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.

  • $\begingroup$ 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. $\endgroup$
    – anon
    Jul 31, 2011 at 18:36
  • $\begingroup$ This follows from the Lusin-Denjoy theorem, because $\sum_{n=1}^{\infty}\rho_n|cos(nx+x_n)|$ when $\rho_n^2=a_n^2+b_n^2$, is finite in every point in a set, and aalso the right series converges absolutely. $\endgroup$ Jul 1, 2023 at 12:31

4 Answers 4


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 $\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} b_n = 0$.


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.


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 $0$.

  • $\begingroup$ 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 $\mathbb{R}$. $\endgroup$ Aug 1, 2011 at 16:29
  • $\begingroup$ I am talking about integration on $S^1$ (or $[0,2\pi]$). $\endgroup$
    – user10676
    Aug 2, 2011 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.

  • $\begingroup$ I like your approach very much. :) $\endgroup$ Aug 5, 2011 at 9:06
  • $\begingroup$ Thanks. Baire category is great! =) $\endgroup$
    – Sam
    Aug 5, 2011 at 15:00

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