Holomorphic Functional Calculus vs Borel Functional Calculus

Question

I am currently learning about different kinds of functional calculus and I was wondering if I could get something cleared up.
The first type of functional calculus we learned about was holomorphic functional calculus, i.e., given a self adjoint operator $H$ and a function $f$ that is holomorphic in a neighbourhood of the spectrum, we can define $$f_\Gamma(H)=\int_{\Gamma}f(\lambda)(\lambda-H)^{-1}d\lambda $$ where $\Gamma$ is a contour contained in region of holomorphy of $f$ that encloses the spectrum.
We also talked about Borel functional calculus, i.e., for a self adjoint operator $H$ we can define $$f(H)=\int_{\sigma(H)} f(\lambda) dP(\lambda)$$ where $f$ is a borel function and we integrate over a projection valued measure with respect to Borel sets.
I've been told that the two functional calculi actually coincide, i.e., $$f_\Gamma(H)=f(H)$$for functions and operators that satisfy the requirements. I was wondering if there was an easy way to show that this was the case. Thank you for any help!

Answer 1

If $H$ is bounded and selfadjoint with spectrum $\sigma$, then $$ (\lambda I-H)^{-1}=\int_{\sigma}\frac{1}{\lambda-\mu}dE(\mu),\;\;\; \lambda\notin\sigma. $$ Suppose $\Gamma$ is a positively-oriented simple closed rectifiable curve in $\mathbb{C}$ which contains $\sigma$ in its interior, and suppose that $f$ is holomorphic on an open neighborhood of $\Gamma$ and its interior. Then $$ \begin{align} f(H)x & =\frac{1}{2\pi i}\oint_{\Gamma}f(\lambda)(\lambda I-H)^{-1}x\,d\lambda \\ & = \frac{1}{2\pi i}\oint_{\Gamma}f(\lambda)\left[\int_{\sigma}\frac{1}{\lambda-\mu}dE(\mu)x\right]d\lambda \\ & = \int_{\sigma}\left[\frac{1}{2\pi i}\oint_{\Gamma}\frac{f(\lambda)}{\lambda-\mu}\,d\lambda\right]dE(\mu)x \\ & = \int_{\sigma}f(\mu)dE(\mu)x. \end{align} $$