0

Let $T:\mathfrak{D}(T)\subseteq X\rightarrow X$ be a linear operator where $X$ is a Banach space. We define the resolvent set of $T$, denoted $Res(T)$ to be the set consisting of all $\lambda \in \mathbb{C}$ such that $T-\lambda I$ is bijective.

Suppose $\lambda \in Res(T)$ , $T$ is closed. Then, $(T-\lambda I)^{-1} : X\rightarrow \mathfrak{D}(T)$ is closed. By why does this imply that $(T-\lambda I)^{-1}$ is bounded?

Is $\mathfrak{D}(T)$ closed in this case? If so, then we can just invoke the Closed Graph Theorem. For then, $\mathfrak{D}(T)$ would be Banach, too.

I suppose to make sense of the above, one regards $(T-\lambda I)^{-1}$ as an operator $X\rightarrow X$. In such a case, it suffices to note the following:

If $T:X\rightarrow Y'$ is a closed linear map and $Y'$ is a subspace of $Y$ then $T:X\rightarrow Y$ is closed too?

  • 1
    As you write, consider $(T-\lambda)^{-1}$ as a map from $X$ to $X$ (a map from $X$ to a subspace of $X$ is naturally also a map from $X$ to $X$). Then the boundedness of $(T-\lambda)^{-1}$ follows from the closed graph theorem. – Janik Apr 04 '22 at 13:47

1 Answers1

1

The closed graph theorem can be applied in the following way: $T$ closed $\implies$ $T - \lambda I$ closed $\implies$ the graph of $T - \lambda I$ is closed. Since $\lambda \in \operatorname{Res}(T)$, $T - \lambda I$ is invertible and hence, $(T - \lambda I)^{-1}$ is defined on all of $X$ and since $T - \lambda I$ has closed graph, then so does $(T - \lambda I)^{-1}$ and the closed graph theorem gives that an operator with domain $X$ and closed graph is bounded.

Regarding your attempt, there exist closed unbounded operators with domains that are not closed.

akkapi
  • 461
  • Yes, $T:C^1[0,1]\rightarrow C[0,1]$ given by $Tf=f'$ is closed but unbounded with the $||.||_{\infty}$ norm –  Apr 04 '22 at 16:04