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I recently began studying Litllewood-Paley Theory trough Grafakos book on fourier analysis, and yet in the beginning of the chapter the author states a motiovational fact about the $L^2$ spaces, being that if a family (and here i'm assuming it is a family as for later applications we use infinite series) of functions $f_j$, that have fourier transforms $\widehat{f_j}$ supported in disjoint sets then $$ \|\sum_j f_j \|_{L^2}^2 = \sum_j \|f_j \|_{L^2}^2 $$

Well this seems like a good motivation as to wanting to extend this sort of comparison into other $L^p$ spaces, but my point is that the natural reason this seems to hold is trough fourier transform being an isometry in $L^2$ only, as it follows $$ \|\sum_j f_j \|_{L^2}^2 = \|\widehat{\sum_j f_j} \|_{L^2}^2 \overset{\ast}{=} \|\sum_j \widehat{f_j} \|_{L^2}^2 = \sum_j \|\widehat{f_j} \|_{L^2}^2 = \sum_j \|f_j \|_{L^2}^2 $$ Using the fourier transform being an isometry, and to take out the summation from the norm we consider the integral over the union of disjoint sets as the sum of the integrals.

My problem is with the $\ast$ step as the fourier transform is linear when we consider finite sums it's fine, but in this case are we indeed considering a family of functions? and if so how to justify this step?

Sidenote: My point on the consideration of infinite sums is the natural decomposition we consider in this theory as the annulus on the frequency space, so its rather fair to ask if this holds for a family of functions.

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  • $\begingroup$ The Fourier transform is bounded and linear from $L^2$ to $L^2$, so it commutes with infinite sums. Unless I am missing something. $\endgroup$
    – rubikscube09
    Jun 4, 2021 at 20:20
  • $\begingroup$ is it direct? i tried opening it via d.c.t, and got stuck, can you remind me what result states this? im a little rusty on functional analysis $\endgroup$
    – Daniel Moraes
    Jun 4, 2021 at 20:24
  • $\begingroup$ The Fourier transform on $L^2$ is an isometry, hence is bounded. As a bounded linear map, it is continuous so you can go through the infinite sum (provided the sum converges in $L^2$). $\endgroup$
    – Nicolas
    Jun 4, 2021 at 20:26
  • $\begingroup$ ooh just taking the limit, ok, that sounds direct, i missed that $\endgroup$
    – Daniel Moraes
    Jun 4, 2021 at 20:27
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    $\begingroup$ Well the infinite sum needs to be an infinite sum in $L^2$ sense, otherwise if it is in some other sense (e.g.) pointwise, then you may need some other conditions or need to apply a convergence theorem. $\endgroup$
    – rubikscube09
    Jun 4, 2021 at 20:27

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