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J. NEUWIRTH AND D. J. NEWMAN The Koebe function z/(l+z)2 is positive everywhere on \z\ =1, z¥" — 1, and lies in the Hardy class Hp for every p < 1/2. We show that this behavior is extreme by proving the following Theorem. If fiz) EH112 and i//(z)^0 a.e. on \z\ =1 then fiz) is a constant. Proof. We may assume that/(z) is not identically 0. If Biz) denotes the Blaschke product for the zeros of/(z) then, as usual, we can write (1) fiz) = 2J(z)F2(z), Fiz) E HK We write the condition /(z)SiO as the equation /(z) = |/(z)| and conclude from (1) that (2) 5(z)F2(z) = ! F2iz) | a.e. on | z\ =1. Since/(z) is not identically 0 it follows that Fiz) is nonzero a.e. on \z\ =1. Thus we may divide (2) by Fiz) and obtain (3) Biz)Fiz) = F~iz) a.e. on | z\ =1. But the left side of (3) is H1 and so has all negative Fourier coefficients 0, the right side is conjugate Hx and so has all positive Fourier coefficients 0!. Thus only the constant term remains and we conclude that both sides are constants. This is to say Biz)Fiz) and F(z) are both constants and so indeed/(z) = (23(z)^(z)). Fiz) is a constant. University of Connecticut and Yeshiva University Received by the editors May 17, 1967.

Positive 𝐻^{1/2} functions are constants