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Given a Hilbert space $\mathcal{H}$.

Consider a normal operator $N:\mathcal{D}(N)\to\mathcal{H}$.

The goal here is to prove: $$\langle\sigma(N)\rangle=\mathcal{W}(N)$$

By a previous result one has: $$\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$ (Rigorous treatment: Normal Operators: Spectrum vs. Numerical Range)

As the numerical range is convex this proves the one inclusion.

Now, for a self-adjoint operator one has: $$\inf\sigma(A)\leq\left(\langle\hat{\varphi},A\hat{\varphi}\rangle=\right)\int_{\sigma(A)\subseteq\mathbb{R}}z\mathrm{d}\nu_{\hat{\varphi}}(z)\leq\sup\sigma(A)$$ As the spectrum is closed this would prove the other inclusion.

But how to adapt this argument to normal operators?

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The key property here is that for probability measures: $$\int_\Omega f\mathrm{d}\rho\in\overline{\langle f(\Omega)\rangle}$$

As the spectrum is closed it holds: $$\int_{\sigma(N)}z\mathrm{d}\nu_{\hat{\varphi}}\in\langle\sigma(N)\rangle$$ (For a unit vector it is a probability measure.)

Concluding that the other inclusion holds, too: $$\overline{\mathcal{W}(N)}\subseteq\langle\sigma(N)\rangle$$


By definition, for bounded $f\geq0$ we have $$ \int_{\sigma(N)}f\,d\nu_\hat\varphi=\sup\left\{\sum_j \alpha_j\,\nu_\hat\varphi(\Delta_j):\ \bigcup_j\Delta_j=\sigma(N), \text{ and }\sum_j \alpha_j\,\chi_{\Delta_j}\leq f\right\}. $$ So we can choose a sequence $\{s_n\}$ of simple functions with $s_n\nearrow f$ uniformly (this is achievable because $f$ is bounded). By choosing a subsequence if necessary, we may assume $f-s_n<2^{-n}$, say. If $s_n=\sum_j\alpha_j\,\chi_{\Delta_j}$, then $f(t)-\alpha_j<2^{-n}$ a.e. on $\Delta_j$. So if we choose one such a $t$ for each $\Delta_j$, we have that $f-\sum_jf(t_j)\,\chi_{\Delta_j}<2^{-n+1}$.

Note that $\sum_j\nu_\hat\varphi(\Delta_j)\,f(t_j)$ is a convex combination of $f(t_1),\ldots,f(t_r)$. So $\int_{\sigma(N)}f\,d\nu_\hat\varphi$ is a limit of convex combinations of points in $f(\sigma(N))$.

For complex $f$, we can write it as $f_1-f_2+i(f_3-f_4)$ with $f_1,\ldots,f_4\geq0$, and we define $$ \int_{\sigma(N)}f\,d\nu_\hat\varphi=\int_{\sigma(N)}f_1\,d\nu_\hat\varphi-\int_{\sigma(N)}f_2\,d\nu_\hat\varphi+i\left(\int_{\sigma(N)}f_3\,d\nu_\hat\varphi-\int_{\sigma(N)}f_4\,d\nu_\hat\varphi\right). $$ So $ \int_{\sigma(N)}z\,d\nu_\hat\varphi(z)$ is a limit of convex combinations of convex combinations $\sum_jt_j\,\nu_\hat\varphi(\Delta_j)$, with $t_j\in\sigma(N)$.

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  • $\begingroup$ You mean of convex sums sort of? The only thing which came to my mind sort of mean value theorem corresponding to convex sets... Can you explain what you meant? $\endgroup$ – C-Star-W-Star Nov 9 '14 at 20:20
  • $\begingroup$ Ok is it true that $\int_\Omega fd\rho\in \langle f(\Omega)\rangle$? $\endgroup$ – C-Star-W-Star Nov 9 '14 at 20:26
  • $\begingroup$ Of course. By definition, $\int_\Omega f\,d\rho$ is a limit of sums $\sum_j\rho(\Delta_j)\,f(t_j)$, with $\sum_j\rho(\Delta_j)=\rho(\Omega)=1$. $\endgroup$ – Martin Argerami Nov 9 '14 at 21:06
  • $\begingroup$ But that is for the Riemann integral, or? $\endgroup$ – C-Star-W-Star Nov 9 '14 at 21:07
  • $\begingroup$ How do you define the Lebesgue integral, then? $\endgroup$ – Martin Argerami Nov 9 '14 at 21:08
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Suppose that $T$ is normal with spectral measure $E$. Let $\lambda_{1},\lambda_{2},\cdots\lambda_{n}\in\sigma(T)$, and suppose that $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$ are real scalars in $[0,1]$ for which $\sum_{j}\alpha_{j}^{2}=1$. Let $\epsilon > 0$ be small enough that the closed spheres of radius $\epsilon$ centered at $\lambda_{j}$ are disjoint. Then there are mutually orthogonal unit vectors $x_{j}$ such that $E(B_{\epsilon}[\lambda_j])x_{j}=x_{j}$, where $B_{\epsilon}[\lambda_{j}]$ is the closed sphere of radius $\epsilon$ centered at $\lambda_{j}$. Then, because every $x_{j}\in\mathcal{D}(T)$, the following expression is well-defined: $$ (T\sum_{j}\alpha_{j}x_{j},\sum_{j}\alpha_{j}x_{j})=\sum_{j}\alpha_{j}^{2}(Tx_{j},x_{j}). $$ One has $(Tx_{j},x_{j})\approx \lambda_{j}$, and the smaller the $\epsilon$ that is chosen at the beginning, the closer this approximation can be made. It follows that $\sum_{j}\alpha_{j}^{2}\lambda_{j} \in \overline{\mathcal{W}(T)}$. Therefore, $$ \overline{\mbox{co}(\sigma(T))}\subseteq\overline{\mathcal{W}(T)}, $$ where $\mbox{co}(\sigma(T))$ is the convex hull of $\sigma(T)$.

