Suppose that for . Prove that for all R, where M(r):=sup{|f(z)|:|z|=r}. I have absolutely no idea where to start with this question. Help starting would be much appreciated.
Last edited by Speed1991; Jan 4th 2012 at 08:45 AM.
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Originally Posted by Speed1991 where M(r):=sup{|f(z)|:|z|=r}. What on earth does that inequality mean?
Sorry, forgot to put in the middle bit
According to the Cauchy inequalities we have , so .
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