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Let $z _1=|z _1 |e^{i \alpha _1 } , ... , z _N=|z _N |e^{i \alpha _N } $ be complex numbers.

Write $\sum _{k=1} ^N |z _k |\cos ^+(\alpha _k - \theta )$, where $\cos ^+ $ is defined to equal 0 if $\cos(\alpha _k - \theta )<0$ .

Chose $\theta $ as to maximize the sum.

Now Rudin states that "this sum is at least as large as the average of the sum over $[- \pi , \pi ]$ , and this average is $\pi ^{-1 } \sum |z _k |$, because

$$\frac {1 } {2 \pi } \int _{- \pi } ^{\pi } \cos ^+ ( \alpha - \theta ) d \theta = \frac {1 } {\pi } $$

What is meant with the sum over $[- \pi , \pi ]$ as this is an uncountable set?

Also, how do I evaluate the integral with $\cos ^+$?, is it correct that the integral is equal to $\int _{\pi /2 } ^{\pi /2 } \cos (x ) d x $ for any $\alpha $, as the function is zero outside this interval?

Thanks in advance!

Alexander
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  • For your last question, yes, the integral is what you say if you put a minus sign on the lower limit. Your reasoning is correct. – Ross Millikan Oct 02 '14 at 14:46

1 Answers1

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He's simply observing that if you have a positive-valued, continuous function $g$ on some interval $I$, then \begin{align} \max_{x \in I} g(x) \geq \frac{1}{|I|} \int_I g(y) dy. \end{align}

'Average of sum' in this context is referring to the average of the function of $\theta$ defined by the sum over $|z_k| \cos^+(\alpha_{k} - \theta)$.

A Blumenthal
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