I know that if $f\in C_{c}^{\infty}(\mathbb{R})$, then the Fourier coefficients obey the inequality $|f_k|\leq \frac{M}{|k|^n}$ where $k$ denotes the $kth$ Fourier coefficient and $M=sup_x|f^{(n)}(x)|$.
Can someone provide a detailed proof as to how we derive the inequalities for the following classes of functions:
Holder continuous: $|f_k|\leq \frac{K}{|k|^\alpha}$ where $K$ is the multiplicative constant, and $\alpha$ is the Holder exponent.
Bounded variation: $|f_k|\leq \frac{var(f)}{2\pi|k|}$
I am not sure exactly how these are computed.
Let $f$ be a function on the circle $\mathbb{R}/\mathbb{Z}$ that is in $L^1$. Let $k > 0$ WLOG. One can compute $$ \begin{aligned} \hat{f}(k) &:= \int_{0}^1 e^{-2\pi i k y} f(y) dy = \sum_{j=0}^{k-1} \int_{k^{-1} j}^{k^{-1} (j+1)} e^{-2\pi i k y} f(y) dy \\ &= \sum_{j=0}^{k-1} \int_{k^{-1} j}^{k^{-1} (j+1)} e^{-2\pi i k y} (f(y) - f(k^{-1} j)) dy \end{aligned}$$ and so $$ \lvert \hat{f}(k) \rvert \leq \sum_{j=0}^{k-1} \int_{k^{-1} j}^{k^{-1} (j+1)} \lvert f(y) - f(k^{-1} j) \rvert dy. $$
Now if $f$ is $\alpha$ Holder continuous with constant $K$, we have $$ \lvert \hat{f}(k) \rvert \leq \sum_{j=0}^{k-1} k^{-1} \cdot K \cdot k^{-\alpha} \leq K k^{-\alpha}. $$ Let $f$ be of bounded variation instead. Use the facts $$ \int_{k^{-1} j}^{k^{-1} (j+1)} \lvert f(y) - f(k^{-1} j) \rvert dy \leq \text{variation of } f \text{ over the interval } [k^{-1}j, k^{-1}(j+1)) $$ and $$ \text{variation of } f \text{ over } [0,1) = \sum_j \text{variation of } f \text{ over } [k^{-1}j, k^{-1}(j+1)) $$ to get the desired bound.