Fourier inversion

It is shown above that the Fourier transform satisfies the identity

$$\varphi (0) = (2 \pi)^{-n} \int_{\mathbb{R}^n} \hat{\varphi} (\xi) d \xi \forall \varphi \in \mathcal{S} (\mathbb{R}^n) .$$ (119)

If $y \in \mathbb{R}^n$ and $\varphi \in \mathcal{S}(\mathbb{R}^n)$ set $\psi (x) = \varphi (x+y)$. The translation-invariance of Lebesgue measure shows that

$$\hat{\psi} (\xi) = \int e^{-ix \cdot \xi} \varphi (x+y) dx$$

$$= e^{iy \cdot \xi} \hat{\varphi} (\xi) .$$

Applied to $\psi$ the inversion formula (8.1) becomes

$$\varphi(y) = \psi (0) = (2 \pi)^{-n} \int \hat{\psi} (\xi) d \xi$$ (120)

$$= (2 \pi)^{-n} \int_{\mathbb{R}^n} e^{iy \cdot \xi} \hat{\varphi} (\xi) d \xi .$$

Proof. The identity (8.2) shows that $\mathcal{F}$ is $1-1$, i.e., injective, since we can remove $\varphi$ from $\hat{\varphi}$. Moreover,

$$\mathcal{G} \psi (y) = (2 \pi)^{-n} \mathcal{F} \psi (-y)$$ (121)

So $\mathcal{G}$ is also a continuous linear map, $\mathcal{G} : \mathcal{S} (\mathbb{R}^n) \to \mathcal{S} (\mathbb{R}^n)$. Indeed the argument above shows that $\mathcal{G} \circ \mathcal{F} =Id$ and the same argument, with some changes of sign, shows that $\mathcal{F} \cdot \mathcal{G} = Id$. Thus $F$ and $\mathcal{G}$ are isomorphisms.

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Proof. Using the inversion formula on $\varphi$,

$$\int \varphi \overline{\psi} dx = (2 \pi)^{-n} \int \left( e^{ix \cdot \xi} \hat{\varphi} (\xi) d \xi \right) \overline{\psi} (x) dx$$

$$= (2 \pi)^{-n} \int \hat{\varphi} (\xi) \overline{\int e^{-ix \cdot \xi} \psi (x) dx} d \xi$$

$$= (2 \pi)^{-n} \int \hat{\varphi} (\xi) \overline{\hat{\varphi}} ( \xi) d \xi .$$

Here the integrals are absolutely convergent, justifying the exchange of orders.

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Proof. Setting $\varphi = \psi$ in (8.5) shows that

$$\Vert \mathcal{F} \varphi \Vert _{L^2} = (2 \pi)^{n/2} \Vert \varphi \Vert _{L^2} .$$ (122)

In particular this proves, given the known density of $\mathcal{S}(\mathbb{R}^n)$ in $L^2 (\mathbb{R}^n)$, that $\mathcal{F}$ is an isomorphism, with inverse $\mathcal{G}$, as in (8.6).

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For any $m \in \mathbb{R}$

$$\langle x \rangle^m L^2 (\mathbb{R}^n) = \left\{ u \in \mathcal{S... ...bb{R}^n) ; \langle x \rangle^{-m} \hat{u} \in L^2 (\mathbb{R}^n) \right\}$$

is a well-defined subspace. We define the Sobolev spaces on $\mathbb{R}^n$ by, for $m \geq 0$

$$H^m (\mathbb{R}^n) = \left\{ u \in L^2 (\mathbb{R}^n) ; \hat{u} = \mathcal{F} u \in \langle \xi \rangle^{-m} L^2 (\mathbb{R}^n) \right\} .$$ (123)

$$\hbox{Thus } H^m (\mathbb{R}^n) \subset H^{m'} (\mathbb{R}^n) \hbox{ if } m \geq m' , H^0 (\mathbb{R}^n) = L^2 (\mathbb{R}^n) .$$

