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.

  • $\begingroup$ Do you know the Radon–Nikodym derivative for the measure? I am also interested in the problem :) $\endgroup$ – 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 $[0,1]$: $$M_f = \int_0^1 \lambda \,d\widetilde{P}_\lambda = \int_{-\infty}^\infty \lambda \,d\widetilde{P}_\lambda$$

  • $\begingroup$ Sorry, could you elaborate a little further on this? $\endgroup$ – mck Jan 20 '13 at 3:47
  • $\begingroup$ @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. $\endgroup$ – user53153 Jan 20 '13 at 4:40

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