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If $f\in C[0,\pi]$ and $f(0)=0$ then $\int_{0}^{\pi}f(x)\cos (nx) dx=0$ for all $n\geq 0$ implies that$f(x)$ is identically $0$ on $[0, \pi]$.
I saw this problem has been solved here before .Here is the link .
$\int_{0}^{\pi}f(x)\cos nx =0$ for all non negative integer n then prove $f(x)$ is identically $0$ on $[0, \pi]$
Now I have some doubts regarding this:-
i) Is the condition $f(0)=0$ necessary here ?
ii) Will $f$ be identically $0$ if $f\in C[0,\pi]$ , $f(0)=0$ and $\int_{0}^{\pi}f(x)\cos (nx)dx =0$ for all $n\geq 1$ (instead of for all $n\geq 0$) ?

Gary
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simu tiyam
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  • Thank you. What about the second doubt? – simu tiyam Apr 01 '21 at 07:48
  • This looks like an application of Fourier theory. – J. De Ro Apr 01 '21 at 07:49
  • I think $f(0) = 0$ would only be required in the case you bring up where we're only given that the integral is zero for $n \geq 1$ instead of $n \geq 0$. And your second case should hold by the same argument they gave, with the added consideration that the $0$th Fourier coefficient must be zero because $f(0) = a_0 = 0$. – Stephen Donovan Apr 01 '21 at 07:49
  • @simutiyam: if continuity of $f$ is to be preserved, then yes, $f(0)=0$ is needed (and implied). – Mittens Apr 01 '21 at 08:29

1 Answers1

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Replace $f$ by $g(x)=f(|x|)$ for $x\in [-\pi,\pi]$.

If $\int_0^\pi f(x)\cos(nx)dx=0$ for all $n\ge 0$ then $\int_{-\pi}^\pi g(x)e^{inx}dx=0$ for all $n$.

If $f(a)\ne 0$ then $\int_{-\pi}^{\pi} g(x) (\frac{1+\cos(x-a)}2)^k dx$ is non-zero for $k$ large enough. $(\frac{1+\cos(x-a)}2)^k$ is a linear combination of $e^{inx},|n|\le k$.

If you only have the condition $\int_0^\pi f(x)\cos(nx)dx=0$ for $n\ge 1$ then replace $f$ by $h=f-\frac{\int_{0}^\pi f(x)dx}{\pi}$ to have it for $n\ge 0$. You'll get that $h=0$ so $f$ is constant.

reuns
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