Fourier Coefficients

Integrating by parts with $\displaystyle u=f(x)$ and $\displaystyle v=\frac{e^{i2\pi kx}}{i2\pi k}$ yields

$$\int_0^1 f(x)e^{i2\pi kx}\,dx=\frac{f(1)-f(0)}{i2\pi k}-\frac{1}{i2\pi k}\int_0^1 f'(x)e^{i2\pi kx}\,dx $$

Repeated integration by parts with $\displaystyle u=f^{(n)}(x)$ and $\displaystyle v=\frac{e^{i2\pi kx}}{i2\pi k}$ reveals

$$\int_0^1 f(x)e^{i2\pi kx}\,dx=\sum_{n=0}^{m-1} (-1)^n\frac{f^{(n)}(1)-f^{(n)}(0)}{(i2\pi k)^{n+1}}+(-1)^m\frac{1}{(i2\pi k)^m}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx \tag 1$$

If $f$ is periodic, then all of the terms in the summation in $(1)$ are zero and we have

$$\int_0^1 f(x)e^{i2\pi kx}\,dx=(-1)^m\frac{1}{(i2\pi k)^m}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx$$

Since the Riemann-Lebesgue Lemma guarantees that $\lim_{k\to \infty}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx=0$, then

$$\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx=o(1)$$

as $k\to \infty$ and we have the desired result.