[proofplan] We prove each property in turn: boundedness from the triangle inequality, uniform continuity from the Dominated Convergence Theorem, and decay at infinity by first computing it for rectangle indicators (explicit Fourier calculation) and then extending to all of $L^1$ by a density argument. [/proofplan] [step:Establish the $L^\infty$ bound $\|\hat{f}\|_\infty \leq \|f\|_1$] For every $u \in \mathbb{R}^d$, \begin{align*} |\hat{f}(u)| = \left|\int_{\mathbb{R}^d} f(x)\,e^{i\langle u, x\rangle}\,d\mathcal{L}^d(x)\right| \leq \int_{\mathbb{R}^d} |f(x)|\,d\mathcal{L}^d(x) = \|f\|_1, \end{align*} since $|e^{i\langle u,x\rangle}| = 1$ for all $u, x$. [/step] [step:Prove uniform continuity via the Dominated Convergence Theorem] [claim:Uniform Continuity Of Fourier Transform] $\hat{f}$ is uniformly continuous on $\mathbb{R}^d$. [/claim] [proof] For any $u, h \in \mathbb{R}^d$, \begin{align*} |\hat{f}(u + h) - \hat{f}(u)| \leq \int_{\mathbb{R}^d} |f(x)|\,|e^{i\langle h, x\rangle} - 1|\,d\mathcal{L}^d(x). \end{align*} The right side is independent of $u$. As $h \to 0$, the integrand $|f(x)|\,|e^{i\langle h, x\rangle} - 1| \to 0$ pointwise and is dominated by $2|f(x)| \in L^1$. By the Dominated Convergence Theorem, the integral tends to $0$. Since the bound is uniform in $u$, this gives uniform continuity. [/proof] [/step] [step:Show $\hat{f}(u) \to 0$ for rectangle indicators by explicit computation] [claim:Decay For Rectangle Indicators] If $R = \prod_{j=1}^d [a_j, b_j]$ and $f = \mathbb{1}_R$, then $\hat{f}(u) \to 0$ as $|u| \to \infty$. [/claim] [proof] The Fourier transform factors: \begin{align*} \hat{f}(u) = \prod_{j=1}^d \frac{e^{iu_j b_j} - e^{iu_j a_j}}{iu_j} \end{align*} for $u$ with all components nonzero. Each factor is bounded by $\min(b_j - a_j,\, 2/|u_j|)$. As $|u| \to \infty$, at least one $|u_j| \to \infty$, driving the corresponding factor to $0$ while the others stay bounded. [/proof] [/step] [step:Extend decay to all of $L^1$ by a density argument] Let $f \in L^1(\mathbb{R}^d)$. Simple functions built from rectangle indicators are dense in $L^1$. Given $\varepsilon > 0$, choose such a simple function $g$ with $\|f - g\|_1 < \varepsilon/2$. By the previous step and linearity, $\hat{g}(u) \to 0$ as $|u| \to \infty$, so there exists $R > 0$ with $|\hat{g}(u)| < \varepsilon/2$ for $|u| > R$. For $|u| > R$, \begin{align*} |\hat{f}(u)| \leq |\hat{f}(u) - \hat{g}(u)| + |\hat{g}(u)| \leq \|f - g\|_1 + \frac{\varepsilon}{2} < \varepsilon. \end{align*} Since $\varepsilon$ was arbitrary, $\hat{f}(u) \to 0$. [/step]