Let $T:H\to \mathbb{C}^m$ a densely defined operator with $H$ Hilbert space. Is it true that if $T$ is closable, then $T$ is bounded?

For example the differential operator $ T:C^1[0,1]\subset C[0,1]\rightarrow C[0,1]$ is closed, densely defined but unbounded.

But what about if the codomain is $\mathbb{C}^m?$ In fact, that space is complete so can I extend $T$ to an operator $T_c$ such that $T_c$ is closed?

## Best Answer

As shown here How common is it for a densely-defined linear functional to be closed? all densely-defined, closed functionals are bounded. This means that all coordinates of $T: H\rightarrow \mathbb{C}^m$ are bounded and hence $T$ itself is bounded.

Now if $T$ is closeable, then its closure is bounded and hence $T$ itself is bounded too.