5
$\begingroup$

Suppose $A$ is a bounded self-adjoint operator on the Hilbert space $\mathcal{H}$. How do I prove that $\sigma(A) \subseteq \overline{\{\langle Ax,x\rangle: x\in \mathcal{H},\; \lVert x\rVert = 1\}}$?

$\endgroup$
3
  • $\begingroup$ Try showing that if $\lambda\in\sigma(A)$, then there is a sequence of vectors $x_n\in \mathcal H$, $\|x_n\|=1$, and $\langle Ax_n,x_n\rangle\to\lambda$. $\endgroup$
    – Aweygan
    Nov 10, 2017 at 18:06
  • $\begingroup$ Is there any example that $\sigma(A) \subsetneq \overline{\{\langle Ax,x\rangle : x\in \mathcal{H},\; \lVert x \rVert = 1\}}$ or $\sigma(A) \not \subseteq \{\langle Ax,x\rangle : x\in \mathcal{H},\; \lVert x \rVert = 1\}$ $\endgroup$
    – maploon
    Nov 10, 2017 at 19:56
  • 1
    $\begingroup$ For both, consider $\mathcal H=\ell_2$. For the first one, let $A$ be the projection onto the space spanned by $e_1$ (all nontrivial projections have spectrum $\{0,1\}$), and the vector $x=\frac{\sqrt{2}}{2}(e_1+e_2)$. For the second, let $A$ be the diagonal operator with diagonal $\frac{1}{n}$. $\endgroup$
    – Aweygan
    Nov 10, 2017 at 20:26

1 Answer 1

5
$\begingroup$

If $\lambda$ is real and $\lambda \notin \overline{\{ \langle Ax,x\rangle : x\in\mathcal{H},\;\|x\|=1 \}}$, then the distance of $\lambda$ to the given set is some $\epsilon > 0$, which leads to the following for all non-zero $x\in\mathcal{H}$: $$ \epsilon \le\left|\frac{\langle Ax,x\rangle}{\|x\|^2}-\lambda\right|, \\ \epsilon \|x\|^2 \le |\langle (A-\lambda I)x,x\rangle| \\ \epsilon \|x\|^2 \le \|(A-\lambda I)x\|\|x\| \\ \epsilon \|x\| \le \|(A-\lambda I)x\|. $$ Therefore, $A-\lambda I$ has a bounded inverse on its range. The range of $A-\lambda I$ is dense because $$ \mathcal{R}(A-\lambda I)^{\perp}=\mathcal{N}((A-\lambda I)^*)=\mathcal{N}(A-\lambda I)=\{0\}. $$ The range of $A-\lambda I$ is closed because if $(A-\lambda I)x_n$ converges to $y$, then $\{ x_n \}$ is a Cauchy sequence by the last inequality given above, which means $(A-\lambda I)x_n$ converges to some $y$ and $\{ x_n\}$ converges to some $x$ and, finally, $$ (A-\lambda I)x=(A-\lambda I)\lim_n x_n = \lim_n (A-\lambda I)x_n = y. $$ Therefore $A-\lambda I$ is continuously invertible for such $\lambda$, which proves that $\lambda$ is in the resolvent set. All spectrum of $A$ is real, which forces $$ \sigma(A)\subseteq \overline{\{\langle Ax,x\rangle : x\in\mathcal{H}, \|x\|=1 \}}. $$

$\endgroup$
3
  • $\begingroup$ I know this is an old answer, but I think it should be mentioned that you are using the bounded inverse theorem to conclude. $\endgroup$
    – J. De Ro
    Feb 1, 2021 at 21:39
  • $\begingroup$ @QuantumSpace : No, I'm not using a theorem. I'm using a direct approach. The bound establishes the existence of an inverse on the range of $A-\lambda I$, and establishes the fact that the inverse is bounded on that range. Then I establish that the inverse is densely defined, and finally that the range of $A-\lambda I$ is closed, which makes the range all of $\mathcal{H}$. $\endgroup$ Feb 2, 2021 at 0:42
  • 1
    $\begingroup$ You are right. You don't need the bounded inverse theorem as the inequality $\epsilon \|x\| \leq \|(A-\lambda)x\|$ expresses that the inverse of $A-\lambda$ is bounded. $\endgroup$
    – J. De Ro
    Feb 2, 2021 at 8:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .