[proofplan] We prove $\widehat{D^\alpha f}(\xi) = (i\xi)^\alpha \hat{f}(\xi)$ by reducing to the case of a single partial derivative via induction on $|\alpha|$, then establishing the single-variable case $\widehat{\partial_{x_j} f}(\xi) = i\xi_j \hat{f}(\xi)$ by separating the $x_j$-integration via Fubini, integrating by parts, verifying the boundary term vanishes (using absolute continuity of $L^1$ functions with $L^1$ derivatives), and reassembling. [/proofplan] [step:Reduce to the case of a single partial derivative by induction] Since $D^\alpha = \partial_{x_1}^{\alpha_1} \cdots \partial_{x_n}^{\alpha_n}$ and $(i\xi)^\alpha = (i\xi_1)^{\alpha_1} \cdots (i\xi_n)^{\alpha_n}$, it suffices by induction on $|\alpha|$ to prove \begin{align*} \widehat{\partial_{x_j} f}(\xi) &= i\xi_j \, \hat{f}(\xi) \end{align*} for each $j \in \{1, \ldots, n\}$, under the assumption $f \in L^1(\mathbb{R}^n)$ and $\partial_{x_j} f \in L^1(\mathbb{R}^n)$. The general case follows by applying this identity $|\alpha|$ times. [/step] [step:Separate the $x_j$-integration via Fubini and integrate by parts] Write $x = (x', x_j)$ where $x' = (x_1, \ldots, x_{j-1}, x_{j+1}, \ldots, x_n) \in \mathbb{R}^{n-1}$. By Fubini's theorem (applicable since $\partial_{x_j} f \in L^1(\mathbb{R}^n)$): \begin{align*} \widehat{\partial_{x_j} f}(\xi) &= \int_{\mathbb{R}^{n-1}} e^{-ix' \cdot \xi'} \left(\int_{\mathbb{R}} \partial_{x_j} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j)\right) d\mathcal{L}^{n-1}(x'). \end{align*} Integrate the inner integral by parts with respect to $x_j$. The boundary term is $[f(x', x_j) \, e^{-ix_j \xi_j}]_{x_j = -\infty}^{x_j = +\infty}$. [guided] We need the boundary term $[f(x', x_j)\,e^{-ix_j\xi_j}]_{x_j = -\infty}^{x_j = +\infty}$ to vanish. Since $f \in L^1(\mathbb{R}^n)$, Fubini's theorem gives $x_j \mapsto f(x', x_j) \in L^1(\mathbb{R})$ for $\mathcal{L}^{n-1}$-a.e. $x'$. Because $\partial_{x_j} f \in L^1(\mathbb{R}^n)$, the function $x_j \mapsto f(x', x_j)$ is absolutely continuous for a.e. $x'$, with representation \begin{align*} f(x', x_j) = f(x', 0) + \int_0^{x_j} \partial_{x_j} f(x', t)\,d\mathcal{L}^1(t). \end{align*} Since $\partial_{x_j} f(x', \cdot) \in L^1(\mathbb{R})$, the integral converges as $x_j \to \pm\infty$, so $\lim_{x_j \to \pm\infty} f(x', x_j)$ exists. But $f(x', \cdot) \in L^1(\mathbb{R})$, and an $L^1$ function cannot tend to a nonzero constant (otherwise $\int |f|\,d\mathcal{L}^1 = \infty$). Therefore \begin{align*} \lim_{x_j \to \pm\infty} f(x', x_j) = 0. \end{align*} Since $|e^{-ix_j\xi_j}| = 1$, the boundary term $f(x', x_j)e^{-ix_j\xi_j} \to 0$ as $x_j \to \pm\infty$, and the integration by parts produces no boundary contribution. [/guided] [/step] [step:Evaluate the remaining integral and reassemble] After integration by parts, the inner integral becomes \begin{align*} \int_{\mathbb{R}} \partial_{x_j} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j) &= -\int_{\mathbb{R}} f(x', x_j) \, (-i\xi_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j) \\ &= i\xi_j \int_{\mathbb{R}} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j). \end{align*} Substituting back into the Fubini decomposition: \begin{align*} \widehat{\partial_{x_j} f}(\xi) &= i\xi_j \int_{\mathbb{R}^{n-1}} e^{-ix' \cdot \xi'} \int_{\mathbb{R}} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j) \, d\mathcal{L}^{n-1}(x') = i\xi_j \, \hat{f}(\xi). \end{align*} [/step]