Why is it that a closed unbounded operator can never have a closed domain?

Question

I have seen a similar question here that tries to answer this. My question is why in practice closed unbounded operators cannot have a closed domain. Is it because a closed domain would imply the domain becomes bounded? I need some clarification.

Yes I’m working with Banach spaces, but not necessarily a symmetric operator — Abel, Jun 22 '22 at 17:50

score 4 · Accepted Answer · answered Jun 22 '22 at 17:37

4

If you look at an unbounded linear operator $T:D\subseteq X \rightarrow Y$ with closed graph, where $X$, $Y$ are Banach spaces, the answer is very easy. $D$ has to be a subspace, otherwise the concept of $T$ being linear would not make sense.

If $D$ is closed, it has to be Banach. Remember that $X$ is Banach.

In that case, the closed graph theorem applies which exactly states that $T$ is continuous.

answered Jun 22 '22 at 17:37

Hyperbolic PDE friend

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That clears my confusion. +1 – Abel Jun 22 '22 at 17:53

Why is it that a closed unbounded operator can never have a closed domain?

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