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I'm reading the example of a continuous function with divergent fourier series from Stein's Fourier Analysis book , 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 :

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Then they define the trigonometric polynomials :

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And show that the partial sums of the fourier series $S_M$ satisfy this :

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Now (this is what I don't understand) , they show the diverence (in red) , but I don't know why is that inequality true. I know I should use the first property of the sequence $N_k$ but i don't understand how.

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Any help would be really appretiated.

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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

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  • $\begingroup$ Isn't $P_N(0)$ just $0$? $\endgroup$ – epimorphic Apr 28 '17 at 7:01
  • $\begingroup$ Hmmm, yeah that seems right. I don't get why he includes $O(1)$ then - in the paragraph after the proof is complete, he goes on to say that the term in the sum corresponding to $N_m$ has absolute value greater than $c\alpha_m\log(N_m)$ because $|\bar{P}_{N_m}(\theta)|=|\bar{f}_{N_m}(\theta)|\geq c\log(N_m)|$ (notice that $0$ has been replaced by $\theta$, see my edit). But I don't think this holds for general $\theta$. $\endgroup$ – John Ryan Apr 28 '17 at 17:51
  • $\begingroup$ It certainly doesn't hold for $\theta = \pi$ since $\lvert \tilde P_N(\pi) \rvert$ is the alternating harmonic series and thus converges to $\log 2$. Don't have the book handy so perhaps I'm missing some further context... $\endgroup$ – epimorphic Apr 29 '17 at 7:43
  • $\begingroup$ math.stackexchange.com/questions/468529/… I think this question/answer implies that it ONLY holds for $\theta=0$. The book is prof.usb.ve/bueno/Libros/… section 2.2 on page 100 of the PDF, aka 83 in the book $\endgroup$ – John Ryan Apr 29 '17 at 17:30

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