# If $|f(z)| \geq |g(z)|$ for $z \in D$ and $E = \{ z \in D : |f(z)| =|g(z)| \}$ has a limit point, then $E=D$.

This is my problem:

Let $D := \{ z \in \mathbb{C} : |z| <1 \}$. Let $f$ and $g$ be analytic functions on $D$. Suppose $|f(z)| \geq |g(z)|$ for all $z \in D$. Define $E = \{ z \in D : |f(z)| =|g(z)| \}$. Show that if $E$ has a limit point in $D$, then $E=D$.

This looks like it should be related to the identity theorem for analytic functions, but I can't seem to get it to work out.

Going for a proof by contradiction, I assume that $E \;$ has a limit point in $D\;$ and that $E \neq D\;$. Then there is a sequence $z_n$ of points in $E$ which converges to a limit point $z_0$ in $D$. By continuity, $z_0 \in E$; that is, $f(z) = re^{i\theta}$ and $g(z) = re^{i\phi}$ for some $r$, $\theta$, and $\phi$.

And here I stop. Any advice on how to proceed?

Thanks.

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I think that you can use the maximum modulus principle to get what you want.

Suppose first that $f$ has no zeros in the unit disk $D$. Then that means that $\dfrac{g(z)}{f(z)}$ is analytic in $D$. Then your condition that $|f(z)| \geq |g(z)|$ for every $z \in D$ translates to

$$\left | \frac{g(z)}{f(z)} \right | \leq 1 \quad \text{for every z \in D}$$

Then since $E$ has a limit point in $D$, there is a sequence of points $z_n \in E$ such that $z_n \to a$ for some $a \in D$ and since for every $z_n \in E$ we have $\left | \dfrac{g(z_n)}{f(z_n)} \right | = 1$

then by continuity also $$\left | \frac{g(z_n)}{f(z_n)} \right | \to \left | \frac{g(a)}{f(a)} \right |$$

so $\left | \dfrac{g(a)}{f(a)} \right | = 1$. Thus we have that

$$\left | \frac{g(z)}{f(z)} \right | \leq \left | \frac{g(a)}{f(a)} \right | \quad \text{for every z \in D}$$

so by the maximum modulus principle we conclude that $g/f$ is constant, say $g/f = c$ and $|c| = 1$. Then this implies that $|f(z)| = |g(z)|$ for all $z \in D$ and this implies that $E = D$ as you wanted to show.

Now, what happens if $f$ has zeros in $D$?

In this case you can use the inequality $|f(z)| \geq |g(z)|$ to conclude that if $f$ has a zero of order $n$ at $a \in D$, then $a$ is also a zero of $g$ of order $m$ and actually $m \geq n$. This implies that the quotient $g/f$ has removable singularities at the zeros of $f$ and then we can redefine the function at those points and again we would have $g/f$ analytic in $D$ and the previous argument applies.

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I have to admit that I'm not entirely convinced by this argument since I'm not really using the full hypothesis that $E$ has a limit point in $D$. For the maximum modulus principle I only need $E$ to be non empty. If someone spots a mistake I would appreciate it if you can point it out please. Thanks. –  Adrián Barquero Jul 14 '12 at 3:53
Well, if you replace the assumption "E has a limit point in $D$" by "$E$ is not empty", then the statement is false : consider $f(z)=z$ and $g(z)=0$ for all $z$, for example. I think the issue here is that you have to rule out the case where $g$ is zero, for example by showing that there exists $z \in E$ with $f(z) \neq 0$. –  Malik Younsi Jul 14 '12 at 4:07
@Malik Thank you very much for pointing that out, you're absolutely right. I guess that now some things have to be modified. I'll try to see if I can fix the argument. –  Adrián Barquero Jul 14 '12 at 4:34
You're welcome! The argument can probably be fixed by something like that : Set $F:= \{ z \in D : f(z)=0 \}$. If $F=D$, then $f$ is identically zero and so is $g$, and the conclusion follows. If not, then $F$ has no accumulation point in $D$. Take a sequence of distinct points $z_n$ in $E$ with $z_n \rightarrow z_0 \in E$. Since $F$ has no accumulation point, for all $n$ sufficiently large we have that $z_n \notin F$. Then $|f(z_0)| = |g(z_0)|$ with $f(z_0) \neq 0$, and now you can apply the maximum modulus principle to conclude that $g/f$ has a maximum at $z_0$... –  Malik Younsi Jul 14 '12 at 13:23