= Solution
Use the <Fourier transform> convention $\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-ix\xi}\,dx$, with complex-linear <distribution> pairings throughout. The <Riemann-Lebesgue lemma> says that, for an <integrable function> $f\in L^1(\mathbb R)$, its <Fourier transform> is continuous and tends to zero as $|\xi|\to\infty$. It also satisfies $\|\widehat f\|_\infty\leq\|f\|_1$. The product rule, or <convolution theorem>, is
$$
\boxed{\widehat{f*g}=\widehat f\,\widehat g\qquad(f,g\in L^1(\mathbb R)).}
$$
Here $(f*g)(x)=\int f(y)g(x-y)\,dy$ belongs to $L^1(\mathbb R)$ by <Tonelli's theorem>, and <Fubini's theorem> proves the displayed identity. The companion <Fourier product formula>, for <Schwartz functions>, is $\widehat{fg}=(2\pi)^{-1}\widehat f*\widehat g$.
A <Schwartz function> is a <smooth function> $\phi:\mathbb R\to\mathbb C$ for which every <Schwartz seminorm>
$$
p_{m,n}(\phi):=\sup_{x\in\mathbb R}|x^m\phi^{(n)}(x)|,\qquad m,n\in\{0,1,2,\ldots\},
$$
is finite. These <seminorms> define the topology of the <Schwartz space> $\mathcal S(\mathbb R)$. In particular,
$$
|\phi(x)|\leq\frac{p_{0,0}(\phi)+p_{2,0}(\phi)}{1+x^2},\qquad
\boxed{\|\phi\|_1\leq\pi\bigl(p_{0,0}(\phi)+p_{2,0}(\phi)\bigr)<\infty.}
$$
Thus every <Schwartz function> is an <integrable function>.
The <Fourier bounds for Schwartz functions> assert that each <Schwartz seminorm> of $\widehat\phi$ is bounded by finitely many <Schwartz seminorms> of $\phi$. Explicitly, <differentiation under the integral sign> and <integration by parts> give
$$
\xi^m\partial_\xi^n\widehat\phi(\xi)
=i^{-m}\mathcal F\!\left[\partial_x^m\bigl((-ix)^n\phi\bigr)\right](\xi),
$$
so the <generalized Leibniz rule> and the preceding integrability estimate give
$$
\boxed{p_{m,n}(\widehat\phi)
\leq\pi\sum_{j=0}^{\min(m,n)}\binom mj\frac{n!}{(n-j)!}
\bigl(p_{n-j,m-j}(\phi)+p_{n-j+2,m-j}(\phi)\bigr).}
$$
In particular the <Fourier transform> maps the <Schwartz space> continuously into itself.
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