Lemma 12.4.6 (Riemann-Lebesgue).label If $f\in L^{1}(\R^{n})$ then $\ha{f}$ is continuous and $\ha{f}(\xi)\to 0$ as $\abs{\xi}\to\infty$.
Proof. Let $\curl{f_j}$ be a sequence of Schwartz functions that converges to $f$ in $L^{1}$. This convergence will then be uniform since
\begin{align*}\norm{\ha{f}-\ha{f_j}}{\infty}\leq\norm{f-f_j}{1}\end{align*}
and uniform limit of continuous functions is continuous. Also observe $\norm{\ha{f}}{\infty}\leq\norm{f}{1}$$\square$
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