Does $\int_{-\pi}^{\pi} f(x) \cos (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is odd?

Question

For $2\pi-$periodic and continuous $f$, does $\int_{-\pi}^{\pi} f(x) \cos (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is odd? Similarly, does $\int_{-\pi}^{\pi} f(x) \sin (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is even?

Motivation. If we are given $f$ is $2\pi$ periodic and we are given both conditions, we can show that $f \equiv 0$ by considering a sequence of trigonometric polynomials which converge uniformly to $f$. Can we get this same result by showing that the first individually shows that $f$ is odd and the second that $f$ is even so that we get $f \equiv 0$?

Hint: Put $g(x)=f(x)+f(-x)$. The integral of $g(x)\sin(nx)$ over $[-\pi,+\pi]$ is zero (this function is odd). Compute the integral of $g(x)\cos(nx)$ over $[-\pi,+\pi]$. — Kelenner, Oct 22 '15 at 16:44

score 0 · Answer 1 · answered Oct 22 '15 at 16:56

0

Let $$s_n(x) = \sum_{n = 1}^{\infty} b_n \sin (nx)$$ then $$s_n(x)\to f(x)\Rightarrow -s_n(x)=s_n(-x)\to f(-x) $$ also $$-s_n(x)=s_n(-x)\to -f(x)$$ uniqueness of limit implies $f(-x)=-f(x)$

answered Oct 22 '15 at 16:56

R.N

4,318

Are you using convergence of fourier series? I don't believe we have that for this particular $f$ guaranteed – MT_ Oct 22 '15 at 17:21
@soke, may i know, why? – R.N Oct 22 '15 at 17:25

Does $\int_{-\pi}^{\pi} f(x) \cos (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is odd?

1 Answers1