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I have tried and failed to prove the Dirichlet–Dini Criterion for Pointwise convergence of Fourier series which is as follows (and is described here: http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Pointwise_convergence) enter image description here

I will appreciate a proof for this theorem or a reference to one - i couldn't do either.

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I'm gonna simplify the problem by assuming that $\ell=x_0=0$: notice that $$ \dfrac{f(x_0+t)+f(x_0-t)}{2}-\ell = \dfrac{(f(x_0+t)-\ell)+(f(x_0-t)-\ell)}{2}. $$ Hence Dini hypothesis is the same as saying that the function $f_1(t) = f(x_0+t)-\ell$ satisfies $$ \int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right|\dfrac{dt}{t}<\infty. $$ As it's proved later in the lemma, translating functions by a complex number $\ell$ and translating the axis $[0,2\pi]$ by $x_0$ acts nicely on the corresponding Fourier series so, if I can find the limit of $S_N(f_1;0)$, I can find the limit of $S_N(f;x_0)$ as well. Thus I'll proof the case $\ell= x_0 =0$ and then use this "nice action" for the general case. Consider $$ g(t)=\dfrac{f(t)+f(-t)}{1-e^{i t}} $$ Let'us see that $g$ is integrable. Notice that $h(t) = \dfrac{t}{1-e^{it}}$ is continuous at $[0,2\pi]$ (using Hôpital for instance) and take $$ K=\max_{[0,2\pi]} h(t)<\infty. $$ Now $$ \int_0^{2\pi}\left|g(t)\right|\,dt= \int^{2\pi}_0\left|\dfrac{f(t)+f(-t)}{t}\right|\left |h(t)\right| \,dt\leq 2K\int^{2\pi}_0\left|\dfrac{f(t)+f(-t)}{2}\right| \,\dfrac{dt}{t} $$ which is finite by hypothesis, thus it's integrable.

Compute now $\hat f(n)+\hat f(-n)$ for every $n$. Notice that \begin{align*} \hat f(n)+\hat f(-n)&=\int ^{2\pi}_0 f(t)e^{-itn}\,dt + \int ^{2\pi}_0 f(t)e^{itn}\,dt\\ &\overset{(*)}{=}\int ^{2\pi}_0 f(t)e^{-itn}\,dt+\int ^{2\pi}_0 f(-t)e^{-itn}\,dt\\ &=\int ^{2\pi}_0 (f(t)+ f(-t))e^{-itn}\,dt\\ &=\int ^{2\pi}_0 g(t)(1-e^{it})e^{-itn}\,dt\\ &=\hat g(n)-\hat g(n-1). \end{align*} I used in $(*)$ the change of coordinates $t\rightarrow t-2\pi$ for the second integral (recall that $e^{it}$ and $f_1(t)$ are $2\pi$- periodic). In particular $$ 2S_N(f;0)= 2 \sum_{|n|\leq N} \hat f (n)=\sum_{|n|\leq N}\hat f(n)+\hat f(-n)=\hat g(N)-\hat g(-N-1). $$ since we got a telescoping sum. Riemann-Lebesgue lemma applies to $g(t)$, hence $$ \lim_{|N|\rightarrow \infty}\hat g(N) = 0 $$ and $$ \lim_{N\rightarrow \infty} 2S_N(f;0) = 0=\ell. $$ For the general case reduce to the first one by considering $f_1(t)=f(t+x_0)-\ell$. Indeed
$$ \int^{2\pi}_0\left|\dfrac{f_1(t)+f_1(-t)}{2}\right| \,\dfrac{dt}{t}=\int^{2\pi}_0\left|\dfrac{f(t+x_0)+f(-t+x_0)-2\ell}{2}\right| \,\dfrac{dt}{t} =\int^{2\pi}_0\left|\dfrac{f(t+x_0)+f(-t+x_0)}{2}-\ell\right| \,\dfrac{dt}{t} $$ which is finite by hypothesis. By the discussion above $$ \lim_{N\rightarrow \infty}S_N(f_1;0) = 0. $$ It remains to point out that

Lemma

$S_N(f_1;0) = S_N(f;x_0) - \ell$ for every $N\geq 0$.

