Existence of invariant states in a $C^*$-algebra

Question

Let $\mathcal{A}$ be a C*-algebra and $\{\tau_t\}_{t\in\mathbb R}$ a weakly-continuous group of *-automorphisms. I've read the claim (without proof) that for any state $\eta$ (that is $\eta$ is a normalized positive functional on $\mathcal A$) $$\omega:A\mapsto\omega(A):=\lim_{N\to\infty}\frac{1}{N}\int_0^N\eta(\tau_t(A))~dt$$ defines a state which is $\tau_t$-invariant. Why is this true? I am not even sure why this limit should exist nor why it should be invariant under the $\tau_t$.

Answer 1

I don't think that the limit, as stated, always exists.

But note that the weak continuity hypothesis implies that the function $t\mapsto \eta(\tau_t(A))$ is continuous, thus integrable. It is also bounded by $\|A\|$. One can check that the maps $\omega_N:A\mapsto\frac1N\,\int_0^N\eta(\tau_t(A))\,dt$ are states.

As the state space is weak*-compact, we can find a convergence subsequence. If we take $\omega$ to be the limit of one of such convergent subsequences, it is indeed invariant.

To simplify notation, I will do the following computations with the index $N$, but in reality it should be some subsequence of the naturals (the one corresponding to the convergent subsequence that we chose). Fix $r\in\mathbb R$. Then $$ \omega(\tau_r(A))=\lim_{N\to\infty}\frac1N\,\int_{0}^N\eta(\tau_t(\tau_r(A)))\,dt =\lim_{N\to\infty}\frac1N\,\int_{0}^N\eta(\tau_{t+r}(A)))\,dt\\ =\lim_{N\to\infty}\frac1N\,\int_{r}^{N+r}\eta(\tau_{t}(A)))\,dt. $$ Now, for the invariance, $$ |\omega(\tau_r(A))-\omega(A)|=\left|\lim_{N\to\infty}\frac1N\,\int_{N}^{N+r}\eta(\tau_{t}(A)))\,dt\right| =\left|\lim_{N\to\infty}\frac1N\,\int_{0}^{r}\eta(\tau_{t+N}(A)))\,dt\right|\\ \leq\lim_{N\to\infty}\frac{r\|A\|}N=0. $$