# Complex Fourier series and its represntation

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?

• what are your $g$ and $h$? In your way they are not real functions. $\sim$ is not used as $=$. Nov 7, 2015 at 19:31
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 $2\pi$.