Proof

\begin{align*} \hat f_1(n)&=\dfrac{1}{2\pi}\int^{2\pi}_0 f_1(t) e^{-itn}\,dt\\ &=\dfrac{1}{2\pi}\int^{2\pi}_0 (f(t+x_0)-\ell) e^{-itn}\,dt\\ &\overset{(*)}{=}\dfrac{1}{2\pi}\int^{2\pi}_0 (f(t)-\ell) e^{-itn+ix_0n}\,dt\\ &=\dfrac{e^{ix_0n}}{2\pi}\int^{2\pi}_0 (f(t)-\ell) e^{-itn}\,dt\\ &=e^{ix_0n} \hat f(n) - \dfrac{e^{ix_0n}\ell}{2\pi}\int^{2\pi}_0e^{-int}\,dt \end{align*} I use at (*) the change of variables $\,t\rightarrow t-x_0$ and that $f$ is $2\pi$-periodic. Thus $$ \hat f_1(n)= \begin{cases} e^{ix_0n}\hat f(n),\quad \text{if $n\neq0$}\\ \hat f(0) -\ell,\quad \text{if $n= 0$} \end{cases} $$ In particular

\begin{align*} S_N(f_1;0)&=\sum_{|n|\leq N} \hat f_1(n) e^{in 0}\\ &=\sum_{|n|\leq N} \hat f_1(n)\\ &=f(0)-\ell+\sum_{0<|n|N} \hat f(n)e^{inx_0}\\ &=S_N(f; x_0)-\ell \end{align*} QED

Hence, $$ \lim_{N\rightarrow \infty}S_N(f;x_0) -\ell=\lim_{N\rightarrow \infty}S_N(f_1;0)=0. $$

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  • $\begingroup$ Can you please elaborate on how does this proof actualy proves the theorem? And why could you say in the beginning that you assume WLOG that x_0 = l = 0? Thanks $\endgroup$ – jon Prime Jan 19 '15 at 0:56
  • $\begingroup$ Edited. I've added the details for the general case. I hope it's clearer now. $\endgroup$ – eduard Jan 19 '15 at 9:48
  • $\begingroup$ Can you please explain why is the inequality ∫ 2π 0 |g(t)|dt≤∫ 2π 0 ∣ ∣ ∣ f(t)+f(−t)t ∣ ∣ ∣ |h(t)|dt correct? $\endgroup$ – jon Prime Jan 19 '15 at 23:27
  • $\begingroup$ In fact it is an equality. Multiply the right Hand Side and simplify $t$. $\endgroup$ – eduard Jan 20 '15 at 12:16
  • $\begingroup$ Can you please elaborate on how you solve it for the general case when l and x_0 aren't 0? $\endgroup$ – jon Prime Jan 20 '15 at 21:57
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See A. Zygmund, Trigonometric Series, Third edition, Volumes I & II combined, Cambridge Mathematical Library, Cambridge University Press, 2002 on page 52.

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I think this article might help you. Pointwise Convergence of Fourier Series, Charles Fefferman, Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 551-571:

http://www.jstor.org/discover/10.2307/1970917?sid=21105651264483&uid=4&uid=2&uid=3737760

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Start with convergence criterion: A necessary and sufficient condition for the Fourier series T(x) of f to converge pointwise to c(x) on E is that there exists a fixed $\delta$ such that $ 0 < \delta < \pi$ and $\int_0^\delta {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $ pointwise on E.

Here ${g_{c(x)}}(u) = \frac{1}{2}\left( {f(x + u) + f(x - u) - 2c(x)} \right)$.

If $\frac{{{g_c}(u)}}{u}$ is integrable,which is your given condition then by the the Riemann Lebesgue Theorem,$\int_0^\delta {{g_{c(x)}}(u)\frac{{\sin \left( {nu} \right)}}{u}du \to 0} $. This proves pointwise convergence. Here you can take any $\delta > 0$.

For details see convergence criterion and Theorem 25 in

Convergence of Fourier Series

https://037598a680dc5e00a4d1feafd699642badaa7a8c.googledrive.com/host/0B4HffVs7117IbmZ2OTdKSVBZLVk/Fourier%20Series/Convergence%20of%20Fourier%20Series.pdf

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