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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.

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

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  • $\begingroup$ Just one question, why does $E_A[m,m+1/n]\neq 0$ imply that there is $x_n$ such that $E_A[m,m+1/n]x_n=x_n?$ $\endgroup$ – L.F. Cavenaghi Oct 22 '18 at 22:44
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    $\begingroup$ @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 $E^2(S)=E(S)$. $\endgroup$ – Disintegrating By Parts Oct 22 '18 at 22:53
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    $\begingroup$ thank you very much! So simple! $\endgroup$ – L.F. Cavenaghi Oct 22 '18 at 22:55
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    $\begingroup$ @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 $\{0\}$. $\endgroup$ – Disintegrating By Parts May 7 '19 at 14:46
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    $\begingroup$ @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. $\endgroup$ – Disintegrating By Parts May 7 '19 at 20:56
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For a selfadjoint operator, every element of the spectrum is an approximate eigenvalue. That shows your equality.

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

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