0
$\begingroup$

Here we have a self-adjoint, densely-defined operator $T$ on a Hilbert space $H$, and $E_M$ is the usual spectral projector for any Borel set $M$, i.e., $E_M=\int_M\text{d}E_t$ (this means, by definition, that $\langle E_Mu,v\rangle=\int_M\text{d}\langle E_tu,v\rangle$ for any $u,v\in H$, where we interpret this integral in the Stieltjes sense). There are various associated properties, such as $T=\int_\mathbb{R}t\text{d}E_t$, $E_t$ is left-continuous and increases as $t$ does (in the sense that $A\leq B \Leftrightarrow 0\leq B-A$), $E_sE_t=E_{\min(s,t)}$, $E_{-\infty}=0$ and $E_\infty=1$.

I would like to show that if $\lambda$ is an isolated eigen-value of $T$ then $E_{\{\lambda\}}$ has range given by the eigen-space of $T$ for $\lambda$, which is simply the null space of $T-\lambda$. I have the inclusion in one direction:

If $v\in E_{\{\lambda\}}H$ then there is $u\in H$ with $v=E_{\{\lambda\}}u$, so, for any $w\in H$, \begin{align}\langle(T-\lambda)v,w\rangle=\int_\mathbb{R}(t-\lambda)\text{d}\langle E_tv,w\rangle=0,\end{align} since \begin{equation} E_tv= E_t(E_{\lambda^+}-E_\lambda)u= \left\{ \begin{array}{lr} 0 &(t\leq\lambda);\\ v &(t>\lambda). \end{array} \right. \end{equation} Hence, $E_{\{\lambda\}}H\subseteq \ker(T-\lambda)$.

What I am stuck with is proving the converse inclusion. Any ideas are welcome!

$\endgroup$
2
$\begingroup$

The statement of the spectral theorem in my notes implies that \begin{equation*} \|(T-\lambda)u\|^2=\int|t-\lambda|^2d\langle E_tu,u\rangle \end{equation*} so that $(T-\lambda)u=0$ if and only if $\langle E_tu,u\rangle$ is constant except at $\lambda$. In other words, if and only if $E_{\{\lambda\}}u=u$.

Christer Bennewitz

PS Thanks for positive comments on my notes.

$\endgroup$
1
$\begingroup$

Solved it. :-) I shall post it here on the off chance somebody else ever has the same problem.

Conversely, suppose $(T-\lambda)v=0$. We wish to show that $E_{\{\lambda\}}v=v$. This is achieved by proving that the following representation of the spectral projector $E_{\{\lambda\}}$ is valid (it is also known as Riesz's formula):

