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?

  • $\begingroup$ yes, that is what I mean. $\endgroup$ May 13 '19 at 11:50

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.


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