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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $H^1$.

  • $\begingroup$ 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

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...

  • $\begingroup$ 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

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.