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I have the following problem to solve.

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My attempt

a. $\int_0^{\pi} x^n f(x) dx =0$ $\forall$ $n \ge 0$ gives $ x^n f(x) dx =0$ almost everywhere in $[0,\pi]$ $\forall$ $n \ge 0$. Putting $n = 0$ we shall get $f(x) = 0$ almost everywhere. As $f(x) \in C[0,\pi]$ we shall say $f(x) = 0$ $\forall$ $x \in [0,\pi]$

b. It is same as a. We shall put $n = 0$ and $\cos(nx) = 1$ $\Rightarrow$ $f(x) = 0$ $\forall$ $x \in [0,\pi]$

c. $\int_0^{\pi} f(x) \sin(nx) dx =0$ $\forall$ $n \ge 1$. Now $f(x)\sin(nx) = 0$ almost everywhere. But I am not getting any more here.

Integrals of b and c are looking like Fourier coefficients of the function $f(x)$. Can we say anything from it?

Thank you for your help.

Supriyo
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  • Both given answers tell you the correct idea, apply the fact that polynomials are dense. Note that the integral being $0$ doesn't imply $f$ is 0 almost everywhere unless $f$ is positive. So you should try to show that $\int f^2 = 0$. – Deven Ware Nov 24 '13 at 02:09
  • Now it is clear to me. Thanks everybody. – Supriyo Nov 24 '13 at 03:32
  • are you preparing for competitive exams? – GA316 Nov 24 '13 at 03:43
  • Yes GA316. Have you any suggestion? – Supriyo Nov 24 '13 at 05:28
  • GAruNkumar is my facebook id. Give me a friend request. we can speak about competitive exams. – GA316 Jan 21 '14 at 18:55
  • @GA316 The first work I did when I returned home from my institution was to deactivate my facebook id as it is very time and data consuming, and attractive. So now a days I have no relation with any of my friends. Do you use your official e-mail id provided by your institution? If yes please give it to me; I shall contact you. – Supriyo Jan 22 '14 at 01:45

3 Answers3

2

a) is true. Hint : use Stone Wierstrass approximation theorem.

b) is false. Hint : consider $f(x) = 1$

GA316
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Your line of reasoning for case a) will only work if we can assume $f \geq 0$. Note that for $f = \cos 2x$, we have $$ \int_0^\pi x^0 \cos 2x = 0 $$ The same is true for your approach to b).

The trick is that you'll need a function generated by polynomial whose sign is identical to that of $f$ at all $x$. One good candidate is the infinite polynomial approximation of $f$.

Ben Grossmann
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Continuous functions can be approximated uniformly both by polynomials and trigonometric polynomials. The condition implies that $$\int_0^{\pi} pf=0$$ for any polynomial. Now choose $p$ such that $\lVert p-f\rVert_\infty<\varepsilon$, and let $M$ an upper bound for $f$ over $[0,\pi]$ and note that $$\int_0^\pi f^2=\int_0^\pi f(f-p)<M\pi\varepsilon $$

For $(b),(c)$ you should be able to find counterexamples if only one of the two conditions hold, but if both hold the above works analogously with $p$ a trigonometric polynomial.

Pedro
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  • Please explain more. – Supriyo Nov 24 '13 at 02:06
  • @pedro Tamaroff if we take $f(x) = 1$ in (b), the integration is zero. so it seems answer false. am I making some mistake? – GA316 Nov 24 '13 at 02:08
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    @GA316 No, you're not. One needs both $(b);(c)$ to hold simultaneously to force $f\equiv 0$. – Pedro Nov 24 '13 at 02:09
  • @PedroTamaroff First of all thank you for your answer and explanation. Please check if my understanding is correct or not. For a When $\int_0^{\pi} f^2(x) = 0$ we can say $f^2(x) = 0$ almost everywhere and always $f^2 \ge 0$ , $f(x)$ is a continuous function thus $f(x) = 0$. For b if we take a function $f(x) = - f(\frac{\pi}{2} +x); 0\le x \le \frac{\pi}{2}$ i.e. $f(x)$ is odd w.r.t. $\frac{\pi}{2}$ the integral will be zero without the function $f(x)$ being zero. Similarly if $f(x)$ be even w.r.t. $\frac{\pi}{2}$ integral of c will be zero without being $f(x) = 0$. – Supriyo Nov 24 '13 at 03:28
  • When b and c hold togther we can approximate the function $f(x)$ using the functions $e^{inx} = \cos(nx) + i\sin(nx)$ and the function $f(x)$ can be shown zero by the method of a. – Supriyo Nov 24 '13 at 03:30
  • @Samprity Exactly. – Pedro Nov 24 '13 at 04:11