why $\lambda \Vert x \Vert \leq \Vert (\lambda-A)x\Vert$ imply $R(\lambda-A)$ is a closed subspace of $\mathscr{X}$ and $\lambda-A$ is 1-1 on $D(A)$?

Question

Suppose $\mathscr{X}$ is a Hilbert space. If $A$ is a densely closed operator and is a generator of a strongly continuous contraction semi-group, if and only if,

$(1)Re<x^*,A>\leq 0$

$(2)\exists\lambda_{0}>0, s.t. R(\lambda_{0}-A)=\mathscr{X}$

I try to prove the sufficiency. But why $\lambda \Vert x \Vert \leq \Vert (\lambda-A)x\Vert$ imply $R(\lambda-A)$ is a closed subspace of $\mathscr{X}$ and $\lambda-A$ is 1-1 on $D(A)$?

Answer 1

First observe that a closed linear operator $T: X \to Y$, $X,Y$ Banachspaces, with $\Vert Tx \Vert \geq C \Vert x \Vert$ has a closed range: Let $Tx_n \to y$. We have to show that $y \in R(T)$. $(x_n)_n$ is a Cauchy-seqeunce in $X$, because $\Vert x_n - x_m\Vert \leq \frac{1}{C} \Vert Tx_n - Tx_m \Vert \to 0$ as $m, n \to \infty$. Let $x := \lim_{n \to \infty} x_n$.
As $T$ is a closed operator we get $y = \lim Tx_n = T(\lim x_n) = Tx$. So $y \in R(T)$ and therefore $R(T)$ is closed.

Back to the question: With the above observation we have shown that $R(\lambda - A)$ is closed in $H$ (If $A$ is closed then $(\lambda - A)$ is also closed). The assumption that $\Vert (\lambda - A)x \Vert \geq \lambda \Vert x \Vert$ implies that $(\lambda - A)$ is injective. The only $x \in \ker(\lambda - A)$ is $x = 0$, otherwise the inequality cannot be fulfilled. So $(\lambda - A)$ is 1-1 on $D(A)$.