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The dual group of a locally compact Abelian group is used as the underlying space for an abstract version of the Fourier transform. If a function $f$ is in $L^{1}(G)$, then the Fourier transform is the function $\hat{f}$ on $\hat{G}$ defined by

$\hat{f}(\chi)=\int_{G}f(x)\overline{\chi (x)}dx$

if $G=\mathbb{R}$, then how can I prove that the Fourier transform is of the form below?

$\hat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-ix\xi}dx$

Ehsan zarei
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  • If $G = \mathbb{R}$, then $\hat{G} = {\chi_{\xi} \colon x \mapsto e^{ix\xi} \mid \xi \in \mathbb{R}}$. – Daniel Fischer Feb 28 '16 at 16:08
  • You can't prove a definition. Also, when you work with the group $\mathbf R$ you have to make a convention about how you think of $\widehat{\mathbf R}$ as $\mathbf R$ in order to let the Fourier transform be regarded as a function on $\mathbf R$ instead of on the abstract $\widehat{\mathbf R}$. Frankly I'd use the factor $e^{2\pi ix\xi}$ in my definition of the Fourier transform. There are many possible conventions, so a "proof" is impossible. – KCd Feb 28 '16 at 16:09
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    Maybe you are asking for a proof that $\widehat{\mathbf R}$ is isomorphic to $\mathbf R$ as topological groups? If so, make that clearer. – KCd Feb 28 '16 at 16:10
  • You asked here for a proof that the dual of $\Bbb R$ is $\Bbb R$… – Watson Feb 28 '16 at 16:27
  • thanks Daniel. if we have $\hat{G}=\left { \chi_{\xi}:x \mapsto e^{ix\xi}\mid \xi\in\mathbb{R} \right }$, how can we complete the proof? – Ehsan zarei Feb 28 '16 at 16:32
  • Yes Watson. I know that $\hat{G}=\left { \chi_{\xi}:x \mapsto e^{ix\xi}\mid \xi\in\mathbb{R} \right }$. but nevertheless i can not complete the proof of this assertion. – Ehsan zarei Feb 28 '16 at 16:41

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As it was said in the comments, $$\Psi : G=\Bbb R \to \hat{G} = \{\chi_{\xi} \,\colon x \mapsto e^{ix\xi} \mid \xi \in \mathbb{R}\}, \xi \mapsto \chi_{_{\xi}}$$ is a group isomorphism.

Then define $$\hat f(\xi) := \hat f(\Psi(\xi)) = \hat f(\chi_{\xi}) = \int_{G}f(x)\overline{\chi_{\xi} (x)} dx$$

We get $$\hat f(\xi) = \int_{G}f(x)\overline{\chi_{\xi} (x)}dx = \int_{G}f(x)\overline{e^{ix\xi}}dx = \int_{G}f(x)e^{-ix\xi}dx$$

as desired.

Watson
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