Resolvent of a Self-Adjoint Operator

Question

Let $\mathbb{H}$ be a Hilbert space and $A$ a self-adjoint operator with domain $D(A) \subset \mathbb{H}$.. Suppose that the spectrum $\sigma(A)$ of $A$ is contained in $[0,\infty)$. Let $R_{A}(z)$ be the resolvent of $A$. I am looking for an elementary proof of the fact that for every $\lambda > 0$: \begin{equation} ||R_{A}(-\lambda)|| \leq \lambda . \end{equation} I found this inequality in Hoslip & Sigal, Introduction to Spectral Theory, Sect. 5.2, where it is stated without proof. Thank you very much in advance for your help.

PS I tried the following line of proof. We have for every $\phi \in D_{A}$: \begin{equation} ||(A+\lambda) \phi||^2 = ||A\phi||^2 + 2 \lambda (A\phi,\phi) + \lambda^2 ||\phi||^2. \end{equation} So the result would immediately follow if we could prove that $(A\phi,\phi) \geq 0$, that is if we could prove that $A$ is positive. Even though I suspect this is true, I cannot see a simple way to prove this.

Answer 1

Suppose $A : \mathcal{D}(A)\subseteq \mathbb{H}\rightarrow\mathbb{H}$ is selfadjoint with $\sigma(A)\subseteq [0,\infty)$. Then, for every $\epsilon > 0$, $(A+\epsilon I)^{-1}=R_A(-\epsilon)$ is a bounded selfadjoint operator with $$ \sigma(R_{A}(-\epsilon)) \subseteq \mbox{closure}\left(\frac{1}{[0,\infty)+\epsilon}\right)=[0,1/\epsilon] \\ \sigma\left(R_{A}(-\epsilon)-\frac{1}{2\epsilon}I\right) \subseteq [-1/2\epsilon,1/2\epsilon]. $$ Norm and spectral radius are the same for a bounded selfadjoint operator. Therefore, $$ \|R_{A}(-\epsilon)-\frac{1}{2\epsilon}I\| \le \frac{1}{2\epsilon}. $$ For a selfadjoint operator such as $R_A(-\epsilon)-\frac{1}{2\epsilon}I$, the norm is determined by the quadratic form $$ -\frac{1}{2\epsilon}\le\left(R_A(-\epsilon)x-\frac{1}{2\epsilon}x,x\right) \le \frac{1}{2\epsilon} ,\;\;\; \|x\| = 1.\\ 0 \le (R_A(-\epsilon)x,x) \le \frac{1}{\epsilon}(x,x) $$ Let $x =(A+\epsilon I)y$. Only the first of the two inequalities is needed in order to conclude that $$ 0 \le (y,(A+\epsilon I)y). $$ This is true for every $\epsilon > 0$. Hence, $(Ay,y) \ge 0$ all $y\in\mathcal{D}(A)$.

Answer 2

Actually, this is a comment to the previous answer, but being too long, I post it as an answer. I realize now that the argument given by TrialAndError can be slightly simplified. From the fact that \begin{equation} \sigma(R_{A}(-\epsilon)) \subseteq [0, 1/ \epsilon], \end{equation} and the equality of norm and spectral radius for a bounded self-adjoint operator, we get \begin{equation} ||R_{A}(-\epsilon))|| \leq 1/ \epsilon, \end{equation} which is equivalent to require that for every $x \in D_{A}$ \begin{equation} ||(A+\epsilon I)x||^2 \geq \epsilon^2 ||x||^2, \end{equation} or developing \begin{equation} ||Ax||^2 +2 \epsilon (Ax,x) \geq 0, \end{equation} and by the arbitrariness of $\epsilon$ we get finally \begin{equation} (Ax,x) \geq 0. \end{equation}