For an unbounded operator $T:\mathcal{H}\_1 \to \mathcal{H}\_2$, if its adjoint $T^\*$ is densely defined, then we know that $T$ is closable. What happens if we replace $\mathcal{H}\_1$ or $\mathcal{H}\_2$ with a general Banach space $\mathcal{B}$? Is there some generalisation of the notion of an adjoint allowing us to analogously conclude closability?



Closable unbounded operators and Banach space adjoints
You can use essentially the same definition. If $T: E\_1 \supseteq D(T)\rightarrow E\_2$ is a linear map between Banach spaces, then we define $x^\*\in D(T^\*)$ with $T^\*(x^\*)=y^\*$ to mean that $y^\*(x) = x^\*(T(x))$ for each $x\in D(T)$.


In terms of the *graph* of the operators, this means that $(x^\*,y^\*)\in G(T^\*)$ exactly when
$$ (x,y)\in G(T) \implies x^\*(y) = y^\*(x). $$
Identify $(E\_1\oplus E\_2)^\*$ with $E\_1^\*\oplus E\_2^\*$ so the annihilator of $G(T)$ is
$$ G(T)^\perp = \{ (x^\*,y^\*) : x^\*(x)+y^\*(y)=0 \ ((x,y)\in G(T)) \}. $$
Thus $JG(T^\*) = G(T)^\perp$ where $J:E\_1^\*\oplus E\_2^\*\rightarrow E\_1^\*\oplus E\_2^\*$ is the map $J(x^\*,y^\*) = (-y^\*,x^\*)$.


We conclude:


* $T^\*$ is the graph of an *operator* when $(0,y^\*)\in G(T^\*)\implies y^\*=0$, equivalently, when $T$ is *densely defined*.
* $T^\*$ is always *closed* in the weak$^\*$-topology.
* So if $E\_1,E\_2$ are reflexive, then $T^\*$ is closed in the weak, and so norm, topology.


One can also reverse this, starting with a weak$^\*$-closed operator $E\_2^\*\rightarrow E\_1^\*$. From this, we see that $T^\*$ is densely-defined if and only if $T$ is closable. (You can also argue this directly).


If you examine the standard proof of these factors for operator on Hilbert spaces, then they are rather similar.


