I would be grateful for a proof of a proposition in Reed Simon I, Chapter VII.3. It states that for a bounded self-adjoint operator $A$ a point $\lambda$ is in the spectrum of $A$ if and only if for every $\epsilon>0$ the spectral projection $P_{(\lambda-\epsilon,\lambda+\epsilon)}$ is nonzero.

As a hint they say that the main ingredient in the proof is the equality $\|(A-\lambda)^{-1}\|=\operatorname{dist}(\lambda,\sigma(A))^{-1}$, but I've got no clue how to use this.


For any $x \in H$ and $\epsilon > 0$, \begin{align} \|(A-\lambda I)P(\lambda-\epsilon,\lambda+\epsilon)x\|^2 & = \int_{(\lambda-\epsilon,\lambda+\epsilon)}(\mu-\lambda)^2d(P(\mu)x,x) \\ & \le \epsilon^2\int_{(\lambda-\epsilon,\lambda+\epsilon)}d(P(\mu)x,x) \\ & = \epsilon^2\|P(\lambda-\epsilon,\lambda+\epsilon)x\|^2. \end{align}

  • Suppose $\lambda\notin\sigma(A)$. Then let $x=(A-\lambda I)^{-1}y$, and notice that \begin{eqnarray*} \|(A-\lambda I)P(\lambda-\epsilon,\lambda+\epsilon)(A-\lambda I)^{-1}y\| &\le \epsilon\|P(\lambda-\epsilon,\lambda+\epsilon)(A-\lambda I)^{-1}y\| \\ \|P(\lambda-\epsilon,\lambda+\epsilon)y\| &\le \epsilon \|(A-\lambda I)^{-1}P(\lambda-\epsilon,\lambda+\epsilon)y\| \\ \|P(\lambda-\epsilon,\lambda+\epsilon)y\| &\le \epsilon \,\mbox{dist}(\lambda,\sigma(A)^{-1} \|P(\lambda-\epsilon,\lambda+\epsilon)y\|. \end{eqnarray*} Therefore $P(\lambda-\epsilon,\lambda+\epsilon)y=0$ for $\epsilon\cdot\mbox{dist}(\lambda,\sigma(A))^{-1} < 1$. Because $y$ was arbitrary, then $$ P(\lambda-\epsilon,\lambda+\epsilon)=0 \mbox{ for } \epsilon < \mbox{dist}(\lambda,\sigma(A)). $$

  • Conversely, suppose $P(\lambda-\epsilon,\lambda+\epsilon) = 0$ some $\epsilon > 0$. Then $$ \int\frac{1}{\mu-\lambda}dP(\mu)=\int_{|\mu-\lambda| \ge \epsilon}\frac{1}{\mu-\lambda}dP(\mu) $$ is a bounded operator $R(\lambda)$ such that $(A-\lambda I)R(\lambda)=R(\lambda)(A-\lambda I)=I$, which proves $\lambda\notin\sigma(A)$.

  • $\begingroup$ Thanks a lot! I think your $\text{dist}(\lambda,\sigma(A))$ should be replaced by $\text{dist}(\lambda,\sigma(A))^{-1}$ $\endgroup$ – Frank Jan 15 '17 at 13:30
  • $\begingroup$ @Frank : Thank you. I corrected that now. $\endgroup$ – DisintegratingByParts Jan 15 '17 at 17:15
  • $\begingroup$ @DisintegratingByParts hey! I'd appreciate it if you can explain two things: 1. how do you get to $\|P(\lambda-\epsilon,\lambda+\epsilon)y\| \le \epsilon \|(A-\lambda I)^{-1}P(\lambda-\epsilon,\lambda+\epsilon)y\| $ from the line above it? second, how do we see that $(A-\lambda I)R(\lambda)=R(\lambda)(A-\lambda I)=I$? I understand that $\int\frac{1}{\mu-\lambda}dP(\mu)$ is notation for the unique operator $B$ s.t $(\phi, B\phi) = \int \frac{1}{\mu-\lambda} d(\phi,P_\lambda \phi)$ but not the operator equality. $\endgroup$ – Mariah Oct 12 '17 at 22:16
  • $\begingroup$ @Mariah : $R(\lambda)=(A-\lambda I)^{-1}$ is the resolvent, which is a bounded operator for all $\lambda\notin\sigma(A)$. And $A$ is a bounded selfadjoint operator in his problem. The integral $\int\frac{1}{\mu-\lambda}dP(\mu)$ can be defined through the operator inner product equality, if you like. $\endgroup$ – DisintegratingByParts Oct 13 '17 at 17:35

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