# Applying functional calculus to the bounded operator $(T \pm iI)^{-1}$

This is the context: What I wish to prove:

2.3 page 16. Let $$T$$ be an essentially self-adjoint operator on a Hilbert space $$H$$. (Now we take its closure) There is a unique homomoprhism of $$C^*$$ algebras from the algebra of continuous, bounded functions on $$\Bbb R$$ into the algebra of bounded operators on $$H$$ which maps the functions $$(x \pm i)^{-1}$$ to the operators $$(T \pm iI)^{-1}$$.

I have proven that $$T$$ satisfying the condition implies $$(T\pm iI)^{-1}$$ is a bounded operator normal operator.

The proof given in the text is as follows.

The spectral theorem is proven by observing $$(T\pm iI)^{-1}$$ generate a commutative $$C^*$$ algebra of operators. By Gelfand Naimark theorem, every commutative $$C^*$$-algebra is isomoprhic to $$C_0(X)$$ for some locally compact space $$X$$.

** In this case $$X$$ may be identified with a closed subset of $$\Bbb R$$ (the spectrum of $$T$$) in such a way that the operators $$(T\pm i I)^{-1}$$ correspond to the functions $$(x \pm iI)^{-1}$$.

I am fine until $$**$$. I don't see how the identification works, especially when we are applying to $$(T+ I)^{-1}$$, so the by GN we should have isomoprhism with $$C(\sigma( (T + iI)^{-1})$$.

Essentially self-adjoint means that the closure of $$T$$ is selfadjoint, and that would not imply that $$(T\pm iI)$$ are surjective, which leaves you dealing with bounded operators $$(T\pm iI)^{-1}$$ that are not everywhere defined. So, that's a bit annoying to deal with, and I'll just assume that $$T$$ is selfadjoint so that $$(T\pm iI)^{-1}$$ are in $$\mathcal{L}(H)$$.
Every $$(T-\lambda I)^{-1}$$ for non-real $$\lambda$$ is defined and bounded, and it lies in the $$C^*$$ algebra generated by $$(T\pm iI)^{-1}$$. For example, if $$|\lambda-i| < 1$$, then $$(T-\lambda I)^{-1}=(T-iI+(i-\lambda)I)^{-1} \\ = (I+(i-\lambda)(T-iI)^{-1})^{-1}(T-iI)^{-1} \\ = \sum_{n=0}^{\infty}(\lambda -i)^n(T-iI)^{-n-1}.$$ Then you can repeat the process to obtain the resolvent for $$|\lambda-2i| < 2$$, and eventually every $$(T-\lambda I)^{-1}$$ for $$\Im\lambda >0$$. The same is true for $$\Im\lambda < 0$$. The same holds for $$(T-\lambda I)^{-1}$$ for $$\Im\lambda < 0$$.
If $$f$$ is a continuous function on $$\mathbb{R}$$ that vanishes at $$\infty$$, then the Poisson integral $$f_{v}(u)=\frac{1}{2\pi i}\int_{-\infty}^{\infty}f(t)\left[\frac{1}{t-u-iv}-\frac{1}{t-u+iv}\right]dt \\ = \frac{1}{\pi}\int_{-\infty}^{\infty}f(t)\frac{(t-u)}{(t-u)^2+v^2}dt$$ converges uniformly to $$f$$ as $$v\downarrow 0$$. Using the results of the previous paragraph, $$f_v(T)$$ is in the $$C^*$$ algebra generated by $$(T\pm iI)^{-1}$$. This is easily extended to deal with functions $$f$$ that have a non-zero limit at $$\infty$$. I don't think general bounded continuous functions $$f$$ can work because of the behavior at $$\infty$$. But everything works if $$f$$ has a non-zero or zero limit at $$\infty$$; that is, $$f(T)$$ is in the $$C^*$$ algebra generated by $$(T\pm iI)^{-1}$$ if $$f$$ is continuous on $$\mathbb{R}$$ and has a limit at $$\infty$$.
• @CL : I have defined the map from $f$ to $f(T)$ for bounded continuous functions with a limit at $\infty$. It's a constructive approach to the problem, but may not be what you want, which is okay. – DisintegratingByParts Apr 13 at 16:01
• @CL : Because it is defined through the resolvent. $f_v(T)=\frac{1}{2\pi I}\int_{-\infty}^{\infty}f(t)\left[\frac{1}{t -iv-T}-\frac{1}{t+iv-T}\right]dt$. Then $f_v(T)\rightarrow f(T)$ as $v\downarrow 0$. – DisintegratingByParts Apr 13 at 16:11