6
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ @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 $E^2(S)=E(S)$. $\endgroup$ – DisintegratingByParts Oct 22 '18 at 22:53
  • 1
    $\begingroup$ thank you very much! So simple! $\endgroup$ – L.F. Cavenaghi Oct 22 '18 at 22:55
  • 1
    $\begingroup$ @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 $\{0\}$. $\endgroup$ – DisintegratingByParts May 7 at 14:46
  • 1
    $\begingroup$ @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. $\endgroup$ – DisintegratingByParts May 7 at 20:56
2
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.