The Schwartz space and the Fourier transform

1 Schwartz functions

Let $𝒮({ℝ}^n)$ be the collection of Schwartz functions ${ℝ}^n\to ℂ$. For $p\ge 0$ and $\varphi \in 𝒮$, write

$${\parallel \varphi \parallel }_p^2=\sum _{|\nu |\le p}{\int }_{{ℝ}^n}{(1+{|x|}^2)}^p{|(D^{\nu }\varphi )(x)|}^2𝑑x.$$

With the metric

$$d(\varphi ,\psi )=\sum _{p\ge 0}{2}^{-p}\frac{{\parallel \varphi -\psi \parallel }_p}{1+{\parallel \varphi -\psi \parallel }_p},$$

$𝒮$ is a Fréchet space.

For a multi-index $\alpha$ and for $\varphi \in 𝒮$, $x\mapsto x^{\alpha }\varphi (x)$ belongs to $𝒮$ and we define $X^{\alpha }:𝒮\to 𝒮$ by $(X^{\alpha }\varphi )(x)=x^{\alpha }\varphi (x)$. $D^{\alpha }\varphi \in 𝒮$ and

$${\parallel D^{\alpha }\varphi \parallel }_p^2=\sum _{|\nu |\le p}{\int }_{ℝ}{(1+{|x|}^2)}^p{|(D^{\nu +\alpha }\varphi )(x)|}^2𝑑x\le {\parallel \varphi \parallel }_{p+|\alpha |}^2.$$

Because $|\{\mu :|\mu |=k\}|=\left(\genfrac{}{}{0pt}{}{n+k-1}{k}\right)$,

$$|\{\mu :\mu \le \nu \}|\le |\{\mu :|\mu |\le |\nu |\}|\le \left(\genfrac{}{}{0pt}{}{n+|\nu |}{|\nu |}\right).$$

The product rule states

$$D^{\nu }(fg)=\sum _{\mu \le \nu }\left(\genfrac{}{}{0pt}{}{\nu }{\mu }\right)(D^{\mu }f)(D^{\nu -\mu }g),$$

and with the Cauchy-Schwarz inequality we obtain for $|\nu |\le p$,

${|D^{\nu }(X^{\alpha }\varphi )|}^2$ $={\left|{\displaystyle \sum _{\mu \le \nu }}\left({\displaystyle \genfrac{}{}{0pt}{}{\nu }{\mu }}\right)(D^{\mu }\varphi )(D^{\nu -\mu }{X}^{\alpha })\right|}^2$

$\le \left({\displaystyle \genfrac{}{}{0pt}{}{n+p}{p}}\right){\displaystyle \sum _{|\mu |\le p}}{\left({\displaystyle \genfrac{}{}{0pt}{}{\nu }{\mu }}\right)}^2{|D^{\mu }\varphi |}^2{|D^{\nu -\mu }{X}^{\alpha }|}^2,$

and with this

${\parallel X^{\alpha }\varphi \parallel }_p^2$ $={\displaystyle \sum _{|\nu |\le p}}{\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^p{|(D^{\nu }(X^{\alpha }\varphi ))(x)|}^2𝑑x$

$\le {\displaystyle \sum _{|\nu |\le p}}{\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^p\left({\displaystyle \genfrac{}{}{0pt}{}{n+p}{p}}\right){\displaystyle \sum _{|\mu |\le p}}{\left({\displaystyle \genfrac{}{}{0pt}{}{\nu }{\mu }}\right)}^2{|D^{\mu }\varphi |}^2{|D^{\nu -\mu }{X}^{\alpha }|}^{2}dx$

$\le C_p{\parallel \varphi \parallel }_{p+|\alpha |}^2.$

For $g,\varphi \in 𝒮$ we have $g\varphi \in 𝒮$, and using the product rule we get

$${\parallel g\varphi \parallel }_p^2\le C_{p,g}{\parallel \varphi \parallel }_p^2.$$

Therefore,

$$\varphi \mapsto D^{\alpha }\varphi ,\varphi \mapsto X^{\alpha }\varphi ,\varphi \mapsto g\varphi$$

are continuous linear maps $𝒮\to 𝒮$.

