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While reading Folland's Real Analysis textbook, I came across the Fourier Analysis section and became troubled by Theorem 8.35. It goes as follows:

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Focusing on part $a.$ of the theorem, I am confused about the sudden change of supposition where we originally considered $f \in L^1 + L^2$, but then in $a.$ suddenly consider $f \in L^p \, (1 \leq p < \infty)$. In particular, I have not seen any mention of the Fourier transform if $f \in L^p$ for $p > 2$ (could we use Schwartz functions?). Furthermore as he begins the proof he lets $f = f_1 + f_2$ where $f_1 \in L^1$ and $f_2 \in L^2$, which to me signifies that he is still using $f \in L^1 + L^2$. The proof of $a.$ makes sense to me, but only for $f \in L^1 + L^2$, how are we suddenly also considering $f \in L^p \, (1 \leq p < \infty)$?

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    $\begingroup$ I think the assumption that $f \in L^1 + L^2$ applies throughout. In part a, $f \in L^p$ is an additional assumption. $\endgroup$
    – user169852
    Nov 3, 2020 at 21:57
  • $\begingroup$ Yes, that's exactly what was throwing me off. I thought Folland was throwing out that assumption. Thank you. $\endgroup$
    – Oreomair
    Nov 3, 2020 at 22:00

1 Answer 1

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If $f=f_1+f_2$ and $f_j\in L^j,$ then we know that $\hat{f_1}\in L^\infty$ and $\hat{f_2}\in L^2$ by the mapping properties of the Fourier transform. In the second sentence of the statement of the theorem, they assume that $f\in L^1+L^2,$ so such a decomposition is perfectly kosher. They assumed that $f\in L^p$ in part (a) to use theorem 8.14.

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  • $\begingroup$ Yes, I am familiar with the mentioned mapping property. I suppose my difficulty is if $f \in L^p$ for $p > 2$, then what exactly is $f^t$ as is defined in the theorem? In particular, what is the $\widehat{f}(\xi)$ term in $f^t$? $\endgroup$
    – Oreomair
    Nov 3, 2020 at 21:54
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    $\begingroup$ The Fourier transform is linear, so it makes perfect sense to take the Fourier transform of $f_1+f_2$ (which equals $f$). The $L^p$ assumption isn't used to take the Fourier transform. $\endgroup$
    – cmk
    Nov 3, 2020 at 21:56
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    $\begingroup$ If you like, $f\in (L^1+L^2)\cap L^p$ in part a. $\endgroup$
    – cmk
    Nov 3, 2020 at 21:56
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    $\begingroup$ Oh gosh, I get it! I understand what you're saying and understand about the linearity principal. I was assuming we disregarded that $f \in L^1 + L^2$ when we said $f \in L^p$. Thank you. $\endgroup$
    – Oreomair
    Nov 3, 2020 at 21:58
  • $\begingroup$ Happens to everyone sometimes :) $\endgroup$
    – cmk
    Nov 3, 2020 at 22:03

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