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As far as I understood, the Fourier decomposition of a function $\boldsymbol{F}\colon\mathbb{R}^{n}\to\mathbb{R}^{m}$ where $\mathbb{R}^{n}$ is endowed with the Euclidean inner product $\left<\cdot,\cdot\right>$ is given by

$\boldsymbol{F}(\bar{x})=\int_{\mathbb{R}^{n}}{\tilde{\boldsymbol{F}}(\bar{\nu})e^{2\pi i \left<\bar{\nu},\bar{x}\right>}}{d\bar{\nu}}$

where $\tilde{\boldsymbol{F}}(\bar{\nu})=\int_{\mathbb{R}^{n}}{\boldsymbol{F}(\bar{x})e^{-2\pi i \left<\bar{\nu},\bar{x}\right>}}{d\bar{x}}$

How does this come about and for which functions does it apply? I'm not even able to find the right framework to work in (Hilbert spaces?). Secondly, could I just replace the Euclidean inner product by the Minkowskian inner product when in Minkowski space?

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"Could I just replace the Euclidean inner product by the Minkowskian inner product when in Minkowski space?" Up to sign you get the same object. As $-\tau t + \xi_1 x_1 + \xi_2 x_2 + \ldots = (-\tau)t + \xi_1 x_1 + \ldots$. But thinking about it as an inner product is perhaps not the best thing: you should think of it as some sort of dual pairing: after all, the physical and frequency spaces are not the same space, and you are not taking products of vectors in the same vector space. –  Willie Wong Apr 4 '12 at 14:42
    
Your first question is perhaps not quite in the scope of this forum. You may want to consider one of the other websites mentioned in the FAQ, such as math.stackexchange.com –  Willie Wong Apr 4 '12 at 14:43
    
Too basic, I get it :-). But could you maybe give me an idea in which direction I should be looking, that would save me a lot of time. –  Wox Apr 4 '12 at 14:50
    
I definitely agree with Willie Wong comment. Take a finite dimensional vector space $E$ with $d\mu$ and Haar measure on it. Then consider the dual space $E^*$ with the dual Haar measure $d\mu^*$. then for $u$ defined on $E$ (say a Schwartz function), $\hat u$ is defined on $E^*$ by $$ \hat u(\xi)=\int_E e^{-2i\pi\langle \xi,x\rangle_{E^*,E}} d\mu (x), $$ and $$ u(x)=\int_E e^{2i\pi\langle \xi,x\rangle_{E^*,E}} d\mu^* (\xi). $$ –  Bazin May 1 '12 at 20:31
    
The second integral above is on $E^*$. –  Bazin May 1 '12 at 20:33

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