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If we define = {exp(-f):f∈}, then |exp(-f)|<1, so is uniformly bounded by 1. Hence Montel's theorem implies that is normal.

To show is normal, let {fn} ⊆ , and let K be a compact subset of the open unit disc.

Consider the sequence {gn}, defined by gn(z)=exp(-fn(z)). So we can extract a subsequence from {gn} that converges uniformly on K, to a holomorphic function, say g.

By assumption, we have g(0)=1/e, so Hurwitz's theorem says g is never zero.

But I'm stuck here, how to use these facts to show that is uniformly bounded on K, or directly show that converges uniformly on K? Could someone help?

eric
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1 Answers1

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Since the open unit disc is a simply connected domain, and each $g_k=e^{-f_k}$ and $g$ are never $0$, you can choose a branch to define $\log g_k(z)$. More precisely, you can pick $z_0=0$ so that $$ \log g_k(z)=\int_{\Gamma}\frac{g'_k(z)}{g_k(z)}\,dz+\log g(z_0) $$ Here $\Gamma$ is a path in the open unit disc that started from $z_0$ and ended at $w$.

Notice then $\log g(z_0)=\log e^{-1}=-1$ is constant for each $k$ and $$ \int_{\Gamma}\frac{g'_k(z)}{g_k(z)}\,dz=\int_{\Gamma} -f'_k(z)\,dz=-f_k(w)+f(z_0)=-f_k(w)+1 $$ We obtain that \begin{aligned} &\int_{\Gamma}\frac{g'_k(z)}{g_k(z)}\,dz-1\longrightarrow\int_{\Gamma}\frac{g'(z)}{g(z)}\,dz-1\\ &\Rightarrow f_k(w)\longrightarrow f(w):=\int_{\Gamma}\frac{g'(z)}{g(z)}\,dz \end{aligned} converges u.c.c. on the open unit disc.

Edit: If you have read the reference post in comment, you may find that they often discuss the case $f_k\to\infty$. However, the main difference is that in your caes, you have an additional assumption $f_n(0)=1$ for all $f_n\in \mathcal F.$ This allows you to avoid the case $f_k\to\infty$ because you can put $z_0=0$ so that $\log g_k(z_0)$ won't change as $k\to\infty$.

León
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