Consider the linear operator $A:D(A)\subset X\to X$. I want to show that $(\lambda I-A)$ is closed given that $(\lambda I-A)$ is invertible. We know that $(\lambda I-A)^{-1}$ is closed.

Now if we let $(x_{n})\subset D(\lambda I-A)$ be such that $\lim_{n\to\infty}x_{n}=0$ and $\lim_{n\to\infty}(\lambda I-A)x_{n}=x\implies x=0$, then $(\lambda I-A)$ is closable.

To me it seems that $(\lambda I-A)$ is closed if and only if $A$ is closed, since we end up with $\lambda\lim_{n\to\infty}x_{n}-\lim_{n\to\infty}Ax_{n}=x\iff\lim_{n\to\infty}Ax_{n}=x=0\iff A$ is closable. But I don't know how to use the aforementioned facts to prove the closedness of $(\lambda I-A)$.

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    $\begingroup$ How are the graphs of $\lambda I - A$ and $(\lambda I - A)^{-1}$ related? $\endgroup$ – Daniel Fischer Feb 15 '16 at 19:03

An operator is closed iff its graph is closed in $X \times X$. (A proof of this can be found below).

So from $(\lambda I - A)^{-1}$ being closed, we get that the graph

$$\left\{\begin{pmatrix}x\\y\end{pmatrix}: x\in X, y = (\lambda I - A)^{-1}x\right \}$$ is closed. This is equivalent to the 'flipped' graph $$\left\{\begin{pmatrix}y\\x\end{pmatrix}: x\in X, y = (\lambda I - A)^{-1}x\right \}$$ being closed. This is actually the graph of $(\lambda I -A)$ since $y = (\lambda I -A)^{-1}x$ implies that $y \in D(A)$ and $(\lambda I - A)y = x$. Equivalence in the initial statement yields closedness of $\lambda I - A$.

Edit: Proof of the statement above.

By definition, an operator $T:X \supset D(A) \rightarrow X$ ($X$ being a Banach space) is closed iff for every sequence $x_n \subset D(T)$ such that $x_n \rightarrow x$ and $Tx_n \rightarrow y$ for some $x,y \in X$ we have that $x \in D(A)$ and $Tx = y$.

This can be rephrased as: For every seqence $\begin{pmatrix}x_n\\Tx_n\end{pmatrix} \subset \Gamma(T)$ which converges in $X \times X$ to some $\begin{pmatrix}x\\y\end{pmatrix}$ we have that $\begin{pmatrix}x\\y\end{pmatrix} \in \Gamma(T)$ (i.e. $x \in D(T)$ and $y = Tx$).

  • $\begingroup$ The domain of $A$ is not $X$ so you can't apply the close graph theorem here. $\endgroup$ – Tsemo Aristide Feb 15 '16 at 19:18
  • $\begingroup$ @TsemoAristide But it is a subset of X, so does it not still hold? $\endgroup$ – Jason Born Feb 15 '16 at 19:20
  • $\begingroup$ It does not hold, for example takes $X=C([0,1])$ and consider the differential operator defined on the subset of differentiable functions $D([0,1])\subset C([0,1])$. $\endgroup$ – Tsemo Aristide Feb 15 '16 at 19:22
  • $\begingroup$ @TsemoAristide: I'm not using the closed graph theorem here. Only the definition of a closed operator. $\endgroup$ – Roland Feb 15 '16 at 19:30

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