Let $f(x)$ is Continuous function on $[0,\pi]$,and for $n=1,2,.....,$ the function $f(x)$ has the following property:$$\int_{0}^{\pi}f(x)\cos{(nx)}dx=0.(n=1,2,......)$$ Proof: $f(x)\equiv C$(C is a constant) on $[0,\pi] .$
In my way of thinking ,applying Weierstrass Approximation Theorem, I could give the proof $f(x)\equiv C,$but I failed. who can help me go futher about this ?Any help will be appreciated!
A typical way to prove that a function $f$ is constant is to show that $$\int_0^\pi f(x)g(x)\,dx=0\tag{1}$$ for every function $g\in C[0,\pi]$ with zero mean (i.e., $\int_0^\pi g(x)\,dx=0$). Indeed, if $f(x_1)\ne f(x_2)$, then let $g$ be the function that is zero on most of the interval and has two triangular peaks at $x_1,x_2$: one positive, one negative. This $g$ satisfies $\int_0^\pi g(x)\,dx=0$ but $(1)$ fails when the peaks are sufficiently narrow.
As Daniel Fischer noted, there are multiple ways to prove that the condition (1) is implied by $$\int_{0}^{\pi}f(x)\cos{(nx)}dx=0,\quad n=1,2,\dots \tag{2}$$ For example, one can use the fact that the cosine system is an orthonormal basis of $L^2[0,\pi]$, and thus excluding $1$ from it leaves us with the orthogonal complement of $1$.
Alternatively, use thw Stone-Weierstrass theorem to uniformly approximate $g$ by a linear combination of $\cos \pi n x$, $n=0,1,2,\dots$; observe that the coefficient of $\cos 0=1$ is small and therefore can be dropped while maintaining uniform approximation.