I'm studying harmonic analysis by myself. One of the online notes gives the following claim as a remark:

For any $N \in \mathbb{Z}^{+}$, let's use $S_{N}$ to denote the partial ($N$ terms) Fourier sum operator, i.e for any function $f$, the function $S_{N}f$ is given by: $$S_{N}f(x) = \sum_{n=-N}^{N}\hat{f}(x)e^{2\pi inx}$$ Then there exists some constant $C>0$, such that for any $f \in L^{1}([0,1])$ and any $\lambda > 0$, the following "analogous" weak-type (1,1) inequality holds: $$\sup_{N}\big|\big\{\theta \in [0,1] \ | \ |S_{N}f(\theta)| > \lambda\big\}\big| \leq \frac{C}{\lambda}\|f\|_{L^1}$$ where $|A|$ denotes the Lebesgue measure of a set $A$. Any ideas on how to prove this inequality? I have tried some approaches with some tricks used to prove the conventional Hardy-Littlewood weak-type maximal inequality, but it doesn't seem to work....Thanks in advance!

Moreover, the notes also says that this inequality is related to a weak form of the almost everywhere convergence of Fourier series on $L^1([0,1])$. More specifically speaking, we can use the weak-type inequality above to show that there exists a subsequence $\{N_{m}\}_{m=1}^{\infty}$ of $\mathbb{Z}^{+}$, such that $S_{N_{m}}f \rightarrow f$ pointwisely and almost everywhere as $m \rightarrow \infty$. Any idea on this?

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    $\begingroup$ The weak 1-1 estimate for the partial sums (uniform in $N$) is usually deduced from the weak 1-1 estimate for the Hilbert transform. With the 1-1- estimates in the hands, one shows convergence in measure, hence a subsequence converges a.e. $\endgroup$ Apr 11, 2021 at 17:09
  • $\begingroup$ Hi Giorgio, thank you so much for your comment! I will take a look into Hilbert transform to see how it works out. However, would you clarify a little bit more on how weak (1,1) estimate induces convergence in measure? Thanks in advance! $\endgroup$ Apr 11, 2021 at 17:15
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    $\begingroup$ Fix $a>0$ and write $\{|S_Nf-f| \geq 3a \} \subset \{|S_N(f-g)| \geq a \} \cup \{|S_Ng-g| \geq a \} \cup \{|g-f| \geq a \}$. Given $\epsilon >0$, choose $g$ smooth such that $\|g-f\|_1 \leq \epsilon$. Then the measure of the first and third set on the RHS are of order $\epsilon$ for any $N$ and the second is empty, for large $N$, since $S_Ng \to g$ uniformly. $\endgroup$ Apr 11, 2021 at 17:23
  • $\begingroup$ Hi Giorgio, thank you so much for your clarification! That's crystal clear. Would you mind elaborating a little bit more on proving the inequality with weak (1,1) estimate of Hilbert transform. I have tried plugging in the expression of Dirichlet kernel and compare it with the Hilbert transform..it seems that we need to compare $\sin(\pi t)$ and $\pi t$ in some neighborhood of $0$? Thank you again for all your help! $\endgroup$ Apr 11, 2021 at 19:19
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    $\begingroup$ It is easier (and usual) to proceed in this way. Assuming that the Hilbert transform (for periodic functions, that is the one yielding the conjugate harmonic function on the circle or also the convolution with $\tan s$) is weak 1-1, you get that the map $f \to \sum_{k=0}^\infty \hat{f}(k) e^{i k s}$ is weak 1-1. Composing with $e^{\pm i Ns}$ you get uniform (with respect to $N$) weak 1-1 estimates for $f \to \sum_{k=N}^\infty \hat{f}(k)k e^{i k s}$ and, by difference, for the partial sums. $\endgroup$ Apr 11, 2021 at 21:53


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