Q: Bounded of fourier coefficients of smooth function

Question

$f(x)$ is a smooth function till $k$ order, and piecewise continuous at $k+1$ order,then we have: $$\left|{\alpha_v}\right|= \left|\frac{1}{2\pi}\int_{-\pi}^{\pi}{f(x)e^{-ivx}dx}\right|= \left|\frac{1}{2\pi iv}\int_{-\pi}^{\pi}f'(x)e^{-ivx}dx\right|= \ldots= \left|\frac{1}{2\pi}\left(\frac{-i}{v}\right)^{(k+1)}\int_{-\pi}^{\pi}f^{(k+1)}(x)e^{-ivx}dx\right| $$ Suppose $\frac{B}{2}$ is a upper bounded of $\left|f^{(k+1)}(x)\right|$, how to get $$\left|a_v\right|,\left|b_v\right|\le \frac{B}{v^{k+1}}$$

Answer 1

Note that we have

$$\begin{align} \left|\int_0^{2\pi}f^{(n+1)}(x)\,e^{-i\nu x}\,dx\right|&\le\int_0^{2\pi}\left|f^{(n+1)}(x)\,e^{-i\nu x}\right|\,dx \tag 1\\\\ &=\int_0^{2\pi }|f^{(n+1)}(x)|\,dx \tag2\\\\ &\le \max_{x\in [0,2\pi]}\left(|f^{(n+1)}(x)|\right)\,\int_0^{2\pi}\,(1)\,dx \tag 3\\\\ &=2\pi B \end{align}$$

In arriving at $(1)$, we invoked the Triangle Inequality for Integrals.

In going from $(1)$ to $(2)$ we used the fact that $|e^{-i\nu x}|=1$.

In going from $(2)$ to $(3)$, we exploited the First Mean Value Theorem for Integrals.