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Let $X$ be a Banach space and let $A \colon X \to X$ be a linear operator. ($D(A) = X$) Prove of disprove that if $A$ is closed then it is necessarly bounded. (I'm having troubles in finding a connection between closedness and boundedness)

Thank you for your time and help.

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What do you mean by closed? If it's that the graph is closed, then it follows from a well-known theorem. –  Davide Giraudo Mar 17 '13 at 19:12
This is the Closed Graph Theorem, which is a quick consequence of the Open Mapping Theorem applied to the map $x \to (x,Ax)$ from $X$ to $X \times X$. –  brom Mar 17 '13 at 19:12
Thank you both, I do not know why I did not figure it out! :D –  user67133 Mar 17 '13 at 19:28

1 Answer 1

up vote 1 down vote accepted

Summarizing the comments:

A partially defined linear operator $A \colon D(A) \subset X \to Y$ is called closed if and only if its graph $\{(x,Ax) \mid x \in D(A)\}$ is closed in $X \times Y$.

Now $D(A) = X = Y$ are assumed to be Banach spaces, so you can apply the closed graph theorem.

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