Relation between the support of spectral measure and proportionality of the identity operator

Question

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?

Answer 1

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 $A=\int_{\sigma}\mu dP_A(\mu)= \lambda P_A\{\lambda\}=\lambda P_A(\mathbb{R})=\lambda I$.

Answer 2

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