Let $f:[0,\pi] \to \mathbb{R}$ be a continuous function such that $$\int_0^\pi f(x)sin(nx) \ dx=0$$ for all $n\geq 1$. Prove that $f$ is identically zero.
I found the problem is the span of $\{\sin(nx)\mid n \ge 1\}$ does not form an algebra and there is no constant function. Then I could not see how can I apply Stone Weierstrass theorem as the conditions are not fulfilled. Please help me out to solve this. Thank you in advance!!
$C^0([0,\pi])\subset L^2([0,\pi])$, hence if we define $g(x)$ as $f(x)$ for any $x\in[0,\pi]$, $-f(-x)$ for any $x\in[-\pi,0)$, we have, for any $x\in(-\pi,\pi)$: $$ g(x)= c_0+\sum_{n\geq 1}c_k \sin(kx),\quad c_k = \frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\sin(kx)\,dx = \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(kx)\,dx = 0.$$ That gives that $g(x)$ is a costant function, hence $f(x)$ is a constant function, but since $\int_{0}^{\pi}f(x)\sin(x)\,dx = 0$, such a constant must be zero. As an alternative, $g(x)$ is constant and odd on $(-\pi,\pi)\setminus\{0\}$, hence $g\equiv 0$ implies $f\equiv 0$.