Parseval's identity holds

Question

Theorem: If $u \in L^2(\mathbb{R}^n)$ then the Fourier transform $\widehat{u} \in S'(\mathbb{R}^n)$ is a $L^2(\mathbb{R}^n)$ function and the Parseval's identity holds: $||\widehat{u}||_{L^2(\mathbb{R}^n)}^2= (2 \pi)^n ||u||_{L^2(\mathbb{R}^n)}^2$ .

Proof: We have that if $\phi, \psi \in S(\mathbb{R}^n)$ then $\int_{\mathbb{R}^n} \phi(x) \widehat{\psi}(x) dx= \int_{\mathbb{R}^n} \widehat{\phi}(\xi) \psi(\xi) d{\xi}$.

We want to show that $\int_{\mathbb{R}^n} \phi(x) \overline{\phi}(x) dx= (2 \pi)^{-n}\int_{\mathbb{R}^n} \widehat{\phi}(\xi) \overline{\widehat{\phi}}(\xi) d{\xi}$.

We pick $\psi(\xi)= \overline{\widehat{\phi}}(\xi)= \int_{\mathbb{R}^n} \overline{\phi}(x) e^{i x \xi} dx=\widehat{\overline{\phi}}(-\xi)=\widehat{\overline{\phi}}^{\vee}(\xi)$.

We want to find $\widehat{\psi}(\xi)$:

$\widehat{\psi}=\widehat{\overline{\widehat{\phi}}}$

$\widehat{\psi}=(2 \pi)^n \overline{\phi} \Rightarrow ||\overline{\phi}||_{L^2(\mathbb{R}^n)}=(2 \pi)^n ||\phi||_{L^2(\mathbb{R}^n)}$

How do we get the last relation?

Or is there maybe a typo?

Because we want to show that $||\widehat{\phi}||_{L^2(\mathbb{R}^n)}^2=(2 \pi)^n ||\phi||_{L^2(\mathbb{R}^n)}^2$.

How can we get this equality?

EDIT:

Let $u \in L^2(\mathbb{R}^n)$, then $\langle \widehat{u}, \phi \rangle=\langle u, \widehat{\phi} \rangle=\int_{\mathbb{R}^n} u \widehat{\phi} dx, \phi \in S(\mathbb{R}^n)$.

$|\langle \widehat{u}, \phi \rangle| \leq ||u||_{L^2(\mathbb{R}^n)} ||\widehat{\phi}||_{L^2(\mathbb{R}^n)}= ||u||_{L^2(\mathbb{R}^n)} (2 \pi)^{\frac{n}{2}} ||\phi||_{L^2(\mathbb{R}^n)}$

So at this point we have used the part about which I asked you before.

So $\frac{|\langle \widehat{u}, \phi\rangle|}{||\phi||_{L^2(\mathbb{R}^n)}} \leq (2 \pi)^{\frac{n}{2}} ||u||_{L^2(\mathbb{R}^n)}$.

From the Hahn-Banach theorem we can expand the linear functional $\phi \to \langle \widehat{u}, \phi \rangle, \phi \in L^2(\mathbb{R}^n)$ and it will hold $||\widehat{u}|| \leq ||u|| (2 \pi)^{\frac{n}{2}}$

The Hahn Banach theorem says that if we have a subspace $M$ of a linear metric space $X$ and if $f$ is a bounded linear functional in $M$, then $f$ can be extended to a bounded linear functional $F$ in $X$ such that $||F||=||f||$.

But why if we extend $\phi$ to $L^2(\mathbb{R}^n)$ will we get $\widehat{u}$ ?

Then from Riesz theorem there is a unique function $v \in L^2(\mathbb{R}^n)$ such that $\langle \widehat{u}, \phi \rangle=\int_{\mathbb{R}^n} \phi \overline{v} dx$.

So is $(\psi, \phi)$ defined like that: $(\psi, \phi)=\int_{\mathbb{R}^n} \phi \overline{\psi} dx$ ?

Then we get $||v|| \leq (2 \pi)^{\frac{n}{2}} ||u||$.

