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For the angular-frequency Fourier transform and Schwartz seminorms $p_{m,n}(\varphi )={\sup }_x|x^m{\varphi }^{(n)}(x)|$, every output seminorm is bounded by finitely many input seminorms. To prove this, differentiation under the integral sign and integration by parts giveThe generalized Leibniz rule, $\Vert \hat{g}{\Vert }_{\infty }\le \Vert g{\Vert }_1$, and $\Vert x^a{\varphi }^{(b)}{\Vert }_1\le \pi (p_{a,b}+p_{a+2,b})$ implyThis proves continuity of the Fourier transform on the Schwartz space and rapid decay of every transformed derivative.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, 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 $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ 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 $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

A distribution $u$ is a $2\pi$ periodic distribution whenIteration gives invariance under all integer multiples of $2\pi$. For a tempered distribution the identity also holds for every Schwartz function, since test functions are dense in the Schwartz space: multiplying by cutoff functions $\chi (x/R)$ equal to one near zero gives such an approximation in every Schwartz seminorm.

For $\varphi \in \mathcal{S}(\mathbb{R})$ define $a_k(x)=\Psi (x+2\pi k)\varphi (x)$. The smooth periodic partition of unity gives $\varphi ={\sum }_k{a}_k$ with convergence in the Schwartz space. Indeed, each $a_k$ is supported where $|x+2\pi k|\le 2\pi$, and the generalized Leibniz rule and arbitrarily rapid decay of the derivatives of $\varphi$ give, for every $m,n,M$,Consequently continuity and periodicity of the tempered distribution giveFor the last interchange, all summands have the fixed compact support of $\Psi$ and their derivatives decrease faster than any power of $|k|$. Hence the series converges in the space of test functions, as well as in the Schwartz space. Notice that $P\varphi$ itself need not be a Schwartz function; the factor $\Psi$ makes the pairing legitimate.

The generalized Leibniz rule follows by mathematical induction. The case $n=0$ is the identity $uv=uv$. Suppose the formula holds for $n$ and differentiate every term using the product rule:For an interior index $r$, the coefficient of $u^{(r)}{v}^{(n+1-r)}$ becomes $\left(\genfrac{}{}{0ex}{}{n}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{r}\right)$ by Pascal's identity. The two endpoint coefficients are one. This provesOnly finitely many derivatives are involved; it suffices that both functions are $n$ times differentiable.

For $u=x(x+1)$ and $v=e^{2x}$, the derivatives of $u$ beyond order two vanish. The three surviving terms of the generalized Leibniz rule giveThe simplified expression works for every integer $n\ge 0$, including the two cases where one or more of the three terms are absent.

For $u=x^2$ and $v=\sin x$, use $d^m(\sin x)/d{x}^m=\sin (x+m\pi /2)$. Again only three terms survive. With $T=x+n\pi /2$, they areSince $\sin (T-\pi /2)=-\cos T$ and $\sin (T-\pi )=-\sin T$, this becomesThe factors $n$ and $n(n-1)$ remove the nonexistent terms when $n=0$ or $1$.

For functions with the required derivatives, the Leibniz rule isAt $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}{g}^{(n+1-k)}$ after regrouping is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$ by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\left(\genfrac{}{}{0ex}{}{n}{p+q}\right)\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)=n!/(p!q!r!)$. Consequentlywhere $p,q,r$ range over nonnegative integers. This is the three-factor generalized Leibniz rule.