# With Khinchine's inequality, prove Fourier basis is unconditional in $L^{p}[0,1]$ only for $p=2$

MathOverflow Asked by Eric Yan on December 3, 2020

I am trying to prove Problem 6.10 in "Classical and Multilinear Harmonic Analysis" by by Camil Muscalu and Wilhelm Schlag.

Problem

Problem 6.10. Let $$1le p < infty$$ and suppose that there exists a constant $$C(p)$$ such that
$$sup_{varepsilon_n=pm1} leftlVert sum_{-N}^N varepsilon_n widehat f(n) e(ntheta) rightrVert le C(p) |f|_p qquad forall fin L^p([0,1]), forall Nge1. tag{6.33}$$
Show that necessarily $$p=2$$. Next show that (6.33) is equivalent to the property that $$sum_{-infty}^infty varepsilon_n widehat f(n) e(ncdot)$$ converges in $$L^p([0,1])$$ for each choice of signs $$varepsilon_n=pm1$$ and $$fin L^p([0,1])$$. This latter property is called unconditional convergence, and this problem therefore amounts to proving that the exponential system is unconditional only for $$p=2$$. In Section 8.4 we shall see that Haar functions are unconditional on $$L^p([0,1])$$ for $$1. Hint: Use Khinchine’s inequality

Here are some attempts,
begin{align*} mathbb{E}Big|sum_{n=-N}^Nepsilon_nhat f(n)e(ntheta)Big|_p&=mathbb{E}Big(int_0^1|sum_{n=-N}^Nepsilon_nhat f(n)e(ntheta)|^pdthetaBig)^{1/p}\ &leqBig(mathbb{E}int_0^1|sum_{n=-N}^Nepsilon_nhat f(n)e(ntheta)|^pdthetaBig)^{1/p}quad(text{since g(x)=x^{1/p} is concave for pgeq1})\ &leq C(sum_{n=-N}^N|hat f(n)|^2)^{1/2}quad(text{Khinchine’s inequality})\ &leq C|f|_2 end{align*}
But
$$mathbb{E}Big|sum_{n=-N}^Nepsilon_nhat f(n)e(ntheta)Big|_pleqsup_{epsilon_n}Big|sum_{n=-N}^Nepsilon_nhat f(n)e(ntheta)Big|_p$$
the inequality seems in the wrong direction. Any suggestion on how to do next? Thanks in advance!