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So this should be a very simple question. The Closed Graph Theorem as stated in Royden is Let $T\colon X \rightarrow Y$ be a linear operator between Banach spaces $X$ and $Y$. Then $T$ is continuous iff it is closed. Now, does this presuppose that the domain of $T$ is either closed, or all of $X$ (which implies it is closed?). In particular, can we only say a continuous linear operator is a closed linear operator if we are considering a closed domain?

For example, consider the identity function in $\Bbb R$ from $(-1,1)\to\Bbb R$. We can take $x_n \rightarrow 1$ such that $x_n \in \Bbb R$ for all $n$, clearly $Tx_n \rightarrow 1 \in\Bbb R$ but $1$ is not in $(-1,1)$ so $T$ is not closed. Is this incorrect?

Basically, all the texts I have looked at have not been explicit about whether we are presupposing the domain to be closed. Also, I keep finding statements like all closed functions are continuous, i.e. that closed operators include continuous operators, but I feel like the above example excludes this. Can somebody verify if I am right? And if I am not, can you explain what is wrong? Thank you!

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  • $\begingroup$ Your example is not correct: the graph of $T$ on $(-1,1)$ is closed in $(-1,1) \times \mathbb R$ (you're probably thinking it isn't closed in $\mathbb R \times \mathbb R$, which is true but irrelevant). Also it isn't a linear operator since it isn't defined on a vector space (you can't verify that $f(0.5)+f(0.5) = f(1)$). $\endgroup$ – Erick Wong Apr 12 '13 at 16:00
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    $\begingroup$ Note that "unbounded operators"--commonly seen in physics--are still usually assumed to have closed graph. But they can be unbounded because the domain is not closed. $\endgroup$ – GEdgar Apr 12 '13 at 16:15
  • $\begingroup$ To Erick, if that is true then I guess my misunderstanding is about what closure is. Take (-1,1)XR. Doesn't (n/(n+1),0) converge to (1,0) which is not in (-1,1)XR? That is a limit point not contained in the set and thus not closed? $\endgroup$ – Fractal20 Apr 12 '13 at 17:09
  • $\begingroup$ Err now I just re-read what you said about it not being closed in RxR, but the closed graph theorem is about it's closure in the XxY, and F:R->R. Perhaps my confusion is that for example I thought a function F:R->R doesn't mean f is defined on all of R. Or else, we need never think of domain F since it would already be given. Is this wrong? $\endgroup$ – Fractal20 Apr 12 '13 at 17:23
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In the Closed Graph Theorem, it is assumed that $T$ is defined on all of $X$. Another quite explicit way of stating the theorem is to say

If $T$ is a linear map defined on a Banach space $X$ and taking values in a Banach space $Y$ for which it is true that the set $$\{(x, Tx) \in X \times Y: x \in X\}$$ is closed in $X \times Y$ then $T$ is bounded.

Sometimes the definition of closed operator is used when stating the theorem. In this case suppose that $X$ and $Y$ are Banach spaces and that $D$ is a subset of $X$. We say that a linear map $T:D \to Y$ is closed if the set $$\{(x,Tx) \in D \times Y: x \in D\}$$ is closed in $X \times Y$.

Then the statement of the Closed Graph Theorem is that

A closed linear map defined on all of $X$ is bounded.

Finally, if an operator is closed, this does not imply that the domain of the operator is closed. As an example take the "weak Laplacian" $\Delta: L^2 \to L^2$ with domain $H^1$.

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  • $\begingroup$ So the definition of a closed linear operator given in my class matches the one here on wikipedia: en.wikipedia.org/wiki/…. Which has as a criteria that $x_n \rightarrow x$ and $x_n \in X$ means that x is in D(T) if T is closed. This is incorrect? Or perhaps my misunderstanding is resulting from restricting a linear operator to be on some arbitrary subset that is not necessarily a vector space. Do we only consider a linear operator if it is defined on a linear space? $\endgroup$ – Fractal20 Apr 12 '13 at 17:18
  • $\begingroup$ So reading your post again, I don't think I have a problem with either of those definitions. I do still have a problem with the Royden representation in my original post. Could you explain that. That is, my attempted counter example I think is fine with the two definitions you gave, but not with the Royden one. That is, could explicitly elaborate on if I am correct that a continuous linear operator is closed only if it's domain is closed? $\endgroup$ – Fractal20 Apr 12 '13 at 17:42
  • $\begingroup$ @Fractal20 No, the the domain must not be closed for the operator to be closed. The answer gives a counterexample. $\endgroup$ – user38355 Apr 14 '13 at 21:28
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In general, it is always true that a continuous function is closed if the codomain $Y$ is a Hausdorff space and $X$ any topological space.

The converse holds when $X$ and $Y$ are Banach spaces. This is precisely the content of the Closed Graph theorem. Nothing more is needed...

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  • $\begingroup$ Well isn't R Hausdorff and (-1,1) a topological space in my example? $\endgroup$ – Fractal20 Apr 12 '13 at 17:19
  • $\begingroup$ So in light of the other replies, is it correct that the we are assuming the continuous functions do be defined on all of X? $\endgroup$ – Fractal20 Apr 12 '13 at 17:53

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