Carleson-Hunt theorem for Wiener algebra case Question:Let $\phi(z)=\sum_{k\in \Bbb{Z}}a_kz^k$, $z\in \mathbb{T}$, with $\sum_{k}|a_k|<+\infty$. Suppose $(c_n)\in l^2$, show $$\sum_{n=-N}^{N}c_n\phi{(z^n)}$$ converges almost every where on $\Bbb{T}$, as $N\to +\infty$.
The case $\phi(z)=z$ is the well known Carleson theorem.

My observation: $$\sum_{n=-N}^N c_n\phi(z^n)=\sum_{n=-N}^N c_n\sum_k a_kz^{nk}=\sum_{k}a_k\sum_{n=-N}^N c_n(z^k)^n.$$ Set $f(z)=\sum_n c_nz^n \in L^2(\Bbb{T})$, and $f_N(z)$ be the partical sum form $-N$ to $N$. Then $$RHS=\sum_{k}a_k f_N(z^k).$$ By Carleson's theorem, $f_N\to f$ a.e. So I want prove above converges to $\sum_k a_k f(z^k)$ (This sum is converges in L^2-norm so well define a.e. on $\mathbb{T}$). If $f\in L^\infty$, this can be easily done by some "$\epsilon-\delta$" argument. But I don't know how to deal with general $f$.

 A: The proof is indeed based on Carleson's theorem, but it should be traced back to the following crucial lemma.

Lemma (about maximal operators, abstract version): Set $1\leq p,q <+\infty$. Let $T_N$ be a sequence of operators mapping measurable functions on $(X,d\mu)$ to measurable functions on $(Y,dv)$. Define their maximal operator $$(T_*f)(y)=\sup_N |(T_Nf)(y)|.$$
If $T_*$ maps $L^p(X,d\mu)$ to $L^q(Y,dv)$ such that $\|T_*f\|_q\leq C\|f\|_p$ for some universal constant $C$ and $T_Nf$ converges a.e.$dv$ when $f$ in some dense subset of $L^p(X,d\mu)$, then $T_Nf$ converges a.e.$dv$ for all $f\in L^p$.Here is a proof in Jose27's answer


Theorem (Carleson): Use the same notation, $$\big|\big|\sup_N|f_N(z)|\big|\big|_2\leq C \|f\|_2$$

As a consequence, set $T_N=f_N$, we see $f_N$ converges a.e. to $f$ because the a.e. convergence holds for smooth functions.
Now back to the question, let $f=\sum_n c_nz^n$ be given, consider $$T_N:\mathscr{l}^1 \to L^2(\mathbb{T}): (a_k)\mapsto \sum_k a_k f_N(z^k).$$
Note that $$\begin{aligned}\big|\big|\sup_N|\sum_k a_k f_N(z^k)|\big|\big|_2 &\leq \big|\big|\sum_k |a_k|\sup_N|f_N(z^k)|\big|\big|_2 \\&\leq \sum_k|a_k| \big|\big|\sup_Nf_N(z^k)\big|\big|_2\\ &\leq \sum_k |a_k| C\big|\big|f_N(z^k)\big|\big|_2 \\ &\leq C\big|\big|f\big|\big|_2\sum_k |a_k| \end{aligned}.$$ Thus $T_*$ is bounded. Now the conclusion follows, because it holds when $(a_k)\in \mathscr{l}^1$ with finite support.
