# Formula for the least element on the spectrum

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?

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

• Just one question, why does $E_A[m,m+1/n]\neq 0$ imply that there is $x_n$ such that $E_A[m,m+1/n]x_n=x_n?$ – L.F. Cavenaghi Oct 22 '18 at 22:44
• @L.F.Cavenaghi If $x_n=E_A[m,m+1/n]x \ne 0$, then $E_A[m,m+1/n]x_n = x_n$ because $E^2(S)=E(S)$. – DisintegratingByParts Oct 22 '18 at 22:53
• thank you very much! So simple! – L.F. Cavenaghi Oct 22 '18 at 22:55
• @RozaTh : $m$ could be an eigenvalue; or it could be an approximate eigenvalue only, in which case $A-mI$ would have null space equal to $\{0\}$. – DisintegratingByParts May 7 at 14:46
• @RozaTh : You know that $E[m,m+1/n] \ne 0$ for all $n > 0$. If $E\{m\}\ne 0$, then $m$ is an eigenvalue; otherwise, if it is $0$, then $E(m,m+1/n]\ne 0$ for all $n$, which makes $m$ and approximate eigenvalue. – DisintegratingByParts May 7 at 20:56

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

• nice comment. Thank you! +1 – L.F. Cavenaghi Oct 22 '18 at 22:56