From a course based on Kreyszig's Introduction to Functional analysis.
Let $X \neq \{0\}$ denote a complex normed vector space, and assume that the operator $T : D(T) ⊂ X → X$ is closed. Let $λ ∈ C$. Show that the operator $T − λI$ is closed. I have tried applying theorem $4.13-3$, but am unsure of how to choose a sequence and showing convergence.
Note that for any $\lambda\in\mathbb{C}$, $D(T)=D(T-\lambda I)$ and both of $T$ and $I$ are closed. Also, for any two closed operators $T, F$ their linear sum is also closed.
For proof, you simply consider $T:D(T)\to X$ and $F:D(F)\to X$ be two closed operators then their corresponding graphs are closed i.e $G_T, G_F$ are closed. We now show $G_{T+F}$ is closed which will imply $T+F:D(T)\cap D(F)\to X$ defined by $(T+F)x=T(x)+F(x)$is closed.
For this let $(x,y)\in \overline{G_{F+T}}$ then $\exists$ a sequence $\{(x_n,y_n)\}_{n\in\mathbb{N}}$ in $G_{F+T}$ converging to $(x,y)$ with $y_n=(T+F)(x_n),\forall n\in\mathbb{N}\implies y_n=T(x_n)+F(x_n),\forall n\in\mathbb{N}$.
Since $G_T, G_F$ are both closed so $T(x_n)\to T(x)$, $F(x_n)\to F(x)$ as $x_n\to x$ as $n\to \infty$. Hence, $y_n=T(x_n)+F(x_n)\to T(x)+F(x)=(T+F)(x)$ as $n\to \infty$, also $y_n\to y$ is clear by definition so $y=(T+F)x$ gives $\overline{G_{T+F}}\subset G_{T+F}$. Hence our claim is proved.