Injectivity of Fourier series in a horizontal strip I'm trying to prove (or give a counter-example, which I couldn't find) the following statement:
Let $g : \mathbb{R} \to \mathbb{C}$ be such that $g(x)=g(x+2 \pi)$ and $g(x)=\sum_{n} c_{n}e^{inx}$, then for points $x, y$ in the horizontal strip $|x-y|<2 \pi$, $g$ is injective.
My reasoning went like this:
Let $g(x)=g(y)$, for $x$, $y$ such that $|x-y|<2 \pi$.
Since $g$ can be writen in complex Fourier series, by Cantor's theorem, it follows that $c_ne^{inx}=c_ne^{iny}$, for at least one $n$. But this implies, since $|x-y|<2 \pi$, that $x=y$. Proving the proposition.
That was what I thought, if someone can help with a counter-example or some comments on the proof, I thank you a lot.
Edited: I'll try to be clearer.
I'm trying to prove (or give a counter-example) to the following statement:
Let $g : \mathbb{R} \to \mathbb{C}$ be such that $g(x)=g(x+2 \pi)$ and $g(x)=\sum_{n} c_{n}e^{inx}$, then there exist points $x, y$ with $|x-y|<2 \pi$ in such a way that $g(x)=g(y)$ implies $x=y$, so for the set of these points $g$ is injective.
What I thought was to let $g(x)=g(y)$, for $x$, $y$ such that $|x-y|<2 \pi$.
Then, by the uniqueness of Fourier series, it follows that $c_ne^{inx}=c_ne^{iny}$, for at least one $n$, (and here is my doubt, because I don't know what is meant by a Fourier series to be the same, I just imagine that if it has the same Fourier coefficients, then at least one term in the expansion must be different for different points, since is the same function.). But this implies, since $|x-y|<2 \frac{\pi}{|n|}<2 \pi$, that $x=y$. Proving the proposition.
My main doubt is this: if a function has a Fourier expansion, is it true that for two different points $x, y$ in its domain, at least one term in the expansion of $x$ will differ from that of $y$? (Excluding the constant function).
Sorry if the first question was confusing. Thank you a lot!

Take a Weierstrass function $f(x)=\sum a^n\cos(b^n\pi x), 0<a<1, b=2k+1, ab>1+3\pi/2$
Then $f$ (given as a Fourier series) is continuous and nowhere differentiable which means there is no interval $(c,d)$ where it is injective as it would be monotonic there (by continuity) and hence differentiable ae there by Lebesgue's theorem
