2.9.1 Fourier transform and energy4C3D61D53E89

Proof. Let $f\in \mathcal S$. We first prove the second rule in (2.9.3). By the mean value theorem, the difference quotients of $\xi \mapsto \mathrm e^{-2\pi \mathrm i x\cdot \xi }$ in the direction of $\xi _ j$ are bounded by $2\pi |x_ j|$, and $x_ jf$ is integrable. Hence dominated convergence allows us to differentiate under the integral sign: $\partial _{\xi _ j}\widehat f$ exists and is the transform of $-2\pi \mathrm i x_ jf$, which is continuous. Since $x_ jf\in \mathcal S$, induction shows that $\widehat f$ is smooth and that $\partial ^\beta \widehat f$ is the transform of $(-2\pi \mathrm i x)^\beta f$.

For the first rule, Fubini's theorem allows us to integrate in $x_ j$ before the other coordinates. For almost every value of the other coordinates, $f$ and $\partial _ jf$ are integrable on the corresponding line, and an integration by parts over $[-a,a]$ followed by $a\to \infty $ gives

because the boundary terms are bounded by $|f(x)|$ at $x_ j=\pm a$, and $f(x)\to 0$ as $|x|\to \infty $. Integrating over the other coordinates proves the first rule. Combining the two rules, we find that $\xi ^\alpha \partial ^\beta \widehat f$ is a constant multiple of the transform of $\partial ^\alpha (x^\beta f)$. This function belongs to $\mathcal S$ and is therefore integrable, so $\xi ^\alpha \partial ^\beta \widehat f$ is bounded by a multiple of its $L^1$ norm. Hence $\widehat f\in \mathcal S$.

For the Gaussian, consider first $d=1$, put $g(x)=\mathrm e^{-\pi x^2}$, which belongs to $\mathcal S(\mathbb R)$, and set $F=\widehat g$. Since $-2\pi \mathrm i xg=\mathrm i g'$, the two rules give $F'=\mathrm i\, \widehat{g'}=\mathrm i(2\pi \mathrm i\xi )F=-2\pi \xi F$, so $F(\xi )\mathrm e^{\pi \xi ^2}$ is constant. Moreover $F(0)=1$: squaring the positive integral $F(0)$ and using polar coordinates in the plane, where $s_1=2\pi $, gives $F(0)^2=2\pi \int _0^\infty r\mathrm e^{-\pi r^2}\, \mathrm d r=1$. Hence $F(\xi )=\mathrm e^{-\pi \xi ^2}$. In dimension $d$, Fubini's theorem writes the transform of $\mathrm e^{-\pi |x|^2}$ as a product of one-dimensional transforms, which proves the formula for $t=1$. For general $t$, the linear substitution $x=y/\sqrt t$ gives

Proof. Let $f\in \mathcal S$ and $\varepsilon >0$. Insert the factor $\mathrm e^{-\pi \varepsilon ^2|\xi |^2}$ into the inversion integral. By Fubini's theorem and lemma 2.9.2 with $t=\varepsilon ^2$,

The functions $g_\varepsilon $ are positive with integral one, and the substitution $y=\varepsilon y'$ shows that their mass outside a fixed ball $B(0,\delta )$ tends to zero as $\varepsilon \to 0$. Since $f$ is bounded, and uniformly continuous because its gradient is bounded, splitting the convolution integral into the regions $|y|<\delta $ and $|y|\geq \delta $ shows that $f*g_\varepsilon \to f$ uniformly. On the left-hand side, $\widehat f\in \mathcal S\subset L^1$ by lemma 2.9.2, so dominated convergence lets $\varepsilon \to 0$ under the frequency integral. This proves the inversion formula.

Write $\mathcal F$ for the Fourier transform on $\mathcal S$, $\mathcal F^*$ for the inverse integral transform $\mathcal F^*g(x)=\int g(\xi )\mathrm e^{2\pi \mathrm i x\cdot \xi }\, \mathrm d\xi $, and $Rf(x)=f(-x)$. The substitution $x\mapsto -x$ shows that $\mathcal F^*=R\mathcal F=\mathcal FR$ on $\mathcal S$. In particular $\mathcal F^*$ maps $\mathcal S$ into itself, and both compositions $\mathcal F^*\mathcal F$ and $\mathcal F\mathcal F^*$ equal $R\mathcal F\mathcal F$, which is the identity on $\mathcal S$ by the inversion formula. For $f,g\in \mathcal S$, applying inversion to $g$ and Fubini's theorem, we obtain Parseval's identity

the interchange of integrals is justified because the double integral of the absolute value is at most $\| f\| _1\| \widehat g\| _1$.

