I'm trying to tackle the following question, but I'm not sure that my solution is correct.

Let $f$ be real-valued $2\pi$ periodic function which is continuous almost everywhere, such that its Fourier series is $\displaystyle \sum_{n=-\infty}^{\infty}c_n e^{inx}$.

A. Write $f$ as a sum of odd function $g$ and even function $h$.

B. Show that Fourier coefficients of $g$ are imaginary and Fourier coefficients of $h$ are real.

C. Assuming that $f$ and $f'$ are continuous, show that Fourier coefficients of $f'$ are $in\cdot{c_n}$.

My solution:

  1. $\displaystyle \sum_{n=-\infty}^{\infty}c_ne^{inx}=\sum_{n=-\infty}^{\infty}c_n\cos(nx)+\sum_{n=-\infty}^{\infty}ic_n\sin(nx)$. If $\displaystyle g\sim \sum_{n=-\infty}^{\infty}ic_n\sin(nx)$ and $\displaystyle h\sim \sum_{-\infty} c_n\cos(nx)$, then $g$ is odd and $h$ is even. Fourier series is additive, hence $f=g+h$, as needed.
  2. I don't know how to explain it...
  3. If $f$ is continuous, then its Fourier series uniformly converges to $f$, therefore we can write that $\displaystyle f'\sim \sum_{n=-\infty}^{\infty}\left(c_ne^{inx}\right)'=\sum_{n=-\infty}^{\infty}in\cdot{c_n}e^{inx}$, i.e, the Fourier coefficients of $f'$ are $in\cdot{c_n}$.

Is my reasoning ok? How should I tackle the second question?

Please help, thank you.

  • $\begingroup$ what are your $g$ and $h$? In your way they are not real functions. $\sim$ is not used as $=$. $\endgroup$
    – Zhanxiong
    Nov 7, 2015 at 19:31

1 Answer 1


For A., you don't need anything related to Fourier analysis. Just write $$f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)] := h(x) + g(x).$$ It is easily checked that $h$ and $g$ such defined are even and odd, respectively.

For B., it can be shown by definition that, the Fourier coefficients for any even real-valued function are real and that for any odd-valued function are pure imaginary. For example, the $n$th Fourier coefficient for $h$ is given by \begin{align} & c_n = \frac{1}{2\pi}\int_{-\pi}^\pi h(x)e^{-inx} dx = \frac{1}{2\pi}\int_{-\pi}^\pi h(x)\cos(-nx)dx \in \mathbb{R}, \end{align} since $\sin(-nx)h(x)$ is odd, and the integration interval is symmetric about $0$.

C. also easily follows by definition (no convergence issues are involved), by definition, the $n$th Fourier coefficient for $f'$ is calculated by (using integration by parts): \begin{align} c_n' = & \frac{1}{2\pi}\int_{-\pi}^\pi f'(x)e^{-inx} dx \\ = & \frac{1}{2\pi}f(x)e^{-inx}\big |_{-\pi}^\pi - \frac{1}{2\pi}\int_{-\pi}^\pi f(x)(-in) e^{-inx} dx \\ = & in\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx} dx \\ = & inc_n, \end{align} where we used that $f$ is of period $2\pi$.

  • $\begingroup$ Excellent answer, thanks a lot! $\endgroup$
    – Galc127
    Nov 7, 2015 at 19:50

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