When does $\mathcal{F}\left ( \sum_{n=1}^{\infty} f_n (x)\right ) = \sum_{n=1}^{\infty} \mathcal{F}(f_n(x))$ where $\mathcal{F}$ the Fourier transform operator.
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- $\begingroup$ You can always do it in $L^2(\mathbb R)$ for convergent series, since $\mathcal F$ is a unitary (hence bounded) operator. $\endgroup$Yurii Savchuk– Yurii Savchuk2013-11-14 18:04:50 +00:00Commented Nov 14, 2013 at 18:04
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1 Answer
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Basically, it is Fubini's theorem. Equality holds at fixed $\lambda>0$ if and only if you have $e^{-i\lambda.x }f(n,x)\in L^1(dx \otimes dn)$ where $dx$ is Lebesgues measures and $dn$ the counting measure over integers, and $dx \otimes dn$ is the product measure.
If you prefer you can state it like this for fixed $\lambda$ :
$\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x).e^{-i.x.\lambda}|dx<+\infty$
We can improve this a little by observing that :
$\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x).e^{-i.x.\lambda}|dx)=\sum_{n=1}^{\infty} \int_{\mathbb{R}}|f_n(x)|dx$ and so the result doesn't depend on (real) values of $\lambda$.
Is that what you want ?
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