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The following is an exercise from Complex Analysis by Stephen Fisher.

Fix a complex number $a$ and a positive real number $R$. Suppose $u$ is a function defined on the circle of radius $R$ centered at $a$. Let $C$ denote this circle.

Show that the average value of $u$ on $C$ is given by $\frac{1}{2\pi}\int_{0}^{2\pi} u(a + Re^{it})dt$.

Any Hints please.

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    What is the definition of the average value? (The integral is usually used as the definition, but I don't know the book you are using.)2012-10-26
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    The exercise is probably meant to relate to Cauchy's integral formula, but as said above: the integral you write is somewhat the definition of "average value" and the goal of the calculation, so the exercise is a bit unclear in its current form. I have the book (2nd edition at least) - which exercise number/page is it, so I can get a context?2012-10-26

2 Answers 2

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We assume Cauchy's integral theorem.

$$(1)\hspace{5mm}\frac{1}{2\pi}\int_{0}^{2\pi} u(a+ Re^{i t})dt = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{u(a+Re^{it})iRe^{it}}{Re^{it}}dt $$

The right side of (1), letting R be the radius of circle $C,$ $u(z)$ the equation of the circle $|z-a|= R $ or $z = a + Re^{it},$ so that $z-a = Re^{it}$ and $dz = iRe^{it},$ is precisely

$$(2)\hspace{5mm}\frac{1}{2\pi i}\oint_C\frac{u(z)}{z-a}dz$$

By Cauchy's integral formula (2) is equal to $u(a).$ $\square $

This exercise is odd because the starting integral is generally taken as the definition of the average value, as noted in a comment above. We would normally derive that form for the average from Cauchy's integral formula.

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    I just tried to answer an essential duplicate of this question at https://math.stackexchange.com/questions/2243917/why-does-the-integral-for-the-average-value-of-a-complex-function-on-a-circle-no/2243933#2243933 and did not even think to apply the Cauchy integral theorem; very elegant!2017-04-20
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    This answer is wrong. No other way to put it... The function $u$ is not supposed to be holomorphic, and so, Cauchy's theorem does not hold. Furthermore, the goal of the exercise is to obtain the average value, rather than the value at the center.2017-04-20
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    @AmitaiYuval: Obviously I assume Cauchy's theorem does hold. The goal of the exercise is to show that the average value has the form given. That follows from Cauchy if it holds. If there is any other way to make sense of the exercise, please give it. And of course if the function is analytic the average value on the circle is the value at the center.2017-04-21
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    @daniel Please have a look at my answer. It does not assume that $u$ is holomorphic. In fact, it works for every $u$ integrable on $C$.2017-04-21
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    @daniel And one more thing. Your answer could be an excellent answer to a different question. You actually proved an alternative formula for $u(a)$, assuming that $u$ is holomorphic. However, it is irrelevant for this post; the goal here is finding the *average value* of a function on a circle. You did not show anything about any average value.2017-04-21
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    @daniel combining your solution with the claim at hand yields the following important observation - if $u$ is holomorphic, then $u(a)$ is equal to the average value of $u$ on a circle centered at $a$. (This follows independently from the fact that both the real and imaginary parts of $u$ are harmonic).2017-04-21
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    @AmitaiYuval: OK, will study your comments and answer in a bit. Thanks.2017-04-21
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As implied in the comments, the most important part in this exercise is the interpretation of "average value". As always, in order to define the average value of a function on $C$ we need a measure on $C$. One should note that different measures evidently yield different average values. However, as $C$ is a smooth curve, it makes sense to take the standard length measure.

Parametrize $C$ by$$\gamma:[0,2\pi R]\to\mathbb{C},\quad s\mapsto a+Re^{is/R}.$$This is an arc-length parametrization, or in other words, pulling back the length measure on $C$ with respect to $\gamma$ yields the standard Lebesgue measure on $[0,2\pi R]$. Hence, the average value of $u$ on $C$ is equal to the average value of $u\circ\gamma$ on $[0,2\pi R]$. The latter is just$$\frac{1}{2\pi R}\int_0^{2\pi R}u\left(a+Re^{is/R}\right)ds.$$The desired expression follows now by substituting $t=s/R$.

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    Hi @Amitai Yuval , can you try answer my question over here? https://math.stackexchange.com/questions/2944389/calculate-the-mean-of-a-function-within-a-part-of-a-circle Its very related to this subject2018-10-27