By lemma 2.5.1(5), applied to Lebesgue measure, $C_ c^\infty (\mathbb R^ d)$, and hence $\mathcal S$, is dense in $L^2$. By Parseval's identity, $\mathcal F$ maps sequences in $\mathcal S$ that are Cauchy in $L^2$ to Cauchy sequences, and $L^2$ is complete; therefore $\mathcal F$ extends uniquely from $\mathcal S$ to a bounded operator on $L^2$, and this extension is an isometry. Since $\mathcal F^*=R\mathcal F$, the inverse integral transform extends in the same way. Both compositions of the two extensions are continuous and equal to the identity on the dense subspace $\mathcal S$, hence everywhere. Thus the extension is unitary, with the stated inverse.

Finally let $f\in L^1\cap L^2$. We choose $f_ j\in C_ c^\infty $ converging to $f$ both in $L^1$ and in $L^2$: first replace $f$ by $f\mathbf1_{B(0,R)}$ with $R$ large, and then mollify; by lemma 2.5.1(2) the mollifications converge in both norms, since their supports lie in a fixed compact set. The integral transforms of the $f_ j$ converge uniformly to that of $f$, because the difference is bounded by $\| f_ j-f\| _1$, while by isometry they converge in $L^2$ to the $L^2$ transform of $f$. A subsequence of an $L^2$ convergent sequence converges almost everywhere: choose it so that the squared $L^2$ errors are summable, and apply Tonelli's theorem to the sum of the squared errors. Hence the two transforms of $f$ agree almost everywhere. $\square$

Proof. Let $g\in L^2$ be the inverse $L^2$ transform of $\widehat\mu $, so that $\widehat g=\widehat\mu $. For $\phi \in \mathcal S$ we have

The first equality follows from the inversion formula for $\phi $ and Fubini's theorem. For the second, we apply Parseval's identity to $\phi $ and $\overline g$; it extends from $\mathcal S$ to $L^2$ because the $L^2$ transform is unitary. We also use that the transform of $\overline g$ at $\xi $ is $\overline{\widehat g(-\xi )}$; this holds for Schwartz functions by definition, and it extends to $L^2$ because the maps $g\mapsto \widehat{\overline g}$ and $g\mapsto \overline{\widehat g(-\, \cdot \, )}$ are continuous from $L^2$ to $L^2$.

Consequently, for every test function $\phi $ the number $\int \phi g\, \mathrm d x=\int \phi \, \mathrm d\mu $ is real when $\phi $ is real and nonnegative when $\phi \geq 0$. By the Cauchy–Schwarz inequality on compact sets, $g$ is locally integrable, so it defines a positive distribution. By proposition 2.5.3(3) its mollifications $g*\rho _\varepsilon $ are nonnegative, and they converge to $g$ in $L^1$ on compact sets by lemma 2.5.1(2). Hence $g$ is real and nonnegative almost everywhere. The measures $\mu $ and $g\, \mathrm d x$ are Radon by proposition 2.2.2, and they define the same distribution; by proposition 2.5.3(4), $\mu =g\, \mathrm d x$. In particular $\int g\, \mathrm d x=\mu (\mathbb R^ d)<\infty $, and $\| g\| _2=\| \widehat\mu \| _2$ because the $L^2$ transform is unitary. $\square$

Proof. For $z\in \mathbb R^ d$, the substitution $v=\pi t|z|^2$ gives

and both sides are $+\infty $ when $z=0$. By lemma 2.9.2 with $t$ replaced by $1/t$, we have $\mathrm e^{-\pi t|z|^2}=t^{-d/2}\int \mathrm e^{-\pi |\xi |^2/t}\mathrm e^{-2\pi \mathrm i z\cdot \xi }\, \mathrm d\xi $. Taking $z=x-y$ and integrating with respect to $\mu \times \mu $, we obtain

Fubini's theorem applies here because the Gaussian in $\xi $ is integrable and $\mu \times \mu $ is finite. Now insert the integral formula for $|x-y|^{-s}$ into $I_ s(\mu )$. All integrands are nonnegative, so Tonelli's theorem allows us to integrate first in $(x,y)$, then use the last identity, and finally integrate in $t$ before $\xi $, even when the integrals are infinite:

For $\xi \neq 0$, the substitution $v=\pi |\xi |^2/t$ evaluates the inner integral as

and the point $\xi =0$ is a Lebesgue null set. Multiplying the constants gives the formula. $\square$

Proof. By lemma 2.9.2 with $\xi =0$ and $t=1$, the Gaussian $\mathrm e^{-\pi |x|^2}$ has integral one. Integrating in polar coordinates and substituting $v=\pi r^2$, we obtain

which proves (2.9.8). By lemma 2.9.2, $\widehat\mu (\xi )=\mathrm e^{-\pi |\xi |^2}$. Hence theorem 2.9.6 and polar coordinates give

The substitution $v=2\pi r^2$ evaluates the radial integral as $\Gamma (s/2)/(2(2\pi )^{s/2})$. Inserting the values of $b_{d,s}$ and $s_{d-1}$ proves the formula. $\square$