Unbounded Linear operator in a closed domain

Question

I am trying to understand the Closed Graph Theorem and so I would like to see an example of an unbounded linear operator $\hat{T}: \mathscr{D}(\hat{T})\rightarrow Y$ where $\mathscr{D}(\hat{T})\subset X$ is a closed domain and $X$ and $Y$ are Banach spaces. In the Closed Graph theorem the domain $\mathscr{D}(\hat{T})$ and the graph $\mathscr{G}(\hat{T})$ must be closed and so I would like to see the necessity of $\mathscr{G}(\hat{T})$ being closed. The main doubt I have is if an unbounded linear operator is allowed to be defined in a closed domain.

Thanks.

Answer 1

Define $T(e_k)=kf$, where $\{e_k\}$ are linear independent in $X$ with $|e_k|=1$, and $f$ is a fixed nonzero vector in $Y$. For other $e$ that cannot represented by $\{e_k\}$, define $T(e)=0$. Then $T$ is defined on the whole space $X$, and it is obviously unbounded.

Answer 2

The closed graph theorem is precisely the reason that it is "hard" to construct an everywhere defined unbounded operator (say acting between Banach spaces). I heared that there is a serious theorem which states that you cannot construct unbounded $T:E \to E$ where $E$ is Banach without axiom of choice (but I don't know where you can find the proof of this fact): please note that in the solution given by Akatsuki you need axiom of choice (since it is equivalent to the existence of linear basis). So in practise I believe that every unbounded operator which you will meet will be not everywhere defined (it's domain would not be closed).