Let
- $f:S^1 \to \mathbb{R}$ be a continuous function
- $\theta \in 2 \pi (\mathbb{R}\setminus \mathbb{Q})$ an irrational multiple of $2 \pi$, so $e^{i\theta}$ is an irrational rotation.
I want to compare the integral average
$$ I^f:=\frac{1}{\ell(S^1)} \int_0^{2 \pi} f(e^{it}) dt $$
and the sequence of averages
$$ E_n^f(z):=\frac{1}{n+1} \sum_{k=0}^{n} f(e^{k\theta i}z) $$ of the values of $f$ at a starting value $z\in S^1$ and its $n$ successive rotations by $\theta$ (for positive integers $n$).
The points in the sum accumulate on the whole unit circle.
Hence, if I weight the term $f(e^{k\theta i}z)$ in $E_{n}$ by the distance to the next point of evaluation in the current sum: $d_{k,n}:=\min \{ k'\theta-k\theta-2\pi m>0\mid0\le k'\le n \}$, I get a sequence of Riemann sums $$ E_{n}'(z):=\frac{1}{2\pi}\sum_{k=0}^{n}d_{k,n}f(e^{k\theta i}z) $$
that converge to $I^f$ as $n\to+\infty$.
(partially solved) Under which conditions is the convergence of $E_n'$ uniform in $z\in S^1$?
Do the non-weighted averages $E^f_n(z)$ always converge pointwise? Do they converge uniformly, when $E_n'(z)$ does?
(solved) If not in general, can I get some estimates for the error of $E^f_n(z)$ if I assume some smoothness for $f$?
Bonus question:
- Are there examples for non-uniform convergence in each case (or even no pointwise convergence for $E_n^f$)?
Partial solution for uniformly Fourier-approximated $f$
I have found an equivalent question here and the accepted answer suggests using Fourier series. In this way, I can show uniform convergence of $E_{n}^{f}$ to $I^{f}$ under the additional condition, that the partial sums $$ f_{m}(z)=\sum_{k=-m}^{m}a_{k}z^{k} $$ of the Fourier series of $f$ converge uniformly to $f$ (e.g. if $f$ is Hölder-continuous):
I first prove the statement for the monomials $e_{m}(z)=z^{m}$, $m\in\mathbb{Z}$. For $m=0$, $$ I^{e_{0}}=1=E_{n}^{e_{0}}(z) $$ for all $z\in S^{1}$, $n\in\mathbb{N}$, so the statement is trivial. For $m\neq0$, \begin{align*} I^{e_{m}}=\frac{1}{\ell(S^{1})}\int_{0}^{2\pi}e^{mit}\,dt & =0 \end{align*} and \begin{align*} |E_{n}^{e_{m}}(z)| & =\bigg|\frac{1}{n+1}\sum_{k=0}^{n}(e^{k\theta i}z)^{m}\bigg|\\ & =\bigg|\frac{z^{m}}{n+1}\sum_{k=0}^{n}(e^{m\theta i})^{k}\bigg|\\ & =\frac{1}{n+1}\bigg|\frac{1-e^{(n+1)m\theta i}}{1-e^{m\theta i}}\bigg|\\ & <\frac{1}{n+1}\bigg|\frac{2}{1-e^{m\theta i}}\bigg|,\tag{1} \end{align*} hence $$ \lim_{n\to+\infty}E_{n}^{e_{m}}(z)=I^{e_{m}} $$ uniformly in $z$.
For $f$ a uniform limit of its Fourier series and any $\varepsilon>0$, there exists $m\in\mathbb{N}$ large enough that $|f-f_{m}|<\varepsilon$. Then (by applying the estimate $(1)$ to the finite number of monomials in $f_{m}$), there exists $N=N(m)$ large enough that for $n\ge N$, $$ |I^{f_{m}}-E_{n}^{f_{m}}|<\varepsilon. $$ Then Hence for all $n\ge N$, we have $$ |I^{f}-E_{n}^{f}|\le|I^{f}-I^{f_{m}}|+|I^{f_{m}}-E_{n}^{f_{m}}|+|E_{n}^{f_{m}}-E_{n}^{f}|\le3\varepsilon. $$ (Note that $N$ has to depend on $m$, since the factor $2/(1-e^{m\theta i})$ in $(1)$ gets arbitrarily large.)
By the same argument, $E_n'$ converges uniformly in this case, so this gives a suitable condition for question 1 and 3.
What remains is question 2 on pointwise convergence in the general case and bonus question 4.