Also given the relation $||\widehat{u}|| \leq ||u|| (2 \pi)^{\frac{n}{2}}$ , can we just take the Fourier transform of it? Do we get that $||\widehat{\widehat{u}}|| \leq ||\widehat{u}|| (2 \pi)^{\frac{n}{2}}$

Answer 1

I read the proof with your notation in "Introduction to Distribution Thoery" by F.G. Friedlander. I try to explain. We assume initially $u=\phi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$, then

$\displaystyle \widehat{\phi}(\xi):=\int e^{i x \cdot \xi} \phi(x) dx$

and by inversion formula is known that

$\displaystyle \phi = \mathcal{F}^{-1} \widehat{\phi}(x) = (2\pi)^{-n}\int e^{i x \cdot \xi} \widehat{\phi}(\xi) d \xi$

Let $\psi \in \mathcal{S}(\mathbb{R}^n)$, from the properties of the $L^1$-Fourier transform, we have

1. $\displaystyle \int \varphi(x) \widehat{\psi}(x) dx = \int \widehat{\phi}(\xi) \psi(\xi) d \xi$

We can choose

$\displaystyle \psi(\xi):=\overline{\widehat{\phi}(\xi)}:=\int \overline{\phi(x)} e^{i x \cdot \xi} dx = (2\pi)^n/(2\pi)^n \int \overline{\phi(x)} e^{i x \cdot \xi} dx = (2\pi)^n \mathcal{F}^{-1}\overline{\phi}$

and then $\widehat{\psi}(\xi)=(2\pi)^n \overline{\phi}$, and $(1.)$ becomes $||\widehat{\phi}||^{2}_2 = (2\pi)^n || \phi||^2$. So for Schwartz functions the Parseval identity is true. Now, if $u \in L^2(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$, the Fourier trasform of $u$ is a function (as tempered distribution, similary to the fact that $L^1_{loc} \subset \mathcal{D}'$). Therefore we have

$\displaystyle <\widehat{u}, \phi > = <u, \widehat{\phi} > =\int u \widehat{\phi} dx , \forall \phi \in \mathcal{S}(\mathbb{R}^n)$

and by Schwartz inequality

2. $\displaystyle |< \widehat{u} , \phi > | \leq ||u||_{2} || \widehat{\phi} |||_{2}=(2\pi)^{n/2} ||u||_2 || \phi||_2, \forall \phi \in \mathcal{S}(\mathbb{R}^n)$

Since $C_{c}^\infty(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n) \subset L^2(\mathbb{R}^n)$, and $C_c(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$, also $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$, and (2.) is true $\forall \phi \in L^2(\mathbb{R}^n)$ by approssimation (it is not difficult to verify formally). Obviously $L : \phi \longmapsto < \widehat{u} , \phi >$ is a continuous functional on $L^2(\mathbb{R}^n)$, i.e. it is an element of the dual space, then, by Riesz rappresentation theorem there is $v \in L^2(\mathbb{R}^n)$ such that

$\displaystyle <\widehat{u}, \phi > = < u, \widehat{\phi} > = \int u \widehat{\phi} dx = \int \phi \overline{v} dx$

and by uniqueness we have $\widehat{u}=\overline{v} \in L^2(\mathbb{R}^n)$. Riesz theorem tells us also that

3. $\displaystyle ||\widehat{u}||_2=||v||_2 \leq (2\pi)^{n/2}||u||_2$ $\forall \phi \in L^2(\mathbb{R}^n)$

it is follow also by definition of the norm of the dual space. Now, $\widehat{u} \in L^2(\mathbb{R}^n)$, and we can apply (3) also for $\widehat{u}$, that is

$\displaystyle ||\widehat{\widehat{u}}||_2 \leq (2\pi)^{n/2}||\widehat{u}||_2 \leq (2\pi)^n ||u||_2$

and by inversion formula $(2\pi)^n u(x)= \int e^{i x \cdot \xi} \widehat{u}(\xi) d\xi := \widehat{u}(-x)$, i.e. $\widehat{\widehat{u}}=(2\pi)^n u^\vee$, then $||u^\vee||_2 = || u||_2$, consequently $||\widehat{u}||_2 =(2\pi)^{n/2} ||u||_2$, also for $u \in L^2(\mathbb{R}^n)$.