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Sep 1, 2010 at 8:32 comment added Matthew Daws Yep, that works! Thanks! For the record, Newman shows that the trig polynomial $(1+z-z^2)/\sqrt5$ works. In my language, $a=5^{-1/2}(\cdots,0,1,1,-1,0,\cdots)$ works. Clearly $\|a\|=3/\sqrt5>1$, Newman shows that $a$ is power-bounded, and the Fourier transform is $(1-\frac45\cos^2\theta)^{1/2}$, which is indeed $<1$ almost everywhere.
Sep 1, 2010 at 8:30 vote accept Matthew Daws
Sep 1, 2010 at 0:21 history answered Yemon Choi CC BY-SA 2.5