Given an operator algebra $\mathcal{A}\subseteq\mathcal{B}(\mathcal{H})$ with $1\in\mathcal{A}$

Consider selfadjoint operators $A=A^*\in\mathcal{A}$.

Define positive elements by: $$A\geq0:\iff\sigma(A)\geq0$$ and positive operators by: $$A\geq0:\iff\mathcal{W}(A)\geq0$$


Do the numerical range and spectrum coincide: $$A=A^*:\quad\langle\sigma(A)\rangle=\overline{\mathcal{W}(A)}$$


For bounded operators one has at least: $$\|A\|<\infty:\quad\sigma(A)\subseteq\overline{\mathcal{W}(A)}$$ So any positive operator is a positive element; but what about the converse?

  • $\begingroup$ I seem to remember that the numerical range and the spectrum have the same convex hull. Is that right? If so, then one is positive iff the other is. $\endgroup$ – GEdgar Jun 29 '14 at 0:11
  • $\begingroup$ Unfortunately they don't as the example matrix shows. Also for self adjoint unbounded operators this is wrong in general if the spectrum is empty. But maybe there is still some hope for self adjoint and bounded operators... $\endgroup$ – C-Star-W-Star Jun 29 '14 at 0:24
  • $\begingroup$ When you're first starting with a $C^{\star}$ algebra, before knowing that such an algebra can be viewed as part of $\mathscr{B}(H)$, a definition of positivity for selfadjoint operators is possible, but one involving numerical range is not. $\endgroup$ – DisintegratingByParts Jun 29 '14 at 3:31
  • $\begingroup$ Did you mean "[...] a definition of positivity for selfadjoint operators 'involving spectrum' is possible, [...]"?? I just remembered right now that a definition involving numerical range is always possible too where the role of normalized vectors is taken by states (positive linear functionals $\omega\in E(\mathcal{A})$). $\endgroup$ – C-Star-W-Star Jun 29 '14 at 12:19
  • 2
    $\begingroup$ It $T$ is a normal bounded operator on a Hilbert space, then $\overline{W(T)} = $ the closed convex hull of $\sigma(T)$, so if $\sigma(T) \subset \mathbb{R}^+$, then $W(T) \subset \mathbb{R}^+$. So the two definitions are equivalent for bounded ops - not sure what happens for unbounded ops. $\endgroup$ – Prahlad Vaidyanathan Jun 29 '14 at 12:20

For unbounded normal operators one has: $$NN^*=N^*N:\quad\langle\sigma(N)\rangle=\overline{\mathcal{W}(N)}$$ (See summary on: Spectrum vs. Numerical Range)

So the notions of positivity agree!


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