Show that $f(x)\equiv 0$.

Question

Let $f:[0,2\pi]\to\mathbb{R}$, which is $2\pi$ periodic and continuous. It is given that for every $n\in\mathbb{Z}$:$$\int_0^{2\pi} f(x)e^{i\left(n+\frac{1}{2}\right)x} = 0.$$ Show that $f(x)\equiv 0$.

So this expression is almost the $n$-th Fourier coefficient $$\hat f(n) = \frac{1}{2\pi} \int_0^{2\pi} f(x)e^{-inx}\ dx$$

I am almost certain there needs to be some substitution in order to express the above with $\hat f(n)$.

I'd be glad for help.

Answer 1

Consider $g(x) = f(x) \cdot e^{ix/2}$ and find out what the assumption tells you about $g$. Then find out what this tells you about $f$.

Answer 2

Hint: if $f(x)$ is $2\pi$-periodic, then $f(x) e^{ix/2}$ is $(\ldots)$-periodic. Fill the dots and use Fourier decomposition with such a periodicity.