$g: [-\pi,\pi]\to R$ cont. with $g(\pi) = g(-\pi)$, $\int_{-\pi}^{\pi}g(t)\sin(nt)dt = 0$ show that $g$ is even.

Question

I have to show that $g: [-\pi, \pi] \to \mathbb R$ is even when: it is continuous, $g(\pi) = g(-\pi)$, and when for all $n \in \mathbb N$:

$$\int_{-\pi}^{\pi}g(t)\sin(nt)\mathrm dt = 0.$$

My attempt: I know that if $g$ is odd:

$$g(x) = \frac{g(x) - g(-x)}{2},$$

then

$$\int_{-\pi}^{\pi}g(t)\sin(nt)\mathrm dt =\int_{-\pi}^{\pi}g(t)\sin(nt)\mathrm dt - \int_{-\pi}^{\pi}g(-t)\sin(nt)\mathrm dt = 0.$$

Hence,

$$\int_{-\pi}^{\pi}g(t)\sin(nt)dt =\int_{-\pi}^{\pi}g(-t)\sin(nt)\mathrm dt$$

But I can not prove that $g$ is even... I was following a hint. Thanks.

Answer 1

Hint: use the Fourier series of the function g(t), it is the sum of even functions and so is even.

Answer 2

HINT: all of the Fourier coefficients of the function $\frac{ g(t) -g(-t)}{2}$ are $0$.