Q:

Modulus of function values on circle is necessarily less than or equal to modulus at centre?

Consider a complex function $f(z)$, a circle $C_\rho$ with radius $\rho$, centred at $z_0$, which can be defined as $z=z_0+\rho e^{i\theta}$ $(\theta \in [0, 2\pi]$). In the book on complex variables which I'm reading, it is stated without clarification that $\left| f(z_0 + \rho e^{i\theta}) \right|\le \left| f(z_0)\right|$. Why would that be the case?
The author says that there is the Gauss's mean value theorem which states that "when a function is analytic within and on a given circle, its value at the center is the arithmetic mean of its values on the circle." (Complex Variables and Applications, Brown & Churchill). But is this theorem relevant to my question above? For if $f(z_0)$ is the average of all values of $\text{ }f(C_\rho)$ then there may be some values on $C_\rho$ which are greater than the average.
[Also, if I understand the statement of the Gauss's mean value theorem correctly then it must be the case that $f(C_\rho)$ must have the same values on the entire region where $f$ is analytic, unless it's $f$ is unbounded.]

A:

My mistake, I wasn't careful enough. The statement in question actually comes from the assumption of the Lemma: "Suppose that $\left| f(z) \right| \le \left| f(z_0) \right|$ at each point $z$ in some neighborhood $\left| z - z_0\right|<\varepsilon$ in which $f$ is analytic. Then $f(z)$ has the constant value $f(z_0)$ throughout that neighborhood.".

