TITLE: Application of Fourier Series and Stone Weierstrass Approximation Theorem
QUESTION [5 upvotes]: If $f \in C[0, \pi]$ and $\int_0^\pi f(x) \cos nx\, \text{d}x = 0$ , then $f = 0$


Define $ g(x) = \begin{cases}
         f(-x) & \text{if } -\pi \leq x < 0;\\
        f(x) & \text{if }  0 \leq x \leq \pi. \end{cases}$
which is an even function
So $g(x)$ can be written as $\sum_{n=0}^\infty a_n\cos (nx)$ for all $x \in [-\pi , \pi]$
$$\therefore \int_0^\pi f^2(x) \, dx = \int_0^\pi f(x) \left(\sum_{n=0}^\infty \cos(nx)\right) \, dx = \sum_{n=0}^\infty \int_0^\pi f(x) \cos(nx) \, dx = 0$$
$$\therefore \int_0^\pi f^2(x) \, dx =0,$$ we get $f(x) = 0$
I think this part is my true 
$$\text{If }f \in C[0 , \pi] \text{ and  } \int_o^\pi x^n f(x) \, dx = 0 \text{ for all } n\geq0, \text{ then } f = 0$$
Since $f$ is continous on a closed interval, then by stone Aproximation Theoren , for each $\epsilon > 0$ , there is a polynomial $p(x)$ such that $|f(x) - p(x)| < \epsilon$.
I want to show that $\int_0^\pi f(x) \, dx = 0$,  please help me how to proceed further and check the first part. Any help would be appreciated.

REPLY [2 votes]: I am usually too lazy to critique answers even when the OP asks for it. It is easier to just snipe
with a comment or two. But @zhw has shamed me into really looking at the answer and maybe offering 
a bit of a tutorial.  Since I gave a sloppy comment I will make amends here I hope.
Here some things that you know or should know judging from the title of the problem and your answer.

A continuous   even function $f$ on $[-\pi,\pi]$ has a Fourier series of the form
$$\sum_{n=0}^\infty a_n\cos nx$$ but this series need not converge pointwise or uniformly to $f$ unless you have
stronger assumptions on $f$. [Here you don't. A course on Fourier series may not prove this negative comment, just proving the positive comment assuming, say, that $f$ is also of bounded variation or continuously differentiable.  It is essential to know why 19th century mathematicians had to fuss so much to get convergence.]
A continuous   even function $f$ on $[-\pi,\pi]$ has a uniform approximation
by a trigonometric polynomial 
of the form
$$\sum_{n=0}^N a_n\cos nx$$ meaning that, for every $\epsilon>0$ you can select at least one such polynomial so that
$$\left| \sum_{n=0}^N a_n\cos nx -f(x)\right| < \epsilon$$
for all $-\pi\leq x \leq \pi$. [Fejer's theorem supplies this as does the Stone-Weierstrass theorem.]
[Dangerous curve ahead!]  If you change $\epsilon$ you may have to choose an entirely different polynomial, so the $a_n$ might change.  Thus statement #2 does not give you a series converging to $f$, it gives you a sequence of trigonometric polynomials converging uniformly to $f$.  In other words Stone-Weierstrass or Fejer's theorem does not give
$$f(x)=\sum_{n=0}^\infty a_n\cos nx$$ 
either pointwise or uniformly.  Don't write it!
If $f(x)=\sum_{i=0}^\infty f_n(x)$ on $[a,b]$ you cannot write  $\int_a^b f   =\sum_{i=0}^\infty \int_a^b f_n $ without claiming uniform convergence (or some more advanced property).  
If $f(x)=\lim_{n\to \infty} f_n(x)$ on $[a,b]$ you can  write  $$\int_a^b fh   =\lim_{n\to \infty}   \int_a^b f_nh $$ 
for any continuous $h$ 
if you are sure you have uniform convergence (or some more advanced property).  

Now we are in a position to tidy up your solution.  Your ideas were fine, just missing some caution.  You tried to use a series but that fails--just use a sequence instead!
For the first problem use Stone-Weierstrass to select an appropriate sequence of functions $p_n \to f$ uniformly on $[-\pi,\pi]$. Check that
$$\int_{-\pi}^\pi f(x)p_n(x)\,dx =0$$
for each $n$ and that
$$\lim_{n\to \infty}  \int_{-\pi}^\pi f(x)p_n(x)\,dx =\int_{-\pi}^\pi [f(x)]^2\,dx.$$
Conclude that $f=0$ since it is continuous. 
For the second problem use Stone-Weierstrass to select an appropriate sequence of functions $p_n \to f$ uniformly on $[0,1]$. Check that
   $$\int_{0}^1 f(x)p_n(x)\,dx =0$$
   for each $n$ and that
   $$\lim_{n\to \infty}  \int_{0}^1 f(x)p_n(x)\,dx =\int_{0}^1 [f(x)]^2\,dx.$$
   Conclude that $f=0$ since it is continuous.