Let $T$ be a linear densely defined operator with $T:H\rightarrow \mathbb{C^m}$ where $H$ is a Hilbert space.
Prove that the operator T is bounded if and only if it is closeable.
What I have tried:
I already have the answer for the implication where you get that T is closeable by implying that T is bounded. Here I used a Theorem that was demonstrated in my Spectral Theory class, which says that if a linear operator T is continuous then it is closable. On the other hand, I am lost with the other implication. This is what I know and the idea I had:
If T is closeable then it has an extension that is closed. Let's call the extension $\overline{T}$. With this, we have that $\overline{T}$ is closed, so by the Closed Graph theorem, we know that $\overline{T}$ is continuous. Here is where I am stuck as I don't know how to proceed and demonstrate that T is also continuous. So if someone can help me I would appreciate it very much, as I am very lost.
P.D. The main questions that I came while thinking about the problem are:
- Is the extension $\overline{T}$ unique ?
- If it is unique could I use somehow the B.L.T Theorem to justify the continuity of T with the continuity of $\overline{T}$?