that for the closed graph theorem to be applied, I only have to show that if $x_n\to 0$ and $Tx_n\to y$, then $y=0$... I am just very confused how I would show this. Sorry if this is a very simple issue, but I am struggling with the more basic aspects of these proofs. @Robert $\endgroup$ – user3784030 Oct 20 '14 at 5:40
  • $\begingroup$ I have edited the above - any suggestions where to go from my edits? @Robert $\endgroup$ – user3784030 Oct 20 '14 at 6:39
  • $\begingroup$ Hint: if $x_n \to x$ and $T x_n \to y$, what do you know about $x_n - x$ and $T(x_n - x)$? $\endgroup$ – Robert Israel Oct 20 '14 at 14:46
  • $\begingroup$ If $x_n\to x$ and $Tx_n\to y$ then $x_n-x\to 0$ and $T(x_n-x)\to T(0)$? @Robert $\endgroup$ – user3784030 Oct 20 '14 at 21:00
  • $\begingroup$ $T(x_n - x) \to y - T(x)$. And the "suppose that..." tells us what? $\endgroup$ – Robert Israel Oct 21 '14 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.