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A version of the Closed Graph Theorem states that if $T: X\rightarrow Y$ is a linear operator between Banach spaces then $T$ is bounded iff the graph of $T$ is closed in $X\times Y$.

To check if the graph is closed we suppose that a sequence $x_n \rightarrow x\in X$ and that $Tx_n \rightarrow y\in Y$ and then have to show that $x\in X$ and $Tx = y$.

My question is, what if $x_n\rightarrow x\in X$, but $Tx_n$ doesn't converge. Then the graph of $T$ is not necessarily closed, but $T$ is necessarily unbounded. Doesn't this contradict the theorem?

I'm sure I'm misunderstanding something simple - any help would be greatly appreciated!

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  • $\begingroup$ If $x_n\to x$ but $Tx_n$ does not converge, then $T$ is discontinuous, correct? $\endgroup$ – Ilya Jul 28 '14 at 15:05
  • $\begingroup$ If $x_n \to x$ and $T x_n$ does not converge, then $T$ is discontinuous, hence its graph is not closed, hence you can find a sequence $(\tilde{x}_n)$ such that $\tilde{x}_n \to \tilde{x}$ and $T\tilde{x}_n \to \tilde{y} \neq T\tilde{x}$. The point of the criterion is that you take a point $(x,y)$ in the closure of the graph, and then need to deduce that the point is in the graph. $\endgroup$ – Daniel Fischer Jul 28 '14 at 15:06
  • $\begingroup$ Thanks for the responses - but why is there necessarily a convergent sequence $(x_n)$ such that $(Tx_n)$ is also convergent? Couldn't it be the case that $(Tx_n)$ does not converge for any convergent $(x_n)$? $\endgroup$ – Sookias Jul 28 '14 at 15:15
  • $\begingroup$ To show that a set $S$ in a Banach space $X$ is closed: Assume $\{ s_{n}\} \subseteq S$ converges to $x\in X$ and show that $x \in S$. Now translate that to your problem of showing that the graph $G(T)\subseteq X\times Y$ is closed in $X\times Y$: Assume $(x_{n},Tx_{n})$ converges in $X\times Y$ to $(a,b)$, and show that $b=Ta$. In either case you start by assuming a sequence in the set converges. When dealing with a sequence in $X \times Y$, the sequence converges iff both coordinates converge. $\endgroup$ – DisintegratingByParts Jul 28 '14 at 16:10
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If there is a sequence $x_n \to x$ such that $Tx_n$ does not converge, then $T$ is unbounded, so the theorem says the graph of $T$ is not closed. And then there is some other sequence $y_n$ such that $y_n \to y$ and $Ty_n \to z$ for some $y$ and $z$ but $z \ne Ty$. There is no contradiction.

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  • $\begingroup$ Many thanks @Robert Israel, but could you please justify the existence of your other sequence $(y_n)$? I thought that $T$ being unbounded simply means that there is some sequence $y_n \rightarrow y$ such that $Ty_n$ does not converge to $Ty$. As opposed to $Ty_n$ converges but not does not converge to $Ty$. $\endgroup$ – Sookias Jul 28 '14 at 15:21
  • $\begingroup$ That's where "the graph is not closed" comes in. If $(y,z)$ is some point in the closure of the graph that is not in the graph, ... $\endgroup$ – Robert Israel Jul 29 '14 at 2:10

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