Proof. By definition, $u \in H^m (\mathbb{R}^n)$ implies that $\langle \xi \rangle^{-m} \hat{u} \in L^2 (\mathbb{R}^n)$. Since $\widehat{{D^{\alpha}} u} = \xi^{\alpha} \hat{u}$ this certainly implies that ${D^{\alpha}} u \in L^2 (\mathbb{R}^n)$ for $\left\vert \alpha \right\vert \leq m$. Conversely if ${D^{\alpha}} u \in L^2 (\mathbb{R}^n)$ for all $\left\vert \alpha \right\vert \leq m$ then $\xi^{\alpha} \hat{u} \in L^2 (\mathbb{R}^n)$ for all $\left\vert \alpha \right\vert \leq m$ and since

$$\langle \xi \rangle^m \leq C_m \sum_{\left\vert \alpha \right\vert \leq m} \left\vert \xi^{\alpha} \right\vert .$$

this in turn implies that $\langle \xi \rangle^m \hat{u} \in L^2 (\mathbb{R}^n)$.

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Now that we have considered the Fourier transform of Schwartz test functions we can use the usual method, of duality, to extend it to tempered distributions. If we set $\eta = \overline{\hat{\psi}}$ then $\hat{\psi} = \overline{\eta}$ and $\psi= \mathcal{G} \hat{\psi} = \mathcal{G} \overline{\eta}$ so

$$\begin{multline*} \overline{\psi} (x) =(2\pi)^{-n}\int e^{-ix \cdot \xi} \overli... ...ix \cdot \xi} \eta (\xi) \, d \xi = (2 \pi)^{-n} \hat{\eta}(x). \end{multline*}$$

Substituting in (8.5) we find that

$$\int \varphi \hat{\eta} dx = \int \hat{\varphi} \eta d \xi .$$

Now, recalling how we embed $\mathcal{S} (\mathbb{R}^n) \hookrightarrow \mathcal{S}' (\mathbb{R}^n)$ we see that

$$u_{\hat{\varphi}} (\eta) = u_{\varphi} (\hat{\eta}) \forall \eta \in \mathcal{S} (\mathbb{R}^n) .$$ (124)

As a composite map, $\hat{u} = u \cdot \mathcal{F}$, with each term continuous, $\hat{u}$ is continuous, i.e., $\hat{u} \in \mathcal{S}' (\mathbb{R}^n)$.

Proof. Since $\hat{u} = u \circ \mathcal{F}$ and $\mathcal{G}$ is the 2-sided inverse of $\mathcal{F}$,

$$u= \hat{u} \circ \mathcal{G}$$ (125)

gives the inverse to $\mathcal{F} : \mathcal{S}' (\mathbb{R}^n) \to \mathcal{S}' (\mathbb{R}^n)$, showing it to be an isomorphism. The identities (8.12) follow from their counterparts on $\mathcal{S}(\mathbb{R}^n)$:

$$\widehat{{D^{\alpha}} u} (\varphi) = {D^{\alpha}} u (\hat{\varphi}) = u ((-1)^{\left\vert \alpha \right\vert} {D^{\alpha}} \hat{\varphi})$$

$$= u (\widehat{\xi^{\alpha} \varphi}) = \hat{u} (\xi^{\alpha} \var... ...lpha} \hat{u} ( \varphi) \forall \varphi \in \mathcal{S} (\mathbb{R}^n) .$$

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We can also define Sobolev spaces of negative order:

$$H^m (\mathbb{R}^n) = \left\{ u \in \mathcal{S}' (\mathbb{R}^n) ; \hat{u} \in \langle \xi \rangle^{-m} L^2 (\mathbb{R}^n) \right\} .$$ (126)

Proof. If $u \in \mathcal{S}' (\mathbb{R}^n)$ is of the form (8.15) then

$$\hat{u} = \sum_{\left\vert \alpha \right\vert \leq -m} \xi^{\alpha} \hat{v}_{\alpha} \hbox{ with } \hat{v}{\alpha} \in L^2 (\mathbb{R}^n) .$$ (127)

Thus $\langle \xi \rangle^m \hat{u} = \sum_{\left\vert \alpha \right\vert \leq -m} \xi^{\alpha} \langle \xi \rangle^{m} \hat{v}_{\alpha}$. Since all the factors $\xi^{\alpha} \langle \xi \rangle^m$ are bounded, each term here is in $L^2 (\mathbb{R}^n)$, so $\langle \xi \rangle^m \hat{u} \in L^2 (\mathbb{R}^n)$ which is the definition, $u \in \langle \xi \rangle^{-m} L^2 (\mathbb{R}^n)$.