If $\gamma$ is a simple, counter-clock-wise, closed contour enclosing only $\lambda$ from $\sigma(T)$ then \begin{equation} ~~~~~~~~~~~~~~~~~~~~~~~~~~~E_{\{\lambda\}} = \frac{1}{2\pi i} \int_\gamma (z-T)^{-1} \text{d} z ~ \big( = - \text{Res}_{z=\lambda}(R_z) \big).~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1) \end{equation} From Stieltjes' inversion formula, for any $\delta>0$ and $u\in H$ we have \begin{equation*} \langle ( E_{\lambda+\delta} - E_{\lambda-\delta} ) u , u \rangle = \lim\limits_{\varepsilon\searrow0} \frac{1}{2\pi i} \int_{\lambda-\delta}^{\lambda+\delta} \langle ( R_{\mu+i\varepsilon} - R_{\mu-i\varepsilon} ) u , u \rangle \text{d} \mu. \end{equation*} By application of the polarisation identity and the uniqueness of the Stieltjes transform, letting $0<\delta<\min_{\mu\in\sigma(T)\backslash\{\lambda\}} |\lambda-\mu|/2$ we have \begin{equation} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~E_{[\lambda-\delta,\lambda+\delta]} = \lim\limits_{\varepsilon\searrow0} \frac{1}{2\pi i} \int_{\lambda-\delta}^{\lambda+\delta} ( R_{\mu+i\varepsilon} - R_{\mu-i\varepsilon} )\text{d} \mu.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2) \end{equation} Let $\varepsilon>0$, and define $\Gamma_1=[\lambda-\delta+i\varepsilon,\lambda+\delta+i\varepsilon]$, $\Gamma_2=[\lambda+\delta+i\varepsilon,\lambda+\delta-i\varepsilon]$, $\Gamma_3=[\lambda+\delta-i\varepsilon,\lambda-\delta-i\varepsilon]$, $\Gamma_4=[\lambda-\delta-i\varepsilon,\lambda-\delta+i\varepsilon]$, and $\Gamma=\Gamma_1+\ldots+\Gamma_4$. Then $\Gamma$ is a simple, closed, clock-wise contour containing only $\lambda$ from $\sigma(T)$. From the residue theorem, it is easy to see that \begin{equation} ~~~\frac{1}{2\pi i} \int_{\lambda-\delta}^{\lambda+\delta} (R_{\mu+i\varepsilon}-R_{\mu-i\varepsilon} ) \text{d} \mu + \text{Res}_{z=\lambda} (R_z) = \frac{1}{2\pi i} \int_{-\varepsilon}^\varepsilon (R_{\lambda+\delta-i\mu}-R_{\lambda-\delta-i\mu}) \text{d} \mu.~~~~(3) \end{equation} Now, let $z\in\Gamma_2\cup\Gamma_4$. If $\mu\in\sigma(T)$, we have $|z-\mu|\geq\delta$. Then it is easily calculated that, for any $u\in H$, \begin{align*} \|R_zu\|^2 = \int_\mathbb{R} \frac{\text{d} \langle E_t u,u \rangle }{|t-z|^2} = \int_{\sigma(T)\backslash\{\lambda\}} \frac{\text{d} \langle E_t u,u \rangle }{|t-z|^2} + \frac{\langle E_{\{\lambda\}} u,u \rangle}{|\lambda-z| ^2} \leq \frac{1}{\delta^2} \int_\mathbb{R} \text{d} \langle E_t u,u \rangle = \frac{1}{\delta^2} \|u\|^2, \end{align*} i.e., $\|R_z\|\leq1/\delta$. Therefore, taking the norm of equation $(3)$ and applying the resolvent identity $R_\alpha-R_\beta=(\alpha-\beta)R_\alpha R_\beta$, we see that the RHS is not greater than \begin{align*} \frac{2\delta\varepsilon}{\pi} \max\limits_{z\in\Gamma_2\cup\Gamma_4} \|R_z\|^2 \leq \frac{2\varepsilon}{\delta\pi}. \end{align*} Sending first $\varepsilon\rightarrow0$, substituting from $(2)$ and then sending $\delta\rightarrow0$ shows that equation $(1)$ holds. The remaining proof is straight-forward: first note $Tv=\lambda v \Leftrightarrow R_z v = (\lambda-z)^{-1}v$, so that by $(1)$ and the residue theorem \begin{align*} E_{\{\lambda\}}v= -\frac{1}{2\pi i}\int_\gamma R_z v \text{d}z = \frac{1}{2\pi i}\int_\gamma \frac{ v \text{d} z}{z-\lambda} = v. \end{align*} The proof is complete.

[For anybody who is interested, this demonstrates the following statement from a set of notes on Spectral Theory written by Christer Bennewitz: Chapter 8, proof of theorem 8.3, "This implies $E_\Delta H$ has finite dimension. In particular eigen-spaces are finite-dimensional." (The notes can be found here, and are in general very lucid and pleasant to read.) Bennewitz has taken a very clean, operator-theoretic approach to the construction of the spectral projectors, using Nevanlinna/Herglotz functions and the inner product on $H$. Unfortunately this approach doesn't immediately give the machinery needed to prove the statement in my original question, whereas the alternative definition of the spectral projectors by contour integration around points in $\sigma(T)$ does. Effectively, I have simply shown that the two definitions are equivalent.]

$\endgroup$

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.