Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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
1  
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
1  
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
show 13 more comments

2 Answers 2

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
Hi, thanks for the effort - but I have to integral the absolute value instead. –  Bombyx mori Dec 31 '12 at 20:06
2  
+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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.