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