Show that $|\hat{f_n}| \leq \frac{C}{|n|}$ for all $n \in \mathbb{Z} \backslash \{0\}$ for a Lipschitz-continuous $f$

I want to show that for a Lipshitz-continuous function $f$, there is some constant $C$ such that $|\hat{f_n}| \leq \frac{C}{|n|}$ for all $n \in \mathbb{Z} \backslash \{0\}$, where $\hat{f}_n$ is the n-th Fourier coefficient of $f$.

I tried to utilize a "typical" estimate of $\hat{f_n}$, i.e.

\begin{equation} |\hat{f_n}| = |\frac{1}{2\pi} \int_0^{2\pi}f(x) e^{-inx}dx| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(x) e^{-inx} dx| \leq ||f|| \end{equation} so it doesn't seem to work. This result seems to be similar to the standard result that if $f \in C^k$, $\hat{f_n} \leq \frac{C}{|n|^k}$, which I know how to prove via integration by parts and Riemann-Lebesgue lemma, and I thought that we could do something similar here - however, I am not sure how to do it.

1. Integration by parts still works if you know that any Lipschitz function $f$ on $[0, 2\pi]$ is absolutely continuous with the derivative $f'$ that is bounded a.e. by the Lipschitz constant of $f$:

$$\|f\|_{\text{Lip}} := \sup_{\substack{x, y \in [0, 2\pi] \\ x \neq y}} \frac{|f(x) - f(y)|}{|x - y|} $$

Then, as in the case of $C^1$ function, integration by parts yields

\begin{align*} \hat{f_n} &= \frac{1}{2\pi} \int_{0}^{2\pi} f(x) e^{-inx} \, \mathrm{d}x \\ &= \frac{1}{2\pi} \left[ -\frac{1}{in} f(x) e^{-inx} \right]_{0}^{2\pi} + \frac{1}{2\pi i n} \int_{0}^{2\pi} f'(x) e^{-inx} \, \mathrm{d}x. \end{align*}

From this, we get the desired bound:

$$ |\hat{f_n}| \leq \frac{1}{\pi n} \|f\|_{\sup} + \frac{1}{n}\|f\|_{\text{Lip}} $$

(Remark. This idea readily extends to the functions of bounded variation.)

2. Alternatively, using the well-known trick,

\begin{align*} |\hat{f_n}| &= \frac{1}{2\pi} \left| \int_{0}^{2\pi} f(x) e^{-inx} \, \mathrm{d}x \right| \\ &= \frac{1}{4\pi} \left| \int_{0}^{2\pi} f(x) e^{-inx} \, \mathrm{d}x + \int_{-\frac{\pi}{n}}^{2\pi-\frac{\pi}{n}} f\left(x+\frac{\pi}{n}\right) e^{-in\left(x+\frac{\pi}{n}\right)} \, \mathrm{d}x \right| \\ &\leq \frac{1}{4\pi} \left(\int_{2\pi-\frac{\pi}{n}}^{2\pi} |f(x)| \, \mathrm{d}x + \int_{-\frac{\pi}{n}}^{0} \left| f\left(x+\frac{\pi}{n}\right) \right| \, \mathrm{d}x + \int_{0}^{2\pi-\frac{\pi}{n}} \left|f\left(x+\frac{\pi}{n}\right) - f(x) \right| \, \mathrm{d}x \right) \\ &\leq \frac{1}{4\pi} \left( \frac{\pi}{n}\|f\|_{\sup} + \frac{\pi}{n}\|f\|_{\sup} + \frac{2\pi^2}{n} \|f\|_{\text{Lip}} \right). \end{align*}

So the desired claim follows.