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.

I have seen the following problem in a test, and there are some elementary solutions to it. I am curious if there is a solution involving Fourier series.

Here it is:

Let $(a_n),(b_n)$ be two sequences of reals such that $$ \lim_{n \to \infty} a_n \cos(nx)+b_n \sin(nx)=0,\ \forall x \in (c,d) $$ where $c<d$ are two real numbers. Prove that $ a_n,b_n \to 0$.

I am interested in a solution using Fourier series. Thank you.

share|improve this question
    
Well, I see how to prove it using standard analysis, FWIW. I imagine a solution of the desired variety would involve constructing a function $f$ with Fourier coefficients $a_n+ib_n $, translating the limit hypothesis into a statement about $f$ (e.g. pointwise convergence), then translating that into something about the real and imaginary parts of $f$, and then deriving the intended conclusions. –  anon Jul 31 '11 at 18:36

4 Answers 4

up vote 7 down vote accepted

I think it's going to be hard to find a Fourier series proof that isn't totally artificial since there aren't any series in the problem. But I can think of a real analysis proof that isn't elementary.

Let $f_n(x) = a_n\cos(nx) + b_n\sin(nx)$. Then there is some $\alpha_n$ such that $f_n(x) = (a_n^2 + b_n^2)\cos(nx - \alpha_n)$. If $n$ is large enough, an entire period of $f_n(x)$ will be contained in $(c,d)$. So the measure of the set of $x$ in $(c,d)$ for which $|f_n(x)| > {1 \over 100}(a_n^2 + b_n^2)$ will be at least ${1 \over 2}(d - c)$ if $n$ is large enough.

But a sequence of functions that converges pointwise on an interval converges in measure to the same limit. So given $\epsilon > 0$, for large enough $n$ you'd also have to have the measure of $\{x \in (c,d): |f_n(x)| > \epsilon\}$ would have to be less than ${1 \over 2}(d - c)$. The only way this is compatible with the above is that for large enough $n$, you have ${1 \over 100}(a_n^2 + b_n^2) < \epsilon$. And this is the same as saying that $\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} b_n = 0$.

share|improve this answer

After my wrong attempt yesterday here is a (hopefully correct) proof that avoids measure theory.

Let $A_n:=\sqrt{a_n^2+b_n^2}$ be the amplitude of $f_n(x):=a_n\cos(nx)+b_n\sin(nx)$. We have to prove that $\lim_{n\to\infty} A_n=0$.

If this is not the case then there is an $\epsilon>0$ such that $A_n\geq 2\epsilon$ for infinitely many $n$. We are going to construct a selection sequence $n_j \to\ \infty$ $(j\to\infty)$ and a sequence of nested closed intervals $I_j\subset\ ]c,d[\ $ $\ (j\geq1)$ of positive length $|I_j|$, such that $$f_{n_j}(x)\ \geq\ \epsilon\quad(x\in I_j)\qquad\qquad(*)$$ for all $j\geq1$. Given these sequences there is a point $\xi$ which belongs to every $I_j$, and for this point $\xi$ we have $f_{n_j}(\xi)\geq\epsilon$ for all $j\geq1$. This would violate the assumption $\lim_{n\to\infty}f_n(\xi)=0$.

To initialize the two sequences $(n_j)_{j\geq1}$ and $(I_j)_{j\geq1}$ we put $n_0:=0$ and $I_0:=\ ]c,d[\ $. Now the recursion step: Given $n_{j-1}$ and $I_{j-1}$ (of positive length) for some $j\geq1$, there is a $n_j>n_{j-1}$ such that $${\rm (a)}\quad {2\pi\over n_j}<|I_{j-1}|\ ,\qquad{\rm (b)}\quad A_{n_j}\geq2\epsilon\ .$$ It follows that $I_{j-1}$ contains a full period of $f_{n_j}$; therefore there exists a closed interval $I_j\subset I_{j-1}$ of positive length such that $(*)$ holds.

share|improve this answer

This argument uses Fourier series, but with the assumption that $(a_n,b_n)$ is bounded (I can't found a way to prove this without somehow proving that it converges to $0$. Anyone is free to suggest a solution).

It is harmless to suppose that $(c,d)=\mathbb{R}$. Since $(f_n)$ converges pointwise to zero and is bounded, the Dominated Convergence Theorem shows that it converges in $L^1(S^1)$-norm. Since the Fourier transform $L^1(S^1) \rightarrow L^\infty(\mathbb{Z})$ is continuous, its Fourier coefficients go to $0$.

share|improve this answer
    
Um... you cannot assume $(c,d) = \mathbb{R}$, can you? The functions $f_n = a_n\cos(nx) + b_n\sin nx$ are not integrable over $\mathbb{R}$. –  Willie Wong Aug 1 '11 at 16:29
    
I am talking about integration on $S^1$ (or $[0,2\pi]$). –  user10676 Aug 2 '11 at 12:45

Tying to prove that the sequences $a_n$ and $b_n$ are bounded (to fill in the gap in user10676's approach), the following argument - using Baire category - turned up:

Let $\epsilon > 0$ be given. Define $F_N = \{x\in (c,d)\;|\;|f_n(x)|\le\epsilon \quad \forall n \ge N \}$. We have

$$(c,d)=\bigcup_{N \in \mathbb N} F_N$$ by pointwise convergence and furthermore the sets $F_N$ are closed by continuity of the functions $f_n$.

By the Baire category theorem there must be $N_0$ such that $F_{N_0}$ has nonempty interior. Choose two points $s,t$ such that $(s,t)\subset F_{N_0}$. Note that then in particular we have that $(s,t) \subset F_N$ for all $N>N_0$.

Supposing $N_0$ to be large enough (making it bigger will do no harm), the congruences \begin{eqnarray*} Nx &\equiv& 0 \pmod{2\pi}\\ Ny &\equiv& \frac\pi 2 \pmod{2\pi} \end{eqnarray*}

have solutions in $(s,t)$ for all $N>N_0$. But then for any $N>N_0$, choosing $x,y$ as above

\begin{eqnarray*} |a_N| &=& |a_N\cos(Nx) + b_N\sin(Nx)| \le \epsilon \\ |b_N| &=& |a_N\cos(Ny) + b_N\sin(Ny)| \le \epsilon \end{eqnarray*}

because $x,y\in (s,t)\subset F_{N_0}$.

This proves that the sequences $a_n$ and $b_n$ converge.

share|improve this answer
    
I like your approach very much. :) –  Beni Bogosel Aug 5 '11 at 9:06
    
Thanks. Baire category is great! =) –  Sam Aug 5 '11 at 15:00

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.