Conversely, suppose $u \in H^m (\mathbb{R}^n)$, i.e., $\langle \xi \rangle^m \hat{u} \in L^2 (\mathbb{R}^n)$. The function

$$\left( \sum_{\left\vert \alpha \right\vert \leq -m} \left\vert \x... ... \right\vert \right) \cdot \langle \xi \rangle^m \in L^2 (\mathbb{R}^n) (m<0)$$

is bounded below by a positive constant. Thus

$$v= \left( \sum_{\left\vert \alpha \right\vert \leq -m} \left\vert \xi^{\alpha} \right\vert \right)^{-1} \hat{u} \in L^2 (\mathbb{R}^n) .$$

Each of the functions $\hat{v}_{\alpha} = \operatorname{sgn} (\xi^{\alpha}) \hat{v} \in L^2 (\mathbb{R}^n)$ so the identity (8.16), and hence (8.15), follows with these choices.

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Proof. The Hilbert space property follows essentially directly from the definition (8.14) since $\langle \xi \rangle^{-m} L^2 (\mathbb{R}^n)$ is a Hilbert space with the norm (8.17). Similarly the density of $\mathcal{S}$ in $H^m (\mathbb{R}^n)$ follows, since $\mathcal{S}(\mathbb{R}^n)$ dense in $L^2 (\mathbb{R}^n)$ (Problem L11.P3) implies $\langle \xi \rangle^{-m} \mathcal{S} (\mathbb{R}^n) = \mathcal{S} (\mathbb{R}^n)$ is dense in $\langle \xi \rangle^{-m} L^2 (\mathbb{R}^n)$ and so, since $\mathcal{F}$ is an isomorphism in $\mathcal{S}(\mathbb{R}^n)$, $\mathcal{S}(\mathbb{R}^n)$ is dense in $H^m (\mathbb{R}^n)$.

Finally observe that the pairing in (8.18) makes sense, since $\langle \xi \rangle^{-m}\hat{u} (\xi)$, $\langle \xi \rangle^{m} \hat{u'} (\xi) \in L^2 (\mathbb{R}^n)$ implies

$$\hat{u} (\xi)) \hat{u'} (- \xi) \in L^1 (\mathbb{R}^n) .$$

Furthermore, by the self-duality of $L^2 (\mathbb{R}^n)$ each continuous linear functional

$$U: H^m (\mathbb{R}^n) \to \mathbb{C} , U(u) \leq C \Vert u \Vert _{H^m}$$

can be written uniquely in the form

$$U(u) = ((u,u')) \hbox{ for some } u' \in H^{-m} (\mathbb{R}^n) .$$

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Notice that if $u,u' \in \mathcal{S} (\mathbb{R}^n)$ then

$$((u,u')) = \int_{\mathbb{R}^n} u(x) u' (x) dx .$$

This is always how we ``pair'' functions -- it is the natural pairing on $L^2 (\mathbb{R}^n)$. Thus in (8.18) what we have shown is that this pairing on test function

$$\mathcal{S} (\mathbb{R}^n) \times \mathcal{S} (\mathbb{R}^n) \ni (u,u') \longmapsto ((u,u')) = \int_{\mathbb{R}^n} u(x) u' (x) dx$$

extends by continuity to $H^m (\mathbb{R}^n) \times H^{-m} (\mathbb{R}^n)$ (for each fixed $m$) when it identifies $H^{-m}(\mathbb{R}^n)$ as the dual of $H^m (\mathbb{R}^n)$. This was our `picture' at the beginning.

For $m>0$ the spaces $H^m (\mathbb{R}^n)$ represents elements of $L^2 (\mathbb{R}^n)$ that have ``$m$'' derivatives in $L^2 (\mathbb{R}^n)$. For $m<0$ the elements are ?? of ``up to $-m$'' derivatives of $L^2$ functions. For integers this is precisely ??.