Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to use the spectral theorem for self adjoint operators to decompose the spectrum of the multiplication operator $f(x) = \frac{1}{1+x^2}$ on $L^2(\mathbb{R}).$ This is a problem in Teschl's "Mathematical Applications to Quantum Mechanics." Here is what I have done so far.

The function $f \in L^\infty(\mathbb{R})$ so it is a bounded operator and its spectrum is equal to the closure of the range of $f$ which is the interval $[0,1].$ There are clearly no eigenvectors since $g = \frac{g}{1+x^2}$ implies that $g=0$ a.e. If $\psi(x) \in L^2(\mathbb{R})$ then the spectral measure is defined by $$\mu_\psi(\Omega) = \langle\psi, \chi_{f^{-1}(\Omega)} \psi \rangle$$ for $\Omega \subset \mathbb{R}$ measurable. Since $f$ is smooth and everywhere 2 to 1, if $\Omega$ is a set of Lebesgue measure $0$ then so is $f^{-1}(\Omega)$ so $\mu_\psi$ is absolutely continuous with respect to the Lebesgue measure for all $\psi.$ Therefore the spectrum is entirely absolutely continuous.

I am having trouble finding a spectral basis so that I can decompose the operator into a direct sum of multiplication operators on finite measure spaces. There doesn't seem to be a general procedure for doing this and I can't think of a good place to start.

share|cite|improve this question
Do you know the Radon–Nikodym derivative for the measure? I am also interested in the problem :) – user59563 Jan 24 '13 at 21:45

Begin with the layer-cake representation of the multiplier: $f(x)=\int_0^1 1_{E_\lambda}\,d\lambda $ where $E_\lambda = \{x: f(x)\ge \lambda\}$. Multiplication by $1_{E_\lambda}$ is a projection operator, say $P_\lambda$: it projects $L^2(\mathbb R)$ onto the subspace of functions that vanish outside of $E_\lambda$. Therefore, the multiplication by $f$ can be written as $M_f=\int_0^1 P_\lambda\,d\lambda $. It remains to transform this integral into the form $\int_{-\infty}^{\infty} t \, d P_t$, as required by spectral decomposition.

Integrate by parts: $$M_f=\int_0^1 P_\lambda\,d\lambda = -\int_0^1 \lambda\, dP_\lambda$$ where the boundary terms do not appear because $P_1$ is the zero operator ($E_1$ is a null set). This is almost what we want, except for the minus sign and the fact that $\lambda\mapsto P_\lambda$ is a decreasing function, rather than increasing. Both are fixed by $\widetilde{P_{\lambda}}=I-P_{\lambda}$. In other words, $\widetilde{P}_\lambda$ is the multiplication by $1_{\{f\le \lambda\}}$. The range of integration can be extended to the whole real line since $\widetilde{P_{\lambda}}$ is constant outside of $[0,1]$: $$M_f = \int_0^1 \lambda \,d\widetilde{P}_\lambda = \int_{-\infty}^\infty \lambda \,d\widetilde{P}_\lambda$$

share|cite|improve this answer
Sorry, could you elaborate a little further on this? – mck Jan 20 '13 at 3:47
@mck Yes, I added some details. And since the edit bumped the post anyway, I put the real angle brackets \langle and \rangle in your post. – user53153 Jan 20 '13 at 4:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.