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Let $D$ be the unit disc, $f$ continuous on $\overline{D} \setminus \{1\}$, holomorphic and bounded on $D$. The problem is to show that for all $z \in D$, $$|f(z)| \leq \sup\limits_{|\zeta| = 1, \zeta \neq 1} |f(\zeta)|$$ I'm stuck. Here are two approaches I tried:

I . $f$ is represented by a convergent power series $\sum\limits_{n=0}^{\infty} a_n z^n$, absolutely and uniformly convergent for $|z| \leq r$, where $r < 1$. I don't know.

II . For each $0 < r < 1$, there is an angle $\theta \in (-\pi/2, \pi/2]$ such that $|f(z)| \leq |f(r e^{i \theta})|$ for all $|z| < r$ (maximal modulus principle). Pick $\theta_r$ to be such that $|\theta_r|$ is the supremum of all $|\theta|$ satisfying the condition I just mentioned. $\theta_r$ would be a limit of the $\theta$, so $|f(z_r)| \geq |f(z)|$ for all $|z| < r$ by continuity, where $z_r$ is defined to be $re^{i \theta_r}$. The idea is I'm trying to get the $\theta_r$ to not be too close to $0$, if possible.

Now pick any sequence of radii $r_1 < r_2 < \cdots $ which converges to $1$. Then $|f(z_{r_1})| \leq |f(z_{r_2})| \leq \cdots$, and we may choose a convergent subsequence of the $z_n$ converging to, say, $z_0 \in \partial D$. If $z_0 \neq 1$, we're clearly done.

To do this, I want to find a subsequence $z_{n_k}$, and a $\delta > 0$, such that $|\theta_{r_{n_{k}}}| > \delta$ for all $k$. I can then pick a convergent subsequence of this subsequence to do the trick. If this is not possible, then it's easy to see that $z_n$ has to converge to $1$ as a result. The numbers $|f(z_n)|$ are nondecreasing and bounded, so $|f(z_n)|$ tends to a limit as $z_n \to 1$. By the way I picked $z_n$, this shows that if $\theta_{r_n}'$ is another sequence of angles such that the maximum of $f$ on $|z| = r_n$ is attained at $r_ne^{\theta_n'}$, then $r_ne^{\theta_n'}$ also goes to $1$.

Not sure if this approach will go anywhere.

zibadawa timmy
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D_S
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1 Answers1

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Suppose $f$ is continuous on $\overline D \setminus \{\zeta _0\}$ for some $\zeta_0 \in \partial D,$ with $f$ bounded and holomorphic in $D.$ Set $M = \sup \{|f(\zeta)|: |\zeta | = 1, \zeta \ne \zeta_0\}.$ For $0<r<1$ set $f_r(z) = f(rz).$ Then by Cauchy,

$$f(0)= f_r(0) = \frac{1}{2\pi i}\int_{\{|\zeta|=1\}}\frac{f_r(\zeta)}{\zeta}\,d\zeta.$$

As $r \to 1^-, f(r\zeta) \to f(\zeta)$ pointwise on $\partial D \setminus \{\zeta_0\}$ from the continuity assumption. By the bounded convergence theorem,

$$f(0)= \frac{1}{2\pi i}\int_{\{|\zeta|=1\}}\frac{f(\zeta)}{\zeta}\,d\zeta.$$

Crash through with absolute values to see $|f(0)|\le M.$

Now consider the automorphisms of the disc $g_a(z) = (a-z)/(1-\bar a z), a \in D.$ Each $f \circ g_a$ satisfies the hypotheses above, with the exceptional point this time being $(g_a)^{-1}(\zeta_0).$ It follows that

$$|f(a)|=|f\circ g_a (0)| \le M.$$

Since $a$ is an arbitray point in $D,$ we're done.

zhw.
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