4

If $f$ belongs to $C[0, \pi]$ and $f(0) = 0$. Determine the cases where the given condition implies that $f$ is identically zero.

Case 1 $\int_{0}^{\pi}\ x^nf(x) =0$ for all non negative integer n

Case 2 $\int_{0}^{\pi}f(x)\cos nx =0$ for all non negative integer n

Case 3 $\int_{0}^{\pi}f(x)\sin nx =0$ for all positive integer n

What I know is if $f(x)$ greater or equal to $0$ in the interval then $\int_{0}^{\pi}\ f(x) =0$ implies $f(x) = 0$.If I use Weierstrass approximation theorem I would get the first case done. But I can not understand how to proceed for last two cases. Can anyone help me out? For your information all the cases are true.

1 Answers1

4

Case 2: Extend $f$ to $F$ on $[-\pi,\pi]$ so that $F$ is even. Then the evenness of $F(x)$ and $\cos nx$ shows

$$\int_{-\pi}^{\pi}F(x)\cos nx\,dx = 2\int_0^{\pi}f(x)\cos nx\,dx = 0$$

for $n=0,1,2,\dots.$ Now use the evenness of $F(x)$ and the oddness of $\sin nx$ to see $$\int_{-\pi}^{\pi}F(x)\sin nx\,dx = 0,$$ $n=1,2,\dots.$ Thus all of $F$'s Fourier coefficients are $0.$ By Parseval's theorem,

$$\int_{-\pi}^{\pi}|F(x)|^2\, dx = 0.$$

The continuity of $F$ then implies $F\equiv 0.$ Because $F=f$ on $[0,\pi],$ we have $f\equiv 0.$

Case 3: See if you can solve this case using ideas similar to the above.

zhw.
  • 105,693
  • Why did u need Parseval's theorem? $a_n = 0$ for every $n\geq 0$ and $b_n =0 $ for every $n\geq 1$...So by fourier series expansion $F(x) = 0$ for all $x\in[-\pi , \pi]$@zhw. –  Oct 05 '17 at 04:48
  • And where $f(0) = 0$ will be used? –  Oct 05 '17 at 05:22
  • @cmi is right : to conclude that 'all of F's Fourier coefficient are 0', you have to mention that $F(0)=0$, so its coefficient $a_0=0$ (otherwise you can only conclude that F is constant) – charmd Oct 05 '17 at 07:03
  • I did not get u...Can u please elaborate? –  Oct 05 '17 at 07:05
  • @cmi The Fourier expansion of a continuous function $g$ does not need to converge to $g$ everywhere. However, Parseval always holds. That's how I get $F\equiv 0.$ – zhw. Oct 05 '17 at 21:58
  • @cmi $f(0)=0$ is used in claiming $F$ is continuous. – zhw. Oct 05 '17 at 21:59
  • @charMD It is given that that the cosine integral is $0$ for $n=0,$ so we're fine. – zhw. Oct 05 '17 at 22:02
  • Can U tell me why u are needing Parseval's theorem? Every $a_n$ and $b_n$ is $0$ , How the fourier Expansion can be divergent for some $x$?@zhw. – anonymous Oct 26 '17 at 03:17
  • Can U tell me why u are needing Parseval's theorem? Every an and bn is 0 , How the fourier Expansion can be divergent for some x?@zhw. – anonymous Oct 30 '17 at 13:37
  • @sani How would you prove this: If $f$ is continuous on $[-\pi,\pi]$ and all of $f $'s Fourier coefficents are $0,$ then $f\equiv 0?$ – zhw. Oct 30 '17 at 14:34
  • Can You please tell me how $f$ can be extended to an even function $F$?@zhw. – cmi Jan 08 '18 at 16:52
  • 1
    For $x\in [-\pi,0],$ define $F(x)= f(-x).$ @cmi – zhw. Jan 08 '18 at 17:11
  • Yes..Can you please explain me what will happen when https://math.stackexchange.com/questions/2597115/f-in-c-pi-pi-and-if-int-pi-pi-fx-sin-nx-mathrm-dx-0-th – cmi Jan 08 '18 at 17:13
  • What will happen in case 3? extending to a even function will not help..@zhw – cmi Jan 08 '18 at 17:57