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Let $H$ be a hilbert space and $T\in L(H)$ a self adjoint operator.

Show that we have in general $\sigma(f(T))\neq f(\sigma(T))$

Any tips?

If I choose a self adjoint operator how the measurable functional calculus looks like?

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Just choose the simplest setting, i.e. the setting of a multiplication operator.

Choose $H=L^2((0,1))$ (with Lebesgue measure),

$$ Tg(x)= x \cdot g(x) $$

and $f(x) = 1$ for $x\neq 1/2$ as well as $f(1/2)=42$.

It is not hard to see that

$$ f(T)g( x) = f(x) g( x) = g( x), $$ where the last equality holds a.e.

Hence, $\sigma(T)=[0,1]$, $\sigma(f(T))={1}$, but $f(\sigma(T))=\{1,42\}$.

EDIT: Further explanation regarding the measurable functional calculus: If $T$ is a multiplication operator on some $L^2$ space as above (multiply with $h$), then $f(T)$ is multiplication with $f\circ h$.

In the general case, the spectral theorem yields a unitary operator $U:H \to L^2(\mu)$ with $T = U^\ast M U$, where $M$ is a multiplication operator as above. Then $f(T) = U^\ast N U$, where $N$ corresponds to the modified multiplication operator (as above).

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  • $\begingroup$ Hello. I got some questions: 1) Why the multiplication operator is self-adjoint? Because $\int_0^1 <Tg(x),g(x)>dx=\int_0^1 xg(x)^2dx$ is real valued for $g\in L^2((0,1))$ ? 2) How you know that $\sigma(T)=[0,1]$ ? Sorry, I am new to that topic :) $\endgroup$ – Duke Feb 8 '15 at 12:05
  • $\begingroup$ @Duke: Yes, that is one way to argue. But it is also easy to see by a direct computation that $\langle f(x) h(x), g(x) \rangle = \langle f(x), g(x) h(x)\rangle$ if $h \in L^\infty$ is real valued. Thus, multiplication with $T$ is self-adjoint. For the second question, try to show that if $T - \lambda {\rm id}$ was invertible, the inverse would be given by $f \mapsto f / (x - \lambda)$, but for $\lambda \in [0,1]$, this is not a bounded linear map. $\endgroup$ – PhoemueX Feb 8 '15 at 22:28
  • $\begingroup$ So, the inverse operator is $f\mapsto f/(x-\lambda)$ as you mentioned. But only for $\lambda \in [0,1]$ the operator is not bounded, because $||(T-\lambda)^{-1}(f)||_X=||f/(x-\lambda)||_X$ goes to infinity if $x\rightarrow \lambda$ and this is possible if and only if because $\lambda \in [0,1]$(and $x\in [0,1]$). But if $\lambda$ is not in $[0,1]$ $(T-\lambda)$ is invertible with a bounded inversion .Not got formulated but correct, right? $\endgroup$ – Duke Feb 9 '15 at 0:41
  • $\begingroup$ @Duke: Yes, that is essentially correct. But you should convince yourselves of that $\Vert x \mapsto f(x)/(x -\lambda)\Vert_{L^2(0,1)}$ is indeed infinite for e.g. $f\equiv 1$ and $\lambda \in [0,1]$. $\endgroup$ – PhoemueX Feb 9 '15 at 6:36
  • $\begingroup$ Alright. Thanks. While thinking about it there are coming more question. I hope its okay for you. Question: (I guess I did not understood the functional calculus very well). But why you are "able" to choose the function f like above? Because we are in the measurable functional calculus and this function is measurable? In the case of the continuous functional calculus you HAVE to choose a continuous f? But these are the only restriction? Beside this you can choose f freely? $\endgroup$ – Duke Feb 9 '15 at 12:37

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