Suppose that $T$ is bounded. Then the spectrum $\sigma(T)$ is closed and bounded. Hence, the closed convex hull of $\sigma(T)$ is the intersection of all closed half planes containing $\sigma(T)$. Every such half plane has the form $\lambda_{0}+e^{i\alpha}\{\lambda : \Re\lambda \ge 0\}$ for some $\lambda_{0}\in\mathbb{C}$ and some $\alpha \in [0,2\pi]$. Such a half plane contains the spectrum of $T$ iff $\sigma(e^{-i\alpha}(T-\lambda_{0}))\subseteq\{ \lambda : \Re\lambda \ge 0 \}$, which is equivalent to $\Re(e^{-i\alpha}(T-\lambda_{0}I))\ge 0$. Hence, $\mathcal{W}((e^{-i\alpha}(T-\lambda_{0}I)) \subseteq \{ \lambda :\Re\lambda \ge 0\}$, which guarantees $$ \mathcal{W}(T)\subseteq\lambda_{0}+e^{i\alpha}\{\lambda :\Re\lambda \ge 0\}, $$ for any bounded normal $T$. Hence, $\mathcal{W}(T) \subseteq\overline{\mbox{co}(\sigma(T))}$ for such $T$ because $\mathcal{W}(T)$ is contained every closed half plane containing $\sigma(T)$. Combining this result with that of previous paragraph gives $$\overline{\mathcal{W}(T)} = \overline{\mbox{co}(\sigma(T))}$$ for any bounded normal $T$. These arguments do not directly extended to the case of unbounded $T$ because of domain issues that can be avoided using these arguments.

To lift the restriction that $T$ is bounded, consider the restriction $T_{r}$ of $T$ to the subspace $\mathcal{H}_{r} = E\{ \lambda : |\lambda| \le r \}$ for $r > r_{0}$, where $r_{0}$ is large enough that $\mathcal{H}_{r_{0}} \ne \{0\}$. For such $r$, the spectrum of $T_{r}$ satisfies $$ \sigma(T)\cap\{ \lambda : |\lambda| < r \} \subseteq \sigma(T_{r}) \subseteq \sigma(T)\cap\{ \lambda : |\lambda| \le r \}. $$ And $\mathcal{W}(T_{r})\subseteq\mathcal{W}(T)$ because $\mathcal{H}_r \subset\mathcal{D}(T)$ for all finite $r$. There is equality of $$ \overline{\mbox{co}(\sigma(T_{r}))}=\overline{\mathcal{W}(T_{r})}. $$ Because every unit vector $x \in \mathcal{H}$ is a limit of $x_{r}=E\{\lambda : |\lambda| \le r\}x$ as $r\uparrow\infty$, then $$ \lim_{r\uparrow\infty} (T_{r}x_{r},x_{r})/\|x_{r}\|^{2} = (Tx,x),\;\;\; x\in\mathcal{D}(T), \\ \implies \mathcal{W}(T) \subseteq \overline{\bigcup_{r > r_{0}}\mathcal{W}(T_{r})} \subseteq \overline{\mathcal{W}(T)}, \\ \implies \overline{\mathcal{W}(T)}=\overline{\bigcup_{r > r_{0}}\mathcal{W}(T_{r})}, \\ \implies \overline{\mathcal{W}(T)} = \overline{\bigcup_{r > r_{0}}\overline{\mathcal{W}(T_{r})}} =\overline{\bigcup_{r > r_{0}}\overline{\mbox{co}(\sigma(T_{r}))}} $$ Because of the first spectral inclusion at the beginning of this paragraph, every convex combination of points in $\sigma(T)$ is in $\sigma(T_{r})$ for some $r > r_{0}$. Therefore, $$ \mbox{co}(\sigma(T))=\bigcup_{r > r_{0}}\mbox{co}(\sigma(T_{r})). $$ It follows that $$ \overline{\mbox{co}(\sigma(T))}=\overline{\bigcup_{r > 0}\overline{\mbox{co}(\sigma(T_{r}))}} = \overline{\mathcal{W}(T)}. $$

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