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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 $C(\sigma( (T + iI)^{-1})$.

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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 $\infty$.

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  • $\begingroup$ Sorry, I am still confused - I have updated my post for the question. I would also be grateful if you may give a reference for this : ) $\endgroup$ – CL. Apr 13 at 7:19
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Apr 13 at 16:01
  • $\begingroup$ @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 $v\downarrow 0$. $\endgroup$ – DisintegratingByParts Apr 13 at 16:11
  • $\begingroup$ Its precisely this "defined through the resolvent " i don't understand, what does it represent? Is there a reference that gives some basic examples? $\endgroup$ – CL. Apr 13 at 16:21
  • $\begingroup$ @CL. Early developments of spectral theory for self-adjoint operators arose out of looking at the singularities of the resolvent. The spectral measure is the "residue" of the resolvent, and that leads to a lot of tricks using complex analysis and the resolvent. $\endgroup$ – DisintegratingByParts Apr 13 at 21:40

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