2 Tempered distributions

For $u:𝒮\to ℂ$, we write

$$⟨\varphi ,u⟩=u(\varphi ).$$

${𝒮}^{\prime }$ denotes the dual space of $𝒮$, and the elements of ${𝒮}^{\prime }$ are called tempered distributions. We assign ${𝒮}^{\prime }$ the weak-* topology, the coarsest topology on ${𝒮}^{\prime }$ such that for each $\varphi \in 𝒮$ the map $u\mapsto ⟨\varphi ,u⟩$ is continuous ${𝒮}^{\prime }\to ℂ$.

For $\psi \in 𝒮$, we define ${\mathrm{\Lambda }}_{\psi }:𝒮\to ℂ$ by

$$⟨\varphi ,{\mathrm{\Lambda }}_{\psi }⟩={\int }_{{ℝ}^n}\varphi (x)\psi (x)𝑑x,\varphi \in 𝒮,$$

and by the Cauchy-Schwarz inequality,

$$|⟨\varphi ,{\mathrm{\Lambda }}_{\psi }⟩|\le {\left({\int }_{{ℝ}^n}{|\varphi (x)|}^2𝑑x\right)}^{1/2}{\left({\int }_{{ℝ}^n}{|\psi (x)|}^2𝑑x\right)}^{1/2}={\parallel \psi \parallel }_0{\parallel \varphi \parallel }_0,$$

whence ${\mathrm{\Lambda }}_{\psi }\in {𝒮}^{\prime }$. It is apparent that $\psi \mapsto {\mathrm{\Lambda }}_{\psi }$ is linear. Suppose that ${\psi }_i\to \psi$ in $𝒮$, and let $\varphi \in 𝒮$. Then

$$|⟨\varphi ,{\mathrm{\Lambda }}_{{\psi }_i}⟩-⟨\varphi ,{\mathrm{\Lambda }}_{\psi }⟩|=|⟨\varphi ,{\mathrm{\Lambda }}_{{\psi }_i-\psi }⟩|\le {\parallel {\psi }_i-\psi \parallel }_0{\parallel \varphi \parallel }_0\to 0,$$

which shows that $\psi \mapsto {\mathrm{\Lambda }}_{\psi }$ is continuous. If ${\mathrm{\Lambda }}_{\psi }=0$, then in particular ${\mathrm{\Lambda }}_{\psi }\overline{\psi }=0$, i.e. ${\int }_{{ℝ}^n}{|\psi (x)|}^2𝑑x=0$, which implies that $\psi (x)=0$ for almost all $x$ and because $\psi$ is continuous, $\psi =0$. Therefore, $\psi \mapsto {\mathrm{\Lambda }}_{\psi }$ is a continuous linear injection $𝒮\to {𝒮}^{\prime }$. It can be proved that $\mathrm{\Lambda }(𝒮)$ is dense in ${𝒮}^{\prime }$.

For a multi-index $\alpha$ and $u\in {𝒮}^{\prime }$, we define $D^{\alpha }u:𝒮\to ℂ$ by

$$⟨\varphi ,D^{\alpha }u⟩={(-1)}^{|\alpha |}⟨D^{\alpha }\varphi ,u⟩,\varphi \in 𝒮.$$

For ${\varphi }_i\to \varphi$ in $𝒮$, because $D^{\alpha }:𝒮\to 𝒮$ and $u:𝒮\to ℂ$ are continuous,

$$⟨{\varphi }_i,D^{\alpha }u⟩={(-1)}^{|\alpha |}⟨D^{\alpha }{\varphi }_i,u⟩\to {(-1)}^{|\alpha |}⟨D^{\alpha }\varphi ,u⟩=⟨\varphi ,D^{\alpha }u⟩,$$

and therefore $D^{\alpha }u\in {𝒮}^{\prime }$.

