The basic idea seems to be that since your function phi(z) is analytic, it must be bounded by the values on the lines Re(z)=0 and Re(z)=1. Also, on each line Re(z)=constant it is maximised at the point Im(z)=0, which gives you the bounds on Re(z)=0 and Re(z)=1. Using almost the same argument, phi(z) is log-convex on [0,1].

However, log-convexity of phi(z) follows directly from the fact that it is an integral over exponential functions of z, which also proves the Holder inequality (and a bit more efficiently imo).

]]>I agree with your remarks. Of course, as you point out, you should be able to reduce the multilinear case to the bilinear case.

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