[proofplan] We integrate by parts $k$ times in the definition of $\hat{f}(n)$, transferring derivatives from $e^{-inx}$ to $f$. The boundary terms vanish by $2\pi$-periodicity of $f, f', \ldots, f^{(k-1)}$. The resulting identity $\hat{f}(n) = (in)^{-k}\widehat{f^{(k)}}(n)$ combined with the [Riemann-Lebesgue Lemma](/theorems/245) gives $\hat{f}(n) = o(|n|^{-k})$. [/proofplan] [step:Perform one integration by parts to relate $\hat{f}(n)$ to $\widehat{f'}(n)$] For $n \neq 0$ and $f \in C^1(\mathbb{T})$: \begin{align*} \hat{f}(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)\,e^{-inx} \, d\mathcal{L}^1(x) = \frac{1}{2\pi}\left[\frac{f(x)\,e^{-inx}}{-in}\right]_{-\pi}^\pi + \frac{1}{2\pi}\int_{-\pi}^\pi \frac{f'(x)\,e^{-inx}}{in} \, d\mathcal{L}^1(x). \end{align*} The boundary term vanishes: $f(\pi)\,e^{-in\pi} - f(-\pi)\,e^{in\pi} = f(\pi)(e^{-in\pi} - e^{in\pi}) = 0$ because $f$ is $2\pi$-periodic ($f(\pi) = f(-\pi)$) and $e^{-in\pi} - e^{in\pi} = -2i\sin(n\pi) = 0$ for $n \in \mathbb{Z}$. Therefore: \begin{align*} \hat{f}(n) = \frac{1}{in}\widehat{f'}(n). \end{align*} [/step] [step:Iterate $k$ times to obtain $\hat{f}(n) = (in)^{-k}\widehat{f^{(k)}}(n)$] Applying the previous step $k$ times (each time the boundary terms vanish by periodicity of $f, f', \ldots, f^{(k-1)}$): \begin{align*} \hat{f}(n) = \frac{1}{(in)^k}\widehat{f^{(k)}}(n). \end{align*} [/step] [step:Deduce the decay bound $\hat{f}(n) = o(|n|^{-k})$] Since $f^{(k)} \in C(\mathbb{T}) \subseteq L^1(\mathbb{T})$: \begin{align*} |\hat{f}(n)| = \frac{1}{|n|^k}\,|\widehat{f^{(k)}}(n)| \leq \frac{1}{|n|^k} \cdot \frac{\|f^{(k)}\|_{L^1(\mathbb{T})}}{2\pi}. \end{align*} The [Riemann-Lebesgue Lemma](/theorems/245) applied to $f^{(k)} \in L^1(\mathbb{T})$ gives $\widehat{f^{(k)}}(n) \to 0$ as $|n| \to \infty$. Therefore $|n|^k\hat{f}(n) = \widehat{f^{(k)}}(n) \to 0$, i.e., $\hat{f}(n) = o(|n|^{-k})$. [/step]