We define $X^{\alpha }u:𝒮\to ℂ$ by

$$⟨\varphi ,X^{\alpha }u⟩=⟨X^{\alpha }\varphi ,u⟩,\varphi \in 𝒮.$$

For ${\varphi }_i\to \varphi$ in $𝒮$,

$$⟨{\varphi }_i,X^{\alpha }u⟩=⟨X^{\alpha }{\varphi }_i,u⟩\to ⟨X^{\alpha }\varphi ,u⟩=⟨\varphi ,X^{\alpha }u⟩,$$

and therefore $X^{\alpha }u\in {𝒮}^{\prime }$.

For $g\in 𝒮$, we define $gu:𝒮\to ℂ$ by

$$⟨\varphi ,gu⟩=⟨g\varphi ,u⟩,\varphi \in 𝒮.$$

For ${\varphi }_i\to \varphi$ in $𝒮$,

$$⟨{\varphi }_i,gu⟩=⟨g{\varphi }_i,u⟩\to ⟨g\varphi ,u⟩=⟨\varphi ,gu⟩,$$

and therefore $gu\in {𝒮}^{\prime }$.

For $\psi \in 𝒮$, integrating by parts yields

$⟨\varphi ,D^{\alpha }{\mathrm{\Lambda }}_{\psi }⟩$ $={(-1)}^{|\alpha |}⟨D^{\alpha }\varphi ,{\mathrm{\Lambda }}_{\psi }⟩$

$={(-1)}^{|\alpha |}{\displaystyle {\int }_{{ℝ}^n}}(D^{\alpha }\varphi )(x)\psi (x)𝑑x$

$={\displaystyle {\int }_{{ℝ}^n}}\varphi (x)(D^{\alpha }\psi )(x)𝑑x$

$=⟨\varphi ,{\mathrm{\Lambda }}_{D^{\alpha }\psi }⟩,$

which implies that $D^{\alpha }{\mathrm{\Lambda }}_{\psi }={\mathrm{\Lambda }}_{D^{\alpha }\psi }$.

$$⟨\varphi ,X^{\alpha }{\mathrm{\Lambda }}_{\psi }⟩=⟨X^{\alpha }\varphi ,{\mathrm{\Lambda }}_{\psi }⟩={\int }_{{ℝ}^n}{x}^{\alpha }\varphi (x)\psi (x)𝑑x=⟨\varphi ,{\mathrm{\Lambda }}_{X^{\alpha }\psi }⟩,$$

which implies that $X^{\alpha }{\mathrm{\Lambda }}_{\psi }={\mathrm{\Lambda }}_{X^{\alpha }\psi }$.

$$⟨\varphi ,g{\mathrm{\Lambda }}_{\psi }⟩=⟨g\varphi ,{\mathrm{\Lambda }}_{\psi }⟩={\int }_{{ℝ}^n}g(x)\varphi (x)\psi (x)𝑑x=⟨\varphi ,{\mathrm{\Lambda }}_{g\psi }⟩,$$

which implies that $g{\mathrm{\Lambda }}_{\psi }={\mathrm{\Lambda }}_{g\psi }$.

Because $\varphi \mapsto D^{\alpha }\varphi$, $\varphi \mapsto X^{\alpha }\varphi$, and $\varphi \mapsto g\varphi$ are continuous linear maps $𝒮\to 𝒮$ and because $\mathrm{\Lambda }:𝒮\to {𝒮}^{\prime }$ is a continuous linear map with dense image, using the above it is proved that

$$u\mapsto D^{\alpha }u,u\mapsto X^{\alpha }u,u\mapsto gu$$

are continuous linear maps ${𝒮}^{\prime }\to {𝒮}^{\prime }$.

