Let $A$ be a self-adjoint operator, (then by the spectral theorem we have: $A = \int_\mathbb R \lambda dP_A(\lambda)$ with $P_A$ a projection-valued measure). I want to show the following: $\lim_{T \to \infty} 1/T \int^T_0 \langle \varphi, e^{-itA} \psi \rangle dt = \langle \varphi, P_A(\{0\}) \psi \rangle$. (which means that in the weak operator topology the Cesaro limits $1/T \int^T_0 e^{itA} dt$ converges to the projection $P_A(\{0\}$)

My calculation: by the spectral theorem we have $\lim_{T \to \infty} 1/T \int^T_0 \langle \varphi, e^{-itA} \psi \rangle dt = 1/T \int^T_0 (\int_\mathbb R e^{-it\lambda} d\mu_{\varphi,\psi}(\lambda)) dt$, where $\mu_{\varphi,\psi}$ is the spectral measure associated to $A$ and to $\varphi,\psi$. By Fubini, this expression equals $\int_\mathbb R (1/T \int^T_0 e^{-it\lambda} dt) d\mu_{\varphi,\psi}(\lambda))$. By dominated convergence, the expression in the inner paranthesis equals $\chi_{\{0\}}(\lambda)$ and thus we get the desired result by taking the limit $T \to \infty$.

My questions: 1) Is my calculation ok?

2) How do I justify the use of Fubini?

3) The calculation doesnt seem to work when I change $e^{-itA}$ to $e^{itA}$, but shouldnt the assertion actually be: $\lim_{T \to \infty} 1/T \int^T_0 \langle \varphi, e^{itA} \psi \rangle dt = \langle \varphi, P_A(\{0\}) \psi \rangle$. My problem: I want to argue that $1/T \int^T_0 e^{it\lambda}dt$ equals $\chi_{\{0\}}(\lambda)$ by dominated convergence. But for $\lambda \neq0$ the pointwise limit is $\lim_{T \to \infty} 1/T(i \lambda e^{i \lambda T} - i \lambda)$, which doesnt tend to $0$.


1 Answer 1


The spectral measure $E$ is a Borel measure. You might be bothered by the complex measures, but you can reduce to positive measures $$ (E(S)x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}(E(S)(x+i^{n}y),x+i^{n}y),\\ \mu_{x,y} = \frac{1}{4}\sum_{n=0}^{3}i^{n}\mu_{x+i^{n}y,x+i^{n}y} $$ The integral with respect to $t$ is a Lebesgue integral. Of course you can treat it as a Riemann integral, and use limits of sums and bounded convergence to justify the interchange of integrals.

The calculation does work for $e^{-itA}$. If you're bothered, swap $A$ for $-A$ because there are no assumptions on $A=A^{\star}$ of positivity, etc.. Anway, $$ \frac{1}{T}\int_{0}^{T}e^{it\lambda}dt = \frac{e^{iT\lambda}-1}{iT\lambda}. $$ The left side is uniformly bounded by $1$ for real $\lambda$. For $\lambda\in\mathbb{R}\setminus\{0\}$, the right side converges pointwise to $1$. For $\lambda=0$, the left side is $1$ for all $T > 0$. So the above converges to $\chi_{\{0\}}(\lambda)$ as $T\rightarrow 0$, and the bounded convergence theorem applies.


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