Given are the almost every where smooth, $2\pi $ periodic, absolutely integrable function $f:\mathbb R \to \mathbb C$. In this case we have finite many points $\gamma_1,\dots ,\gamma_n\in [-\pi,\pi)$ where $f$ is not continuous, but every where else is $f$ smooth. We can also assume that the limits $\lim_{h\to 0} f(x\pm h),\: \lim_{h\to0}f'(x\pm h)$ always exists for every $x\in \mathbb R$.
Now I want to show that $$c_k(f):=\frac 1 {2\pi} \int_{-\pi}^{\pi} f(x)e^{-ikx}dx$$
does not decrease faster than $\tfrac 1{|k|}$.
So this is what I tried:
We can define the "sawtooth" function $$V_\gamma(x):=\cases {\tfrac 1 {2\pi}(\gamma -\pi-x)\quad,-\pi<x<\gamma\\\tfrac 1 {2\pi} (\gamma+\pi-x)\quad,\gamma\leq x\leq\pi}$$
which has a jump at $\gamma$ of magnitude $1$. The Fourier coefficients of $V_\gamma$ are given by
$$c_0(V_\gamma)=0,\qquad c_k(V_\gamma)=\tfrac 1 {2\pi i k}e^{-ik\gamma},\quad k=\pm1,\pm2,\dots$$
Let $A_j$ the magnitude of the jump of $f$ at $\gamma_j$, then we can define the $2\pi$ periodic, almost everywhere smooth, absolutely integrable function $g:\mathbb R\to\mathbb C$
$$g(x):=f(x)-\sum_{j=1}^n A_j V_{\gamma_j}(x).$$
We have now that $g$ is also continuously, so we know that $c_k(g)$ decrease faster to zero than $\tfrac 1{|k|}$. We get now by using the Fourier operator (and some small equation manipulation)
$$c_k(f)=c_k(g)+\tfrac 1 {2\pi i k} \sum_{j=1} A_j e^{-ik\gamma_j}.$$
My Problem is now, that I don't have a good way to show that $\sum_{j=1}^nA_j e^{-ik\gamma_j}$ does not vanish for big $k$. It should be clear since $e^{-ik\gamma _j}$ is a kind of periodic and the $\gamma_j$ are all different to each other, but I don't want to go this way.
So does anyone have an idea how I could show that $\sum_{j=1}^nA_j e^{-ik\gamma_j}$ does not vanish or maybe a total different way to show this?
Best Answer
Take $$H(x)=\sum_k c_k(H) e^{i kx}$$ $2\pi$-periodic, $C^1$ and non-zero at $\gamma_j$ only for $j=1$.
Let $h_m(x)=e^{i m x}H(x)$ so that $$h_m(x)=\sum_k e^{ikx} c_{k-m}(H)$$
If $\lim_{k\to \infty}\sum_{j=1}^nA_j e^{ik\gamma_j}=0$ you'd get that $$|A_1 H(\gamma_1)|=\lim_{m\to \infty}|\sum_{j=1}^n A_j h_m(\gamma_j)|=\lim_{m\to \infty} |\sum_k c_{k-m}(H) \sum_{j=1}^nA_j e^{ik\gamma_j}|=0$$ which is a contradiction.