Infimum of numerical range = infimum of spectrum?

Question

Let $\mathcal{H}$ be a Hilbert space and $A: D(A) \to \mathcal{H}$ a densely defined self-adjoint operator. Its numerical range is defined to be the set $\{\langle x, Ax\rangle: \|x\| =1\}$. My question is: does the following identity hold? $\inf\{\langle x, Ax\rangle: \|x\| =1\} = \inf \sigma(A)$, where $\sigma(A)$ is the spectrum of $A$? I believe it does, but couldn't prove myself. I found a lot of results on the internet about the relations between numerical range and spectrum, but did not find the one I am asking about.

Answer 1

One inequality is relatively easy, as it can be proved without relying on the the spectral theorem.

Assume $$ m=\inf\{\langle Ax,x\rangle \,:\, x\in D(A),\,\|x\|=1\}$$ and $m\neq -\infty.$ For $x\in D(A)$ we have $$\|(A-\lambda I)x\|\|x\|\ge \langle (A-\lambda I)x,x\rangle\\ \ge (m-\lambda)\|x\|^2$$ Thus $$\|(A-\lambda I)x\|\ge (m-\lambda)\|x\|$$ Hence $A-\lambda I$ is invertible for $\lambda<m,$ i.e. $\inf\sigma(A)\ge m.$ Clearly the inequality holds if $m=-\infty.$

The converse inequality can be proved by relying on the spectral theorem. We have $$A=\int\limits_a^bt\,dE(t),$$ where $a=\inf\sigma(A)$ and $b=\sup\sigma(B).$ Then $$\langle Ax,x\rangle=\int\limits_a^bt\,d\langle E(t)x,x\rangle \ge a \int\limits_a^b\,d\langle E(t)x,x\rangle =a\|x\|^2$$ Hence $m\ge a.$

Remark For bounded self-adjoint operators the proof can be based on the property that the spectral radius of $A$ is equal $\|A\|$ or by using the notion of the square root of positive operators. So the spectral theorem can be avoided.