# Closable unbounded operators and Banach space adjoints

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?

• yes, that is what I mean. – Dave Shulman May 13 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.