If a function $f(x)$ is periodic with $p$ such that $f(x)=f(x+p)$, then suppose I look at its Fourier co-efficents $C_k$, with $$C_k=\frac{1}{p}\int\limits_{p}f(x)e^{-2\pi kx/p}\mathrm{d}x$$
Is it possible that a function $f(x)$ exists $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $C_1=C_{-1}=0$?
So as far as my analysis goes, $$\frac{1}{p}\int\limits_{p}f(x)e^{-2\pi x/p}\mathrm{d}x=0$$ $$\Rightarrow \int\limits_{p}f(x)\cos{\frac{2\pi x}{p}}\mathrm{d}x=0$$ and simultaneously $$\Rightarrow \int\limits_{p}f(x)\sin{\frac{2\pi x}{p}}\mathrm{d}x=0$$ I don't know of any function that would fulfill the previous two criterion and would also be periodic with a period of $p$. Maybe, looking at the transform in the reciprocal domain would help because I would end up convoluting stuff with delta functions which I guess would be simpler, but will try it and see, but I think that I can't have function which behave in this manner or from a first look at the functions, my intuitions tell me that they would end up having periods of $\frac{p}{2}$