It is still false. Take $$A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right).$$ Then it is definite positive, the operator norm of $A$ is strictly larger than $1$, but the operator norm of $$A^n = \left(\begin{array}{cc} 1 & n \\ 0 & 1 \end{array}\right).$$ is equivalent to $n$ as $n$ goes to infinity.

Since $n >m$, you may have $A^m=0$ whereas $A^n \ne 0$, so the proposed inequality cannot be true in generality.

Fourier analysis on groups by Walter Rudin for the general theory. Here, on can note that $$f_n(e^t) = \int_\mathbb{R} \varphi_n(e^s) f(e^{t-s}) ds$$ and recognise a usual convolution product of $s \mapsto \varphi(e^s)$ and $r \mapsto f(e^r)$.

Sorry, I wrote that too quickly. I hope it is fine now. Pearson Theorem follows from central limit Theorem by the application of a suitable quadratic form.

You are right. I corrected my answer. Pearson theorem can be found in statistic books, since it is the theoretical foundation of the $\chi^2$ test. Yet, I did not find a reference on internet.

@Vokram. I put an addendum at the top to answer your question.

What is the definition of the transient part?

Yes, you are right, I answer a different question. I will think more about that.

Haar measure on $\mathbb{U}$ is also the uniform measure on $\mathbb{U}$, namely the image of the uniform measure on $[0,2\pi[$ by $\theta \mapsto e^{i\theta}$. In my example, we have $$E\big[ \sup_{f\in \mathcal{F}} U_f \big| \mathcal{S} \big] = 1$$ whereas $$\sup_{f\in \mathcal{F}} E \big[U_f \big| \mathcal{S}\big] d\pi = 0.$$