Lemma 11

Let $\varphi \in \mathscr{S}(\mathbf{R}^n)$. The function $\mathscr{F}(\varphi )$ is indefinitely differentiable on $\mathbf{R}^n$ and one has

for every $\alpha$ in $\mathbf{N}^n$.

One may suppose that $n\geqslant 1$. Let $\varphi \in \mathscr{S}(\mathbf{R}^n)$. The function defined by $(x, y)\mapsto \varphi (x)$ exp(2$i\pi x\cdot y$) from $\mathbf{R}^n\times \mathbf{R}^n$ into $\mathbf{C}$ satisfies the hypotheses of Corollary 1 of IV, p. 198 for every integer $k$. The Fourier transform of $\varphi$ is therefore indefinitely differentiable and satisfies

$\partial^{\alpha}(\mathscr{F}\varphi )(y) = (-2i\pi )^{|\alpha|}\int_{\mathbf{R}^n}x^{\alpha}\varphi (x)$ exp($-2i\pi x\cdot y$)$d\mu(x)$

for every $y\in \mathbf{R}^n$, which implies the first formula.

Let us prove the second formula. By induction on $|\alpha |$, it is enough to do so when $|\alpha |= 1$, and one is easily reduced to the case $\alpha = (1,0, . . . ,0)$. Write every $x\in \mathbf{R}^n$ in the form $x= (x_1, x')$ with $x'\in \mathbf{R}^{n-1}$, and denote by $\mu_1$ (resp. $\mu'$) the Lebesgue measure on $\mathbf{R}^{n-1}$ (resp. $\mathbf{R}$). By the Lebesgue-Fubini theorem (INT, V, p. 96, § 8, n$^o4$, th. 1), for every $y= (y_1, y')\in \mathbf{R}\times \mathbf{R}^{n-1}$, we get

$\mathscr{F}(\partial_1\varphi )(y) =\int$ exp($-2i\pi x'\cdot y'$)

$\times^{\mathbf{R}^n}(\int^{^{-1}}_{\mathbf{R}}(\partial_1\varphi )(x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1))d\mu'(x')$.

For every compact interval $[a, b]$ in $\mathbf{R}$ and every $x'\in \mathbf{R}^{n-1}$, one has

$\int_a^b(\partial_1\varphi )(x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1) =$

$[\varphi (x_1, x')$ exp($-2i\pi x_1y_1$)$]^b_a$

$+ 2i\pi y_1\int_a^b\varphi (x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1)$ by integration by parts (FVR, II, p. 10, formula (10)). When $a$ tends to $-\infty$ and $b$ tends to $+\infty$, the first term of the second member converges to 0 since $\varphi \in \mathscr{S}(\mathbf{R}^n)$. The second term converges, by the Lebesgue theorem (INT, IV, p. 137, § 3, n$^o7$, th. 6), to

$2i\pi y_1\int_{\mathbf{R}}\varphi (x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1)$,

since the mapping $x_1\mapsto \varphi (x_1, x')$ is integrable on $\mathbf{R}$. As $x_1\mapsto \partial_1\varphi (x_1, x')$ is also integrable on $\mathbf{R}$, one deduces that

$\int_{\mathbf{R}}\partial_1\varphi (x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1)$

$= 2i\pi y_1\int_{\mathbf{R}}\varphi (x_1, x')$ exp($-2i\pi x_1y_1$)$d\mu_1(x_1)$ and finally, applying the Lebesgue-Fubini theorem again, one concludes that

$\mathscr{F}(\partial_1\varphi )(y) = 2i\pi y_1\int_{\mathbf{R}^n}\varphi (x)$ exp($-2i\pi x\cdot y$)$d\mu(x)$,

as desired.