Confusion about Mean Value Theorem stated in a textbook

Question

There is a Lemma in chapter 1 (section 1.11) of the textbook: Graduate text in mathematics, functional analysis, Conway which states:

If $f$ is analytic in a neighborhood of $\overline B(a;r)$, then

$$f(a) = \frac{1}{\pi r^2} \int \int_{B(a;r)} f, $$

where $\overline B(a;r) = \{z: |z-a| \leq r\}$.

For the proof it uses the mean value property as: if $0 < t \leq r$, $f(a) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(a+te^{i\theta}) d\theta. $

I am confused about mean value theorem here?

Answer 1

Let $c(\theta)=a+te^{i \theta}$ for $\theta \in [- \pi, \pi]$.

Then, by Cauchy:

$f(a)=\frac{1}{2 \pi i} \int_c \frac{f(w)}{w-a}dw$.

Now use the definition of the integral $ \int_c$ to get the desired formula.

Answer 2

Let $\varphi(t)=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f\left(a+te^{i\theta}\right)d\theta$, then $\varphi'(t)=\dfrac{1}{2\pi}\dfrac{1}{t}\displaystyle\int_{-\pi}^{\pi}f\left(a+te^{i\theta}\right)te^{i\theta}d\theta=0$. So $\varphi(t)=C$, and we have $\varphi(0)=f(a)$.