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Let $f$ be an analytic on the set $C^r_0=\{z\in\mathbb{C}||z|

Then $$|f(a)|\leq\frac{1}{\pi r^2}\int_{C^r_a}|f(z)|dxdy$$ where $C^r_a=\{z\in \mathbb{C}||z-a|

I have been taught the Cauchy-integral formula which seems similar to this. But, it is not the same. Does anyone know how this is done? (Because I do not where to begin)

My version of the Cauchy theorem is this:

If $D \subset \mathbb{C}$ is a simply connected space. $f:D\to\mathbb{C}$ an analytic function in $D$ and $\gamma\subseteq D$ a closed smooth curve that includes $z$ then: $$f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\mu)}{\mu-z}d\mu$$

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    Your hypotheses are strange: $f$ is defined on $D(0,r)$ (I prefer this standard notation), but then somehow $f$ is defined on $D(a,r)?$ – 2017-01-13

2 Answers 2

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This might be better stated in the following way. Suppose $U\subset \mathbb C$ is open and $f$ is analytic on $U.$ If $\overline {D(a,r)} \subset U,$ then

$$|f(a)| \le \frac{1}{\pi r^2}\int_{D(a,r)} |f|\, dA.$$

Here $A$ is area measure in the plane.

Proof: First, if $0\le s \le r,$ then

$$\tag 1 f(a) = \frac{1}{2\pi}\int_0^{2\pi} f(a + se^{it})\, dt.$$

There are two ways to see this: i) it falls right out of Cauchy's theoerm; ii) $f$ is analytic, hence harmonic. $(1)$ is just the mean value property for harmonic functions.

Let's now look at

$$\int_{D(a,r)} f\, dA.$$

This can be evaluated by using polar coordinates. If you do that and use $(1),$ you'll find

$$f(a) = \frac{1}{\pi r^2}\int_{D(a,r)} f\, dA.$$

(This is the area version of the mean value property.) The desired inequality is then obtained by taking absolute values, then moving the absolute value signs inside the integral.

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    I don't see how $(1)$ falls right out of Chauchy's theorem and also how this integral is evaluated my polar coordinates, could you give further detail on this. This might be because my theorem is of a different shape. We have not learned the mean value property harmonic functions. – 2017-01-14
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    Also I don't know where the $\frac{1}{2\pi}$ comes from in the integral, because I thought the substitution $z=a+se^{it}$ gave the integral just without that $\frac{1}{2\pi}$ part. – 2017-01-14
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    @BozoVulicevic To see the $1/2\pi$ has to be there, take $f\equiv 1.$ Cauchy says $f(a) = 1/(2\pi i) \int_{|z-a|=r} f(z)/(z-a)\,dz.$ This contour integral turns directly into $(1).$ I'm not sure what problem you're having with polar coordinates. The polar coordinates formula gives $$\int_{D(a,r)} f\,dA = \int_0^r\int_0^{2\pi} f(a+se^{it})\, dt\,s\,ds\, .$$ – 2017-01-14
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This says

$|f(x)| \leq \frac 1 {m(D)}\int_D |f(x+iy)|dxdy$

for disks $D\subset \mathbb R^2$ centered at x (here $m$ is the area measure).

The result is thus that the point is dominated by the disk average. You can expect this to be true in light of the maximum modulus theorem.

Using polar coordinates, you may transform this expression; remember to use $re^{i\theta}$. In particular, you want to rewrite the expression into the Cauchy integral formula for polar coordinates.