Let's consider function $f(x)$ that is $2\pi$-periodic, analytical and $f(x) \ge m > 0$. And consider its Fourier series $f(x) = \sum\limits_{k=-\infty}^{\infty}a_k e^{ikx}$. Then consider $\frac{1}{f(x)}$ and its Fourier series $1/f(x) = \sum\limits_{k=-\infty}^{\infty}b_k e^{ikx}$. Then I claim that $\sum\limits_{k=-n}^{n}b_{-k}a_k$ tends to $1$. How to prove it?
Really I have no idea how to prove it. Maybe apply something like summation by parts? I tested it numerically for some functions - it works. But I really don't know even how to start the proof.
Great thanks for any help or ideas!
Both $f(x)$ and $g(x)=\frac{1}{f(x)}$ are functions in $L^2(0,2\pi)$. In particular, their (symmetrically) truncated Fourier series are converging to $f(x)$ and $g(x)$ in $L^2$. The term $$ \sum_{n=-N}^{N}a_n b_{-n} $$ is the mean value, over $(0,2\pi)$, of the product between two truncated Fourier series. Since the product of such truncations converges to $f(x)\cdot g(x)=1$ in $L^2$, $$ \lim_{N\to +\infty}\sum_{n=-N}^{N}a_n b_{-n} = \frac{1}{2\pi}\int_{0}^{2\pi}f(x)\,g(x)\,dx = 1 $$ as wanted. It is interesting to apply such lemma to $f(x)=\frac{1}{3+2\cos(x)}$, for instance. In such a case we find that the Fourier coefficients of $f(x)$ are deeply related with the continued fraction of $\sqrt{5}$.