3 The Fourier transform

For Borel measurable functions $f,g:{ℝ}^n\to ℂ$, for those $x$ for which the integral exists we write

$$(f*g)(x)={\int }_{{ℝ}^n}f(x-y)g(y)𝑑y={\int }_{{ℝ}^n}f(y)g(x-y)𝑑y,x\in {ℝ}^n,$$

and for those Borel measurable $f,g:{ℝ}^n\to ℂ$ for which the integral exists we write

$${⟨f,g⟩}_{L^2}={\int }_{{ℝ}^n}f(x)\overline{g(x)}𝑑x.$$

For $\xi \in {ℝ}^n$ we define

$$e_{\xi }(x)=e^{2\pi i\xi \cdot x},x\in {ℝ}^n,$$

and for $\varphi \in 𝒮$ we calculate, integrating by parts,

$$(D^{\alpha }\varphi )*e_{\xi }={(2\pi i\xi )}^{\alpha }\varphi *e_{\xi }.$$

We define $ℱ\varphi :{ℝ}^n\to ℂ$ by

$$(ℱ\varphi )(\xi )={⟨\varphi ,e_{\xi }⟩}_{L^2}={\int }_{{ℝ}^n}\varphi (x)\overline{e_{\xi }(x)}𝑑x={\int }_{{ℝ}^n}{e}^{-2\pi ix\cdot \xi }\varphi (x)𝑑x,\xi \in {ℝ}^n,$$

which we can write as

$$(\varphi *e_{\xi })(0)={\int }_{{ℝ}^n}\varphi (y)e_{\xi }(-y)𝑑y={\int }_{{ℝ}^n}\varphi (y)\overline{e_{\xi }(y)}𝑑y=(ℱ\varphi )(\xi ).$$

By Fubini’s theorem,

$ℱ(\varphi *\psi )(\xi )$ $={\displaystyle {\int }_{{ℝ}^n}}\psi (y)\left({\displaystyle {\int }_{{ℝ}^n}}\varphi (x-y)\overline{e_{\xi }(x)}𝑑x\right)𝑑y$

$={\displaystyle {\int }_{{ℝ}^n}}\psi (y)\left({\displaystyle {\int }_{{ℝ}^n}}\varphi (x)\overline{e_{\xi }(x+y)}𝑑x\right)𝑑y,$

whence

$$ℱ(\varphi *\psi )=(ℱ\varphi )(ℱ\psi ).$$

We calculate

$$ℱ(D^{\alpha }\varphi )(\xi )=((D^{\alpha }\varphi )*e_{\xi })(0)=({(2\pi i\xi )}^{\alpha }\varphi *e_{\xi })(0)={(2\pi i\xi )}^{\alpha }(ℱ\varphi )(\xi ),$$

whence

$$ℱ(D^{\alpha }\varphi )={(2\pi i)}^{|\alpha |}{X}^{\alpha }ℱ\varphi .$$

It follows from the dominated convergence theorem

$(D^{\alpha }ℱ\varphi )(\xi )$ $={\displaystyle {\int }_{{ℝ}^n}}{(-2\pi ix)}^{\alpha }{e}^{-2\pi ix\cdot \xi }\varphi (x)𝑑x$

$={(-2\pi i)}^{|\alpha |}{\displaystyle {\int }_{{ℝ}^n}}{e}^{-2\pi ix\cdot \xi }{x}^{\alpha }\varphi (x)𝑑x$

$={(-2\pi i)}^{|\alpha |}ℱ(X^{\alpha }\varphi )(\xi ).$

Therefore

$$ℱD^{\alpha }={(2\pi i)}^{|\alpha |}{X}^{\alpha }ℱ,D^{\alpha }ℱ={(-2\pi i)}^{|\alpha |}ℱX^{\alpha }.$$ (1)

Using the multinomial theorem,

${(1+{|\xi |}^2)}^p{|(D^{\nu }ℱ\varphi )(\xi )|}^2$ $={\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){|\xi |}^{2k}{|(D^{\nu }ℱ\varphi )(\xi )|}^2$

$={\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){\displaystyle \sum _{|\alpha |=k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\alpha }}\right){\xi }^{2\alpha }{|(D^{\nu }ℱ\varphi )(\xi )|}^2$

