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Rudin asked me to bound $$\int_{-\pi}^{\pi} |D_{n}(t)|\,dt$$ from above. I need a bound at the level of $o(\log(n))$.

The background is:

If $s_{n}$ is the $n$-th partial sum of the Pourier series of a function $f\in C(T)$, prove that $$\frac{s_{n}}{\log[n]}\rightarrow 0$$ uniformly. That is, prove that $$\lim_{n\rightarrow \infty}\frac{|s_{n}|_{\infty}}{\log(n)}=0$$On the other hand, if $\lambda_{n}/\log[n]\rightarrow 0$, prove that there exist an $f\in C^(T)$ such that sequence $s_{n}(f,0)/\lambda_{n}$ is unbounded.

Update: a numerical evaluation for $n=10^{800}$ is inconclusive.

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Is not the integral of the Dirichlet kernel equal to $2\pi $, according to its representation as a trigonometric sum? Or did you mean the integral of $|D_n|$? – user53153 Dec 31 '12 at 18:55
The Dirichlet kernel. – Bombyx mori Dec 31 '12 at 19:03
Be sure to rewrite your question to make this clear. As for the estimate, my suggestion is to focus on the intervals where $|\sin (n+1/2)t|\ge 1/2$ and use the bound $|\sin t/2|\le t/2$ there. This way you get a sum of integrals of $\int t^{-1}dt$ over a bunch of intervals, which should lead to the desired bound. – user53153 Dec 31 '12 at 19:27
Think again. Up to a multiplicative constant, there is no waste in restricting to the set where $|\sin (n+1/2)t|>1/2$. There is no waste in $|\sin t/2|\le t/2$ either. From there on, write it out carefully. "$C\sum \frac{1}{n}$" is not being careful. // Also, I checked with the book and it asks $>C \log n$". The notation $o(\log n)$ means something different. See here but do not overuse this notation where an ordinary inequality can be written just as easily. – user53153 Dec 31 '12 at 20:17
OK, this makes things clear. You are referring to Real and Complex Rudin. Exercise 19 does not say that $\|D_n\|_1=o(\log n)$; this particular statement is yours, not Rudin's. And, as noted above, a false one. – user53153 Dec 31 '12 at 20:38
up vote 1 down vote accepted

I'll replace $|\sin t/2|$ by $|t|$ since they are comparable: $$\frac{|t|}{\pi}\le \left|\sin\frac{t}{2}\right| \le \frac{|t|}{2} \ \text{ for }t\in [-\pi,\pi]$$ Claim: for all $\lambda\ge 1$ $$\frac{1}{3}\log \lambda \le \int_0^\pi \frac{|\sin \lambda t|}{t}\,dt \le \log \lambda +\log \pi +1.$$

Proof. For the upper bound, split the integral into "small $t$" part and the rest: $$\int_0^{\lambda^{-1}} \frac{|\sin \lambda t|}{t}\,dt \le \int_0^{\lambda^{-1}} \frac{\lambda t}{t}\,dt =1$$ and $$\int_{\lambda^{-1}}^\pi \frac{|\sin \lambda t|}{t}\,dt \le \int_{\lambda^{-1}}^\pi \frac{1}{t}\,dt = \log\lambda + \log \pi$$

The lower bound needs a bit more work. Since the integrand is nonnegative, we can restrict the region of integration to the set $|\sin \lambda t|\ge 1/2$. This set contains the intervals $I_k=[\pi \lambda^{-1} (k+1/6), \pi \lambda^{-1} (k+5/6)]$ for all integers $k$ such that $0\le k \le \lambda-1$. The integral over $I_k$ is at least $$ \int_{I_k} \frac{1/2}{t}\,dt \ge |I_k| \frac{1/2}{\pi \lambda^{-1} (k+1)} = \frac{1/3}{k+1} $$ Therefore, the integral is bounded from below by $$\frac13 \sum_{k=0}^{\lfloor \lambda-1\rfloor }\frac{1}{k+1} \ge \frac{1}{3} \log \lambda.$$

Remark. If $\|D_n\|_{L^1}=o(\log n)$ were true, the estimate in #19 would hold for the Fourier series of any finite measure on $[-\pi,\pi]$. This is not the case. The fact that $s_n$ comes from a continuous function should be used.

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This is not really needed, because Rudin already showed this in his book. – Bombyx mori Dec 31 '12 at 23:55
@user32240 You are welcome. – user53153 Dec 31 '12 at 23:56
But thanks for the help nevertheless; I need to think about the problem myself for a while. Since it is in 3rd edition I assume Rudin must had been serious about it. – Bombyx mori Dec 31 '12 at 23:58

Well, you can get the value of the integral very explicitely. As you already noted $$ \int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt = \int_{-\pi}^\pi \frac{\sin(nt)}{\sin\frac t2}\cos\frac t2dt = \int_{-\pi/2}^{\pi/2} \frac{\sin(2nt)}{\sin t}2\cos t\,dt\,. $$ Now the fraction can be transformed into $$ \frac{\sin(2nt)}{\sin t} = \frac{e^{2nti}-e^{-2nti}}{e^{ti}-e^{-ti}} = \frac{e^{4nti}-1}{e^{2ti}-1} = 1+e^{2ti}+e^{4ti}+\cdots+e^{(4n-2)ti} $$ We multiply this with $2\cos t=e^{-ti}+e^{ti}$ and get for the integrand $$ \frac{\sin(2nt)}{\sin t}2\cos t = e^{-ti}+2e^{ti}+2e^{3ti}+\cdots+2e^{(4n-5)ti}+2e^{(4n-3)ti}+e^{(4n-1)ti}. $$ Furthermore, $$ \int_{-\pi/2}^{\pi/2} e^{kti}dt = \frac2k\sin\frac{k\pi}2, $$ and hence we finally get $$ \int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt =2+4-\frac43+\frac45-\frac47\pm\cdots+\frac{4}{4n-3}-\frac{2}{4n-1} $$ which is obviously convergent by the Leibniz criterion and hence bounded by a constant.

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Hi, thanks for the effort - but I have to integral the absolute value instead. – Bombyx mori Dec 31 '12 at 20:06
+1 and please don't delete the answer: combining it with one of my comments, one can get a proof of $\pi = 4-4/3+4/5-4/7+\dots$ which is neat by itself. – user53153 Dec 31 '12 at 20:10
True. Nice. Unfortunately, I need to go now. Maybe I'll solve the problem another time. – Ralph Tandetzky Dec 31 '12 at 20:15

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