I'm reading an example of a continuous function with divergent Fourier series from Stein-Shakarchi's Fourier Analysis. However, in the construction, I don't understand the last step (where they actually show the divergence).
They start showing this, no problem so far :
Then they define the trigonometric polynomials :
And show that the partial sums of the Fourier series $S_M$ satisfy
Now (this is what I don't understand), they show the divergence (in the red box), but I don't know why that inequality is true. I think I should use the first property of the sequence $N_k$ but I don't understand how.
Any help would be really appreciated.
The idea is that each $P_{N_k}$ corresponds to a sum of $2N_k$ frequencies (terms that look like $c_ne^{in\theta}$). These are the frequencies between $N_k$ and $3N_k$. Correspondingly, each $\bar{P}_{N_k}$ corresponds to a sum of $N_k$ frequencies, namely those between $N_k$ and $2N_k$.
The partial sum $S_{2N_m}(f)(\theta)$ is concerned with the first $2N_m$ frequencies of $f(\theta)$. This is more explicitly written as \begin{align} S_{2N_m}(f)(\theta) = \alpha_m\bar{P}_{N_m}(\theta) + \sum_{k=1}^{m-1}\alpha_kP_{N_k}(\theta) \end{align} The first term contains all frequencies between $N_m$ and $2N_m$, and the sum contains all of the rest of the smaller frequencies in $f(\theta)$.
Note (and this may take a brief review of the whole problem) that all the terms in the sum are positive, so that the following is true: \begin{align} | S_{2N_m}(f)(0)| = |\alpha_m||\bar{P}_{N_m}(0)| + \sum_{k=1}^{m-1}|\alpha_k||P_{N_k}(0)| \end{align} For the first term, we have \begin{align} |\alpha_m||\bar{P}_{N_m}(0)|=|\alpha_m||\bar{f}_{N_m}(0)|\geq \alpha_mc\log(N_m) \end{align}
(EDIT: For $\theta=0$, the sum is $0$ and the result is clear. For general $\theta$, the sum is $O(1)$ for the following reason. I don't know why the author included $O(1)$, perhaps someone can clear this up?)
Note that $P_{N_k}(\theta)$ is uniform and bounded in $\theta$ and $N_k$, so there is a constant $M$ independent of $\theta$ and $N_k$ such that \begin{align} |P_{N_k}(\theta)|<M \end{align} and so, remembering that $\alpha_k = \frac{1}{k^2}$, we have \begin{align} \sum_{k=1}^{m-1}|\alpha_k||P_{N_k}(\theta)| \leq M \sum_{k=1}^{m-1}\frac{1}{k^2} = O(1) \end{align} And the result follows.
EDIT: Here is the subsequent paragraph in the text, for context