Spectrum of inverse is contained

Question

Let $A$ be a positive, self-adjoint operator in a Hilbert space such that $$\|A\|\leq 1/2.$$ Let $A^{-1}$ be its inverse (we suppose it exists). Now I want to show that $$\operatorname{spec}\left(A^{-1}\right)\subset [2, \infty).$$ From the above I know that $$\left\|A^{-1}\right\|\geq 2,$$ but then I am stuck. Can someone help?

Answer 1

First, let's not even suppose that $A$ is positive, then specialize in the obvious way. It's easy to see that if $\lambda\neq 0,$ $$\lambda\in \rho(A^{-1})\iff \frac{1}{\lambda}\in\rho (A),$$ where $\rho(A)$ is the resolvent set of $A$. By the spectral theorem, if $A$ is bounded and self-adjoint on $H$, then the spectrum is both real and contained in $[-\|A\|,\|A\|]$. So, \begin{align*}\sigma (A)&\subset [-\|A\|,\|A\|]=[-1/2,1/2]\\ &\implies (-\infty, -1/2)\cup (1/2,\infty)\subset\rho (A)\\ &\implies (-2,2)\subset\rho(A^{-1})\\ &\implies \sigma(A^{-1})\subset (-\infty,-2]\cup [2,\infty). \end{align*}

If $A$ is positive, then we clearly get what we want.