Resolvent of a Self-Adjoint Operator 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.
 A: 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 $y\in\mathcal{D}(A)$.
A: 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}
