# $X,Y$ are Banach spaces, $T$ is linear, $x_n\to 0$ and $Tx_n\to y$, then $y=0$ and $T$ is continuous.

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?

• I don't understand the stuff after "Then, ...": you seem to be showing that $y=0$ but that is one of the assumptions. – Rudy the Reindeer Oct 20 '14 at 5:27
• The title is misleading. – Did Oct 20 '14 at 6:40
• I tried to summarize what the question I am trying to prove was - sorry if it seems vague. – 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.
• 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 – user3784030 Oct 20 '14 at 5:40
• Hint: if $x_n \to x$ and $T x_n \to y$, what do you know about $x_n - x$ and $T(x_n - x)$? – Robert Israel Oct 20 '14 at 14:46
• 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 – user3784030 Oct 20 '14 at 21:00
• $T(x_n - x) \to y - T(x)$. And the "suppose that..." tells us what? – Robert Israel Oct 21 '14 at 1:07