If $A:D(A)\to X$ is a closed operator then $\lambda I -A$ is closed?

Question

It is said in the text that it is "easily checked" but unfortunately I don't see how this is so. I probably am missing something obvious, so it'd be great if someone could point it out!

$X$ is a Banach space, $A:D(A)\to X$ is a linear operator, then $A$ closed $\implies \lambda I - A$ closed

So let's say $\|x_n- x\|\to 0$ and $\|\lambda x_n - Ax_n -y\|\to 0$. I want to show $x\in D(A)$ and $y=\lambda x - Ax$. But I'm kind of lost at this point, since I can't see how I'd cook these up to arrive at the conclusion. Any hints are appreciated

Answer 1

As $\|x_n- x\|\to 0$ and $\|\lambda x_n - Ax_n -y\|\to 0$ we have $\|Ax_n-(\lambda x-y)\|\to 0$. So, $$\left\{\begin{align} x_n&\to x\\ Ax_n&\to \lambda x-y \end{align}\right.$$ As $A$ is closed we conclude that $x\in D(A)$ with $Ax=(\lambda x-y)$, i.e. $y=\lambda x-Ax$, as you want.

Answer 2

We have $\lVert \lambda x - Ax_n-y\rVert \leq \lVert \lambda(x-x_n)\rVert + \lVert \lambda x_n - Ax_n-y\rVert$ which implies $\lambda x-Ax_n\to y$.

Can you show that $Ax_n\to Ax$?