# Continuous version of mean ergodic theorem with spectral theorem

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 $0$.

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.