$T$ unbounded operator not closed $\implies$ resolvent of T is empty?

Question

I am studying the subject of unbounded operators and I'm wondering why if an operator is not closed than his resolvent is empty. Thanks !

Answer 1

Your definition of "resolvent" apparently says that $\lambda$ is in the resolvent if $\lambda I - T$ is injective and surjective, and has a bounded inverse.

We have the following lemma:

Let $A$ be an unbounded operator with domain $D(A)$ and image $R(A)$. Suppose $A$ is injective, and consider $A^{-1}$ as an unbounded operator with domain $D(A^{-1}) = R(A)$. Then $(A, D(A))$ is a closed operator iff $(A^{-1}, D(A^{-1}))$ is.

You can give a very quick proof by looking at the graphs of $A, A^{-1}$ as subsets of $H \oplus H$ and seeing how they relate to one another.

Now in our example, if you want $\lambda$ to be in the resolvent set of $T$ then we need to have $\lambda I - T$ injective and $R(\lambda I - T) = H$. Moreover the inverse $(\lambda I - T)^{-1}$, whose domain is $H$, is a bounded and everywhere defined operator, so in particular it is a closed operator. By the lemma, $\lambda I - T$ is closed as well, and this implies $T$ is closed.