2
$\begingroup$

I'm studying right now spectral theory of unbounded self-adjoint operators. A corollary of spectral theorem states the following: let $H$ be a (separable) Hilbert space and $(D_T, T)$ a self-adjoint operator on H. There exists a unique map $$ L^\infty(\mathbb{R};\mathbb{C}) \longmapsto L(H),\ f \longrightarrow f(T) $$ continuos with norm $\le 1$ with the following properties:

(i )This is a ring homomorphism

(ii) $f(T)^* = \overline{f}(T)$ and $f \ge 0$ iff $f(T) \ge 0$

(iii) If $f_n$ converges pointwise to $f \in L^\infty (\mathbb{R};\mathbb{C}$) and $\sup_n ||f_n||_\infty < +\infty $ then $f_n(T) \mapsto f(T)$ strongly.

The author starts the proof setting $H=L^2_\mu(X)$ with $(X, \mu)$ is a measure space with $\mu < \infty$ and $(D_T, T) = (D_g, M_g)$ where $D_g := \{ \phi \in L^2_\mu | g\phi \in L^2_\mu \}$, $M_g \phi = g \phi$. Then he set

$f(T) := M_{f \circ g}$ for every $f \in L^\infty$.

My question is: why $f(T)$ is well defined? I haven't understood why if $f_1$, $f_2$ are in the same equivalence class $f \in L^\infty(\mathbb{R}; \mathbb{C})$ (remember we're using the Lebesgue measure now) then $M_{f_1 \circ g} = M_{f_2 \circ g}$, that is $f_1 \circ g = f_2 \circ g\ \mu$ a.e.

I'm attending a master degree in Math, and I read this on lecture notes of a Spectral Theory course made in ETH, Zurich. This is the corollary 4.43.

$\endgroup$
  • 2
    $\begingroup$ To deal with general $T$, you need bounded Borel measurable functions with the sup norm, not an essential sup norm. That wouldn't even work for the identity operator $I$ because the functions need a definite value at $1$ in that case. $\endgroup$ – DisintegratingByParts Jun 21 '15 at 14:24
  • 1
    $\begingroup$ Yes, it was exactly my problem; why the lecture notes wrote this map on $L^\infty$? (Wikipedia makes this mistake too: en.wikipedia.org/wiki/…) Are they mistakes, or notations, or something else? $\endgroup$ – StefanoG Jun 21 '15 at 14:28
  • 1
    $\begingroup$ They've probably explained it somewhere in your lecture notes or class that you're dealing with a sup norm and not an essential sup norm. They did explain on Wikipedia that the functions are bounded Borel functions, but the use of $L^{\infty}$ is wrong. I suggest using $B^{\infty}$ or some other notation to make the distinction. It's good you caught that; the use of $L^{\infty}$ for the notation is entirely inappropriate. $\endgroup$ – DisintegratingByParts Jun 21 '15 at 14:33
  • 1
    $\begingroup$ Thank you, I thought my head would have blown up... However, the identity operator is surely a counterexample to every use of $L^\infty$, I should have thought before. $\endgroup$ – StefanoG Jun 21 '15 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.