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Let $f : [−π, π] → \mathbb{R}$ be continuous. Pick out the case(s) which imply that $f ≡ 0$.

(a) $\int_{-\pi}^{\pi}{x^nf(x)\,dx} = 0$, for all $n ≥ 0$.

(b)$\int_{-\pi}^{\pi}{\cos (nx)\, f(x)\,dx} = 0$, for all $n ≥ 0$.

(c)$\int_{-\pi}^{\pi}{\sin (nx)\, f(x)\,dx} = 0$, for all $n ≥ 1$ .


If we take $\sin nx$ in (b) and $\cos nx$ in (a), then the integral becomes zero. So, (b) and (c) are not true.
But I can not verify (a). Can I get some help?
Also, in the answer it is given that all are true.
Where did I make a mistake?

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2 Answers 2

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For $(a)$, use Weierstrass' approximation theorem. You know $$\int_{-\pi}^\pi fp=0$$ for any polynomial $p$. Pick $\varepsilon >0$; and approximate $f$ uniformly by a polynomial $p$, that is make $M(p)= \sup\{|f(t)-p(t)|:t\in[-\pi,\pi]\}<\varepsilon$. Now, observe that $$\int_{-\pi}^\pi f^2=\int_{-\pi}^\pi f^2-\int_{-\pi}^\pi fp=\int_{-\pi}^\pi f(f-p)$$

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  • $\begingroup$ from here it follows that (a) is true?Am I right? $\endgroup$
    – ami
    Commented Jan 17, 2014 at 5:25
  • $\begingroup$ and what about (b) and (c)? $\endgroup$
    – ami
    Commented Jan 17, 2014 at 5:25
  • $\begingroup$ @ami Since you got them right, I didn't address it. $\endgroup$
    – Pedro
    Commented Jan 17, 2014 at 5:52
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I just complete the answer of others. This is the approximation theorem related to the polynomial and the trigonometric polynomial. And $(b)$ and $(c)$ is related to the approximation of trigonometric polynomial.

$f(x)=\frac{a_0}{2}+\sum (a_ncons(nx)+b_nsin(nx))$ and $a_n,b_n$ is above in the choice $(b)(c)$

So the $(b)(c)$ just provide the half information of $f$, so it is wrong.

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