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Let $f$ be a continuously differentiable $2π$ -periodic real valued function on the real line. Let $a_n =\displaystyle \int_{-\pi}^{\pi} f (t) \cos nt\; dt$ where $n$ is a non-negative integer.

Pick out the true statements:
(a) The derivative of $f$ is also a $2π$-periodic function.
(b) $|a_n| ≤ C/n$ for all n, where $C > 0$ is a constant independent of n.
(c) $a_n → 0$, as $n → ∞$.

Totally stuck on this problem. Can somebody help me?

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They are all right.

$(a)$:$$f'(x+2π )=\lim_{dx\to0}\frac{f(x+2π+dx)-f(x+2π)}{dx}=\lim_{dx\to 0}\frac{f(x+dx)-f(x)}{dx}=f'(x)$$

$(b)$ Use integration by parts and the fact that continuous is bounded on a compact interval.

$(c)$ Follows easily by $(b)$, or you can use the Riemann-Lebesgue Theorem.

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Where are you stuck? For example (a), have you tried out any examples to get an intuition? For (b) and (c), did you calculate the integral? That would be a good start..

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HOW CAN I ABLE TO CALCULATE THE INTEGRAL AS $f(t)$ is not known. please explain in detail. – UUUU Jan 12 '13 at 15:15
There is a method called integration by parts (, or check out the wolfram alpha solution for the integral here:… – omar Jan 12 '13 at 15:20

We can also deduce c) from Bessel's inequality

We have that $<f,e_n> = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} dx = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \cos(-nx) dx + \frac{i}{2\pi} \int_{-\pi}^{\pi} f(x) \sin(-nx) dx$.

Bessel's inequality tells us that $\displaystyle \sum_{n=-\infty}^{\infty} |<f,e_n>|^2 \le \frac{1}{2\pi}\int_{-\pi}^{\pi} |f(x)|^2 dx < \infty$, from which we conclude that $|<f,e_n>| \to 0 $ as $|n|\ \to \infty$, and hence it follows that for nonnegative $n$, $\Re(<f,e_n>) = a_n \to 0$ as $n \to \infty$ as well.

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