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I'm trying to prove that $\forall \delta >0, \, \exists C>0 \, \,s.t.\, |\int_{x}^{\pi}D_n(t)\, dt| \leq \frac{C}{n}$ for all $x \in [0,\delta]$, where $D_n$ denotes the Dirichlet kernel. I'm reading this from Giaquinta's Linear and metric structures (p. 446). The author gives two hints. One way is to do integration by parts. I think he means to integrate by parts the following: $$ \int_{x}^{\pi} {\sin((n+1/2)x) \over \sin{(x/2)}}dx. $$ But the two obvious ways to integrate this by parts give me back some horrible expressions that i don't know how to use in order to prove the result. The other hint is to use the following: $$ \lim_{n \rightarrow \infty}\int_{\delta <|s|\leq \pi} f(t) D_n(t) \,dt =0 $$ for any integrable periodic $f$, and $\delta >0$. I've tried this last expression with $f=1$, but I don't see how the conclusion would follow from this. By the way, the primitive of the Dirichlet kernel, which I've calculated with Mathematica, is too complicated to work with. To make things worse, I don't know how to prove the statement of this last hint (although the preliminary theorem that appears in the book I do understand). Any suggestions would be greatly appreciated.

PD: I wrote the second hint as it appears in the book, but I guess that where it says "$|s|$" it should say "$|t|$".

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We learned yesterday (!) that $\int_0^\pi D_n(t) \,dt$ is bounded w.r.t. $n$. If you choose $f$ so that $f(t)=n$ on a small neighborhood of zero and $f(t)=1$ away from zero, you can probably deduce that $n \int_\delta^\pi D_n(t)\, dt$ is bounded... –  Siminore Aug 9 '12 at 15:58
    
Thank you Siminore, once again, i will think about this. Regarding the proof of the last hint, i think it should follow from math.stackexchange.com/questions/152763/dirichlet-kernel . –  Pani Aug 9 '12 at 16:28

2 Answers 2

I couldn't prove Siminore's last suggestion, but I could arrive at $n\int_{\delta}^{\pi}\leq C$, for a certain constant $C$, doing an integration by parts. Applying $$ \int_{\delta}^{\pi}u dv= uv\big|_{\delta}^{\pi} - \int_{\delta}^{\pi}vdu, $$ with $$ u=1/\sin[t/2], $$ we get that $$ \int_{\delta}^{\pi}{\sin[ (n + 1/2)t] \over \sin[t/2]} \, dt = -{\cos[(n + 1/2)t] \over (n + 1/2)\sin[t/2] }\big|_{\delta}^{\pi} + \int_{\delta}^{\pi}{ \cos[(n + 1/2)t] \cos[t/2] \over 2(n + 1/2)\sin^2[t/2]} dt. $$ The result is pretty obvious from this expression. Now the inital problem reduces to show that this implies $$ \Bigg| \int_{x}^{\pi} D_n(t)\, dt \Bigg | \leq {C^{*} \over n} $$ for all $x \in [0,\delta]$, for a certain $C^{*}$. I am still having trouble with this last implication though. By the way, a primitive of the Dirichlet Kernel is easy to express if we consider the expression $$ D_n(t)= 1+ 2\sum_{k=1}^{n} \cos[kt] \, dt. $$

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We have a problem for $x=0$, since $$ \int_{0}^{\pi} D_n (t) dt = \pi, $$ and this cannot be bounded by $c(\delta)/n$, wich tends to zero.

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Well, I think the author probably meant $x \in [\delta,\pi]$. Just another typo. I only needed this to prove the Dirichlet-Jordan theorem, but what i have shown in my first answer is enough to do so. I hope this helps somebody someday. –  Pani Oct 2 '12 at 13:28

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