3
$\begingroup$

Let $f_n(x)=u_n\sin(nx)$ where $\displaystyle\sum f_n$ converges pointwise, and $ \displaystyle x \mapsto \sum_{n=0}^{+\infty} f_n(x)$ is continuous.

Prove that $ u_n\rightarrow 0$ when n tends to $+\infty$

My 'attempt':

We know that $S_n=\displaystyle\sum_{k=1}^{n}u_k\sin(kx)$ converges to a real $a$, and $$ S_{n+1}-S_n=u_{n+1}sin((n+1)x) $$ converges to $0$. By induction we can prove that for all $x=(\frac{\pi}2)^k$ then $u_n$ tends to $0$.

I really don't know how can I continue.

Thank you in advance,

$\endgroup$
  • $\begingroup$ This is probably a sledgehammer: If $f$ is the limit function, then the $u_n$ are the Fourier $\sin$ coefficients of $f$. Since $f$ is continuous, these tend to $0$ by the Riemann-Lebesgue Lemma. $\endgroup$ – David Mitra Feb 20 '14 at 13:58
  • $\begingroup$ @DavidMitra Thanks but I don't understand why $u_n$ are necessarily the fourier sin coefficients of f ? $\endgroup$ – user117932 Feb 20 '14 at 14:58
  • $\begingroup$ $f$ is integrable on finite intervals (presumably). I may be misremembering, but I think this implies that if $f$ is expressed as a trigonometric sum, then that sum must be the Fourier series for $f$. $\endgroup$ – David Mitra Feb 20 '14 at 15:05
  • $\begingroup$ @DavidMitra In fact, it works only if the convergence is uniform $\endgroup$ – user117932 Feb 20 '14 at 16:51
  • $\begingroup$ The result in my last comment "if a trigonometric series converges everywhere to an integrable function $f$, then it is the Fourier series of $f$" is more general than that. It can be found in Zygmund's Trigonometric Series, Theorem IX 3.1. The proof of this doesn't seem easy; I would guess your problem admits a relatively simple solution. What results have you seen so far? $\endgroup$ – David Mitra Feb 20 '14 at 17:05
2
$\begingroup$

It's much simpler than my comments above indicate.

All you need is that the sum converges pointwise on a non-degenerate interval $[a,b]$ (in fact, all you need is that the sum converges on a set of positive measure; see my last comment above). Assume this is the case.

Suppose $\lim\limits_{n\rightarrow\infty} u_n\ne0$. Choose $\delta>0$ and a subsequence $(u_{n_k})$ such that $|u_{n_k}|>\delta$ for all $k\in\Bbb N$.

Note we must have $$\lim_{k\rightarrow\infty}u_k\sin(k x)= 0\ \text{ for all }\ x\in[a,b]\tag{1}$$

Given any closed interval $I\subset [a,b]$, we can find $n_k$ with $k$ as large as desired and $n_k|I|>2\pi$. With $n_k$ selected as such, we can find a closed subinterval of $J$ of $I$ so that $ \sin(n_k x)>1/2$ for all $x\in J$.

Using this, we inductively find a sequence of closed intervals $[a,b]\supset I_1\supset I_2\supset\cdots$ and positive integers $n_1<n_2<\cdots$ such that $$ |u_{n_k}|\sin(n_kx)\ge \delta/2\tag{2} $$ for $x\in I_k$, $k\in \Bbb N$.

The set $I=\bigcap\limits_{k=1}^\infty I_k$ is non-empty. Pick $x\in I$. Then $x\in[a,b]$ and satisfies $(2)$ for all $k\in\Bbb N$. But this contradicts $(1)$.

It follows that $\lim\limits_{n\rightarrow\infty} u_n=0$.


(I don't see how to obtain a simpler proof that takes advantage of your stronger hypothesis.)

$\endgroup$
  • $\begingroup$ Wah! It's excellent. Thank you very much. $\endgroup$ – user117932 Feb 20 '14 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy