# Pointwise convergence of Fourier series from Rudin

This theorem about pointwise convergence of Fourier series. And some moments from here little bit confuses me.

1) How rigorously show that the last two integrals tends to zero? How to use that $|f(x+t)-f(x)|\le M|t|$ for $t\in (-\delta,\delta)$?

I would be very grateful for your help!

• The integrals tend to $0$ by the Riemann-Lebesgue lemma, which follows because the functions in square brackets are absolutely integrable, and that follows because $g$ remains bounded near $t=0$. Dec 19, 2015 at 14:35
• Why functions in brackets are absolutely untegrable? Can you show it in detail please? Because all this is little bit hard to understand. Dec 19, 2015 at 14:45
• Do you see that $g(t)$ is bounded near $t=0$ because of the assumption on $f$ at $x$? Dec 19, 2015 at 16:58
• @TrialAndError, Yes i see that $g(t)$ is bounded in some small neightborhood of zero ( let it's $[-\delta/2, \delta/2]$. Right? What about outside of this segment i.e. $[-\pi, -\delta]\cup [\delta, \pi]$? Dec 19, 2015 at 17:11
• Outside that interval $1/\sin(t/2)$ is bounded. Dec 19, 2015 at 17:35

A) The assumption that $|f(x+t)-f(x)|\le M|t|$ is used to infer that $g(t)$ is bounded.

In more detail, assuming $\delta$ sufficiently small, for $0<|t|<\delta$ we have $|\sin(t/2)|\geq |t|/4$ by the Taylor series. Thus, if $|f(x+t)-f(x)|\le M|t|$ for $|t|\le\delta$ (so that $|f(x-t)-f(x)|\le M|t|$ also), we get $$|g(t)|=\left|{f(x-t)-f(x)\over \sin(t/2)}\right|=\left|{f(x-t)-f(x)\over t}\cdot {t\over \sin(t/2)}\right|\leq 4M|t|\leq 4M\delta$$

For $\delta < |t| <\pi$, we have that $|1/\!\sin (t/2)|$ is bounded because it is $\leq 1/\!\sin(\delta/2)$, a fixed number. $f$ is assumed Riemann integrable on $[-\pi,\pi]$ and periodic (Rudin 8.13); it is therefore bounded (Rudin 6.1, intuitively because there has to be an upper sum for each partition). This lets us bound $g(t)$ for $\delta<|t|<\pi$.

B) The fact that the integrals go to zero is deduced from $(74)$ which says that the Fourier coefficients of a Riemann-integrable function go to zero. To apply it we need to verify that $g(t)\cos t$ and $g(t)\sin t$ are Riemann-integrable. To do this easily I jump ahead to Rudin 11.33, which just requires me to show that these functions are bounded and continuous almost everywhere, and that in turn follows from $g$ being bounded and from $f$ and $\cos$ and $\sin$ being continuous almost everywhere. (Apologies for jumping ahead; my Riemann integrability skills are rusty.)

• Very nice answer! Confuses only one moment that I don't know theorem 11.33. I know it's statement but it's based on measure theory. Why functions $g(t)\cos t$ and $g(t)\sin t$ is continuous almost everywhere? Dec 19, 2015 at 19:02
• $g(t)$ is the product of $f(x-t)-f(x)$, which is discontinuous exactly where $f(x-t)$ is, i.e., at the same places that $f$ is discontinuous, and $1/\sin(t/2)$, which is discontinuous at $t=2n\pi$ for $n$ an integer, a countable set. $g(t)$ is thus continuous almost everywhere (= except at a set of measure zero, except maybe you haven't learned what that is yet). Given that $\cos t$ and $\sin t$ are continuous everywhere, we get the desired result. Dec 19, 2015 at 20:34
• 1) "And $1/\sin(t/2)$, which is discontinuous at $t=2n\pi$ for $n$ an integer, a countable set." - I think that we consider here $n=0$ since we consider $[-\pi,\pi].$ 2) Also you said that $f$ may be discontinuous. The set of discontinuity is the set of zero measure since $f$ is bounded and integrable (theorem 11.33). Right? Dec 20, 2015 at 6:21
• Yes, the set of discontinuity for $f$ is a set of measure zero since $f$ is Riemann integrable. Dec 20, 2015 at 8:56
• I’m not sure I see why you have that it’s less than 4M|t|. Doesn’t the absolute value of t cancel so it should only be less than 4M? I know that doesn’t change the result that it’s bounded. Thanks for any insight! Feb 20, 2021 at 7:22