Q:

Is a linear closed and everywhere defined operator bounded?

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.

A:

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.

