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I have a problem: Suppose $f$ is analytic on closed disk $(\bar{\mathbb{D}} = \{z \in \mathbb{C} : |z| \leq 1 \})$. Assume that $|f(z)| = 1$ for $|z| = 1$, and that $f$ is not constant. Show that image of f contains entire open unit disk $(\mathbb{D} = \{z \in \mathbb{C} : |z| \le 1 \})$.

I know that I should use Rouché's theorem and maximum modulus principle but I do not know how to apply them.

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3 Answers 3

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$f$ has a zero in the interior of the unit disc, otherwise $\frac1{f(z)}$ would be analytic, contradicting the maximum modulus principle (as f is not constant). let this zero be $\zeta$.

the function $\phi(z) = \frac{\zeta-z}{1-\bar \zeta z}$ maps the unit disc to itself, preserving the boundary and interchanging the points $0$ and $\zeta$

write $$ g(z) = f \circ \phi (z) $$ then $g$ is analytic on and inside the unit disc, has a zero at the origin, and satisfies: $$ |g(z)|=1 $$ on the boundary.

for a complex number $s$ with $|s| \lt 1$, regarded as a (constant) function we have $$ |-s| \lt |g(z)| $$ on the boundary, so by Rouché's theorem the function $g(z)-s$ has the same number of zeroes as $g(z)$, so some point $z'$ must satisfy $g(z')-s=0$. now set $z''=\phi(z')$ and we have $$ f(z'')=s $$

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I think this will solve without Rouché's Theorem.

STEP 1: Lets prove, first, that $f(\Bbb{D}) \supset D(0,1/2)$. Indeed, for $w \in D(0,1/2)$, if we suppose that $f(z)-w$ has no roots in $\Bbb{D}$, then the function $G_w(z) = \frac{1}{f(z)-w}$ is analytic on $\Bbb{D}$ and well-defined. By the Maximum Modulus Principle, we must have that (as $|f| = 1$ on the boundary)

$$ \frac{1}{|f(z)-w|} \le \frac{1}{1-|w|} < \frac{1}{|w|} $$

And, from the same Theorem, we have that, if equality is attained on $\Bbb{D}$,then $G_w$, and consequently $f$, must be constant. But $f$ has at least one zero, because $\frac{1}{|f|} \le 1$ on the boundary, which shows that, as $f$ is nonconstant, $f(z_0)=0$ for some $z_0 \in \Bbb{D}$.

Taking this $z_0$ above, we see that we must have a contradiction, and, thus, the claim follows.

STEP 2: We iterate the process. Using the same argument, if $w \in D(0,1- 2^{-k-1})$, then

$$ \frac{1}{|f(z)-w|} \le \frac{1}{1-|w|} < 2^{k+1} $$

For $z$ on the boundary, and, by induction hypothesis, as $f(\Bbb{D}) \supset D(0, 1 - 2^{-k})$, then there exists $z_0, w_0$ such that $f(z_0) = w_0 $ and $w_0-w < 2^{-k-1} $. Repeating the argument given above, we have a contradiction.

Therefore, we may conclude that, $\forall \; k \in \mathbb{N}$, we have that $f(\Bbb{D}) \supset D(0,1-2^{-k}) \Rightarrow f(\Bbb{D}) \supset \Bbb{D}$.

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The analyticity of $f$ on $\bar{D}$ and condition $|f(z)|=1$ for $|z|=1$ imply that $f$ has at most finite number of zeros in $\mathbb{D}$. Call them $z_k$, $k=1,\ldots,n$ and consider the finite Blaschke product with those zeros $$ B(z)=\prod_{k=1}^n\frac{z-z_k}{1-\bar z_k z}. $$ We know that $|B(z)|=1$ on the unit circle, so the function $f/B$ has no zeros inside $\mathbb{D}$ and the modulus $1$ on the boundary. The function $u(z)=\text{Re}\,\ln\frac{f}{B}$ is harmonic in $\mathbb{D}$ and zero on the boundary, hence, the identical zero, which gives us that $f/B=c=\text{constant}$ with $|c|=1$ and, finally, $f=cB(z)$, i.e. an automorphism of $\mathbb{D}$.

Edit (an alternative proof): Assume that $w\in\mathbb{\bar D}$ and $w\not\in f(\mathbb{\bar D})$, then the function $g(z)=f(z)-w$ is never zero inside the closed disc, i.e. $|g(z)|\ge\epsilon$ in $\mathbb{\bar D}$. Using the Schwarz reflection principle for the unit circle, we get the analytical continuation of $g$ as $\frac{1}{\bar f(1/\bar z)}-w$. It is symmetrically never zero outside the disc and has a finite nonzero limit at infinity. Thus the function $1/g(z)$ is entire and bounded, hence, a constant due to the Liouville theorem. So $f$ is constant, which is a contradiction.

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