Check the following statements about closed Operator $A:X \rightarrow X$

Question

Let $X$ be a Banach space and $A:X\rightarrow X$ be a linear and closed Operator. Are the following statements true?

1. If for some $c \in \mathbb{C}$ the Operator $(A-c)$ is injective, then $(A-c)^{-1}$ is closed.

2. $\forall c \in \mathbb{C},~~cA$ is closed.

Here are my thoughts so far: I found that both statements are true.

1. Since $A$ is closed it follows that $(A-c)$ must be closed. Now since $(A-c)$ is supposed to be injective, its inverse $(A-c)^{-1}$ exists. Now per definiton $(A-c)^{-1}$ is closed if its graph $\mathcal{G}$ is closed, but $\mathcal{G}$ is nothing else than the transposed of the graph of $(A-c)$ which is closed.

2. Let $(x_{n})_{n\in \mathbb{N}}$ be a sequence in $X$ with limit $x$ and $\lim_{n\to \infty} Ax_{n} =y$. Then since $A$ is closed we have: $Ax=y$. Now consider $\lim_{n \to \infty} (cA)x_{n} = c\lim_{n\to \infty} Ax_{n} = cy=:z$. Then one finds: $(cA)x = c(Ax) = cy = z$. So the operator $cA$ is closed.

Can someone please check if my reasoning is correct? Thank you!

Answer 1

Your answer for 1) is correct but there is a logical mistake in 2): You have start with $x_n \to x$ and $(cA)(x_n) \to y$. If $c=0$ then $cA=0$ is closed. if $c \neq 0$ then $x_n \to x$ and $A(x_n) \to \frac 1 cy$. So $Ax=\frac 1 c y$ or $y =cAx$, as required.