It is known that the $L^2$-norm of a Fourier series equals the $l^2$-norm of the coefficients. Are there similar results in the case of $L^p$-norm for $p\neq 2$? Can it be expressed explicitly in terms of the coefficients?
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$\begingroup$ isn't any operation on the Fourier series expressible "in principle" in terms of an operation on the coefficients? it won't be a simple expression (except for p=2), but why would it fail "in principle"? $\endgroup$– Carlo BeenakkerCommented Sep 27, 2013 at 16:47
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2 Answers
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Consider a function $f \in L^2$ and $f \notin L^{p}$ (for $p>2$). Now multiple the Fourier coefficients by random signs. Almost surely, the new function, $g$, will be in $L^{p}$ (by Khinchin's inequality and Fubini's theorem). Thus we have two functions, $f$ and $g$, both of whose Fourier coefficients have the same absolute values, one of which has finite $L^{p}$ norm and the other of which has infinite $L^{p}$ norm. Thus, no expression involving only the absolute values of the Fourier coefficients can compute (or even bound!) the $L^p$ norm of a function.
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The answer is "not really, except in the sense that Carlo Beenaker has mentioned".
I suggest you look at the discussion of the Hausdorff–Young inequality in Chapter IV Section 2 of Katznelson's Introduction to Harmonic Analysis (2nd edition, Dover). Probably there will also be the same cautionary remarks and salutary counterexamples in Edwards's book Fourier Series.
(Could someone please do me the favour of flagging this as CW?)
