Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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 –  user01123581321345589144... 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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.