$={\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){\displaystyle \sum _{|\alpha |=k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\alpha }}\right){|({\xi }^{\alpha }{D}^{\nu }ℱ\varphi )(\xi )|}^2.$

Applying (1),

$$|({\xi }^{\alpha }{D}^{\nu }ℱ\varphi )(\xi )|={(2\pi )}^{|\nu |}{(2\pi )}^{-|\alpha |}|(ℱD^{\alpha }{X}^{\nu }\varphi )(\xi )|.$$

Then

${\parallel ℱ\varphi \parallel }_p^2$ $={\displaystyle \sum _{|\nu |\le p}}{\displaystyle {\int }_{{ℝ}^n}}{(1+{|\xi |}^2)}^p{|(D^{\nu }ℱ\varphi )(\xi )|}^2𝑑\xi$

$={\displaystyle \sum _{|\nu |\le p}}{\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){\displaystyle \sum _{|\alpha |=k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\alpha }}\right){|({\xi }^{\alpha }{D}^{\nu }ℱ\varphi )(\xi )|}^{2}d\xi$

$={\displaystyle \sum _{|\nu |\le p}}{(2\pi )}^{2|\nu |}{\displaystyle \sum _{k=0}^p}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){(2\pi )}^{-2k}{\displaystyle \sum _{|\alpha |=k}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{\alpha }}\right){\displaystyle {\int }_{{ℝ}^n}}{|(ℱD^{\alpha }{X}^{\nu }\varphi )(\xi )|}^2𝑑\xi .$

Applying the Plancherel theorem, the product rule, and the Cauchy-Schwarz inequality yields

${\displaystyle {\int }_{{ℝ}^n}}{|(ℱD^{\alpha }{X}^{\nu }\varphi )(\xi )|}^2𝑑\xi$ $={\displaystyle {\int }_{{ℝ}^n}}{|(D^{\alpha }{X}^{\nu }\varphi )(\xi )|}^2𝑑\xi$

$={\displaystyle {\int }_{{ℝ}^n}}{\left|{\displaystyle \sum _{\beta \le \alpha }}(D^{\beta }{X}^{\nu })(D^{\alpha -\beta }\varphi )\right|}^2𝑑\xi$

$\le {\displaystyle {\int }_{{ℝ}^n}}{\displaystyle \sum _{\beta \le \alpha }}{|(D^{\beta }{X}^{\nu })(\xi )|}^2\cdot {\displaystyle \sum _{\beta \le \alpha }}{|(D^{\alpha -\beta }\varphi )(\xi )|}^2.$

This yields

$${\parallel ℱ\varphi \parallel }_p\le C_p{\parallel \varphi \parallel }_p,$$

whence $ℱ:𝒮\to 𝒮$ is continuous.

For $p>n/2$, using the Cauchy-Schwarz inequality and spherical coordinates we calculate

$|(ℱ\varphi )(\xi )|$ $\le {\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^{-p/2}{(1+{|x|}^2)}^{p/2}|\varphi (x)|𝑑x$

$\le {\left({\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^{-p}𝑑x\right)}^{1/2}{\left({\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^p{|\varphi (x)|}^2𝑑x\right)}^{1/2}$

$={\left({\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle {\int }_{S^{n-1}}}{(1+r^2)}^{-p}𝑑\sigma r^{n-1}𝑑r\right)}^{1/2}{\left({\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^p{|\varphi (x)|}^2𝑑x\right)}^{1/2}$

$={\left({\displaystyle \frac{{\pi }^{n/2}\mathrm{\Gamma }\left(p-\frac{n}{2}\right)}{\mathrm{\Gamma }(p)}}\right)}^{1/2}{\left({\displaystyle {\int }_{{ℝ}^n}}{(1+{|x|}^2)}^p{|\varphi (x)|}^2𝑑x\right)}^{1/2}$

$\le {\left({\displaystyle \frac{{\pi }^{n/2}\mathrm{\Gamma }\left(p-\frac{n}{2}\right)}{\mathrm{\Gamma }(p)}}\right)}^{1/2}{\parallel \varphi \parallel }_p.$