I have proved the 'resolution of the identity' for a normal operator, namely that there is a unique spectral measure E such that $\int_{{\sigma}(T)} {\lambda}\,dE=T$

If (${\lambda}_{n}$) is the sequence of eigenvalues of $T$, How do I prove that

a) $\sum_{n=1}^\infty \int_{\{\lambda_n\}}\lambda\,dE(\lambda)=\sum_{n=1}^\infty \lambda_n E(\{\lambda_n\})$

and b) that $E({\lambda}_{n})$ orthogonal projection to the eigenspace of $\lambda_n$?

  • $\begingroup$ If the underlying Hilbert space is not separable, then the set of distinct eigenvalues of a normal operator may be uncountable, and not a sequence. $\endgroup$ – DisintegratingByParts Mar 19 '14 at 3:33
  • $\begingroup$ I guess the operator must be also bounded, doesn't it? $\endgroup$ – C-Star-W-Star Sep 28 '14 at 23:50

a) Since $\{\lambda_n\}$ is a single point, the identity function is constant there: so $$ \int_{\{\lambda_n\}}\lambda\,dE(\lambda)=\lambda_n\,E(\{\lambda_n\}) $$ by definition of integral.

b) For $n\ne m$, using that the spectral integral is multiplicative, $$ E(\{\lambda_n\})\,E(\{\lambda_m\})=\int_{\{\lambda_n\}}\lambda\,dE(\lambda)\,\int_{\{\lambda_m\}}\lambda\,dE(\lambda)=\int_{\sigma(T)}1_{\{\lambda_n\}}(\lambda)\,dE(\lambda)\,\int_{\sigma(T)}1_{\{\lambda_m\}}(\lambda)\,dE(\lambda) =\int_{\sigma(T)}1_{\{\lambda_n\}}(\lambda)\,1_{\{\lambda_m\}}(\lambda)\,dE(\lambda)=0. $$ A shorter version (implied by the argument above) would be $$ E(\{\lambda_n\})\,E(\{\lambda_m\})=1_{\{\lambda_n\}}(T)\,1_{\{\lambda_n\}}(T)=1_{\{\lambda_n\}\cap\{\lambda_m\}}(T)=0. $$


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