Showing that a linear operator is closed 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 $(\lambda I-A)$.
 A: 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 $y = Tx$). 
