Formula for the least element on the spectrum

Question

Let $A$ be a self-adjoint operator defined on a dense subset of an Hilbert space $\mathcal{H}$. Assume that $A$ is bounded below in the sense there is $m \in \mathbb{R}$ such that $$\langle Ax,x\rangle \geq m,~\forall x : \|x\| = 1.$$

I want to show that: $$ m = \inf\{\lambda : \lambda \in \sigma(A)\} = \inf \{\langle Ax,x\rangle : \|x\| = 1\}.$$

I know that if $E_A$ denotes the unique spectral measure that represents $A$, then $\mathrm{supp}~E_A = \sigma(A),$ from which follows the first equality. So, it is only left to prove the last equality. Any hints?

Thanks in advance.

Answer 1

Let $m=\inf\;\{ \lambda : \lambda\in\sigma(A) \}$. Then, for every positive integer $n$, $E_{A}[m,m+1/n] \ne 0$. So there exists a unit vector $x_n\in\mathcal{D}(A)$ such that $E_{A}[m,m+1/n]x_n = x_n$, which gives \begin{align} 0 & \le \langle (A-mI)x_n,x_n\rangle \\ & = \int_{m}^{m+1/n}(\lambda-m) d\langle E(\lambda)x_n,x_n\rangle \\ & \le \frac{1}{n}\langle E[m,m+1/n]x_n,x_n\rangle \\ & \le \frac{1}{n}\langle x_n,x_n\rangle = \frac{1}{n}. \end{align}

Therefore, $\lim_n \langle A x_n,x_n\rangle = m$.

Answer 2

For a selfadjoint operator, every element of the spectrum is an approximate eigenvalue. That shows your equality.