Can someone tell me if the following is true.

Let A be a self-adjoint operator on a Hilbert space $\mathcal{H}$, and let $P_A$ be the projection valued measure (spectral measure) obtained from the spectral theorem, such that

$A=\int t\ dP_A(t)$

How do I show that

$A=\lambda I$ $\ \ \ $ $\Leftrightarrow$ $\ \ \ $ $\text{supp}P_A=\{\lambda\}$

I have a strong feeling that it is true. I think one direction is clear due to the definition of the spectral measure, since it is defined on the Borel algebra of the spectrum of A. But for the other direction I don't know what to do?

  • $\begingroup$ Hints. What is $P_A(\{\lambda\})$? What is the value of the spectral integral $\int\lambda dP_A$? Is it equal to $A$ ? Then use the uniqueness of the spectral measure. $\endgroup$ – Yurii Savchuk Mar 12 '18 at 8:50
  • $\begingroup$ Well $P_A(\{\lambda\}=I$ if the support is $\{\lambda\}$, and by construction the integral is equal to $A$. But I cannot see why $\int t\, dP_A(t)=I$?? (i can see that I made a bad choice of name for the variable in the integral in my forst post since $\lambda$ is fixed. $\endgroup$ – Simon Mar 12 '18 at 9:01
  • $\begingroup$ Check the definition of the spectral integral for step-functions. $\int\lambda dP_A=\lambda I$ $\endgroup$ – Yurii Savchuk Mar 12 '18 at 9:27
  • $\begingroup$ This is why I commented on the bad choice of variable name. Because I do know that $\int\lambda\,dP_A(t)=\lambda I$ but i do not know how to calculate $\int t\,dP_A(t)$. $\endgroup$ – Simon Mar 12 '18 at 9:29

If $A=\lambda I$, then $$ 0= \left\|\int_{\sigma} (\mu-\lambda)dP_A(\mu)x\right\|^2=\int_{\sigma} |\mu-\lambda|^2d\|P_A(\mu)x\|^2,\;\;\; x\in H. $$ If $d\|P_A(\mu)x\|^2$ has support outside of $\{\lambda\}$, then the above integral is non-zero. So $P_{A}(S)=P_{A}(S\cap\{\lambda\})$ for all Borel sets $S$ if $A=\lambda I$. Conversely, if $P_A$ is supported on $\{\lambda\}$, then $A=\int_{\sigma}\mu dP_A(\mu)= \lambda P_A\{\lambda\}=\lambda P_A(\mathbb{R})=\lambda I$.

  • $\begingroup$ How do you get $\int_\sigma \mu\,d P_A(\mu)=\lambda P_A\{\lambda\} $?? $\endgroup$ – Simon Mar 13 '18 at 8:38
  • $\begingroup$ It is a nice answer though!! $\endgroup$ – Simon Mar 13 '18 at 8:45
  • $\begingroup$ If $P_A$ is supported on $\{\lambda\}$, then $\int_{\sigma} = \int_{\{\lambda\}}$. $\endgroup$ – DisintegratingByParts Mar 13 '18 at 12:46
  • $\begingroup$ Oh yes I see. Thanks! $\endgroup$ – Simon Mar 13 '18 at 12:52
  • $\begingroup$ Now I think I am comfusing myself again.. How do you get the equality $\left\Vert\int_\sigma \mu\,dP_A(\mu)x\right\Vert^2=\int_\sigma\vert\mu-\lambda\vert^2\,d\Vert P_A(\mu)x\Vert^2$?? $\endgroup$ – Simon Mar 16 '18 at 19:21

You are almost done. On the support of $P_A$, $t=\lambda$.


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