# Demonstrating that a non-constant holomorphic function attains its maximum on the boundary of its region?

Suppose that $f$ and $g$ are holomorphic in a region containing the disc $|z| \leq 1$. Suppose that $f$ has a simple zero at $z = 0$ and vanishes nowhere else in $|z| \leq 1$. Let

$$f_{\epsilon}(z) = f(z) + \epsilon g(z)$$

Show that if $\epsilon$ is sufficiently small, then $f_{\epsilon}(z)$ has a unique zero in $|z| \leq 1$.

Rouche's theorem states that if $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior, and if $|f(z)| > |g(z)|$ for all $z \in C$, then $f$ and $f + g$ have the same number of zeroes inside $C$.

I want to show that $|f(z)| > |\epsilon g(z)|$. Then $f(z) = \epsilon g(z)$ and $f_{\epsilon}(z)$ have the same number of zeroes in $C$ (i.e. one zero), and I'm done.

But how do I show that $|f(z)| > |\epsilon g(z)|$? Assuming $g(z)$ is not the constant $0$ function, a solution I've seen says that $|f(z)|$ and $|g(z)|$ attain maxima $M_f$ and $M_g$ respectively on $|z| = 1$. How precisely is this true?

I'm aware of the maximum modulus principle, which states that if $f$ is a non-constant holomorphic function in a region $\Omega$, then $f$ cannot attain a maximum in $\Omega$. Can I alter this definition in some way to conclude that if $f$ is a non-constant holomorphic function, $f$ achieves its maximum on the boundary of the region? Is it as simple as saying that since it cannot attain a maximum inside of $\Omega$, it has to obtain a maximum on $\Omega$'s boundary? How come the function can't just stretch all the way to $\infty$ inside the region? -- why must it attain a maximum on the boundary?

Thanks.

In general, it's possible for a function that is holomorphic on an open domain $\Omega$ to have a pole on the boundary $\partial \Omega$, as you suggest. For instance, $(z-1)^{-1}$ is holomorphic on the open unit disk but goes to $\infty$ at $z = 1$.

But recall the assumption from the very beginning: $f$ and $g$ are holomorphic in a region containing the closed unit disk $\{|z| \leq 1\}$. In particular, they are continuous on the closed disk and therefore obtain a maximum modulus somewhere on the closed disk. The maximum modulus principle ensures that can't occur in the interior (provided they are not constant), so it must be on the boundary.

• That makes sense, thanks -- however, why precisely does $f$ and $g$ being continuous on the closed disk guarantee they obtain a maximum modulus somewhere on the closed disk? Oct 9, 2014 at 2:43
• The extreme value theorem: a continuous function from a compact set to $\mathbb R$ attains a minimum and a maximum (more generally speaking, the continuous image of a compact space must be compact; in $\mathbb R$ this means closed and bounded, which implies a min and a max). The closed unit disk is compact, so we have a maximum. Oct 9, 2014 at 2:45
• Note that an important point in this question is that $|f(z)|$ also attains a non-zero minimum on the unit circle. Oct 9, 2014 at 2:53

Let $m = \min_{|z| = 1} |f(z)|$, $M = \max_{|z| = 1} |g(z)|$. We have $m>0$. Choose $\epsilon >0$ such that $\epsilon M < m$.

If $|z|=1$, then $|f_\epsilon(z)| = |f(z)+\epsilon g(z)| \ge |f(z)|-\epsilon |g(z)| \ge m-\epsilon M >0$, so $f_\epsilon$ has no zeroes on the unit circle.

Furthermore, if $|z| = 1$, we have $|f_\epsilon(z) - f(z)| = \epsilon |g(z)| \le \epsilon M < m \le |f(z)|$, hence by Rouchés theorem, $f_\epsilon$ and $f$ have the same number of zeros (and poles) in the open unit disk.