Let $D = \{ z \in \mathbb{C}; |z| < 1 \}$ be the open unit disk, and let $f$ be a one-to-one holomorphic function such that $f(0) = 0$ and $\bar{D} \subseteq f(D)$, where $\bar{D}$ is the closure of $D$, i.e. the closed unit disk. This may turn out to be a trivial question but, is it necessarily the case that $|f'(0)| >= 1$? I am wondering if some kind of lower bound version of Schwarz's lemma may hold. There may be an easy counterexample, but I just don't know many examples of such functions...

Edit: can we just apply the usual Schwarz lemma to $f^{-1}$? I will have to think a bit...

  • $\begingroup$ as noted applying Schwarz to the inverse of $f$ gives the result; if you want a fancier proof, note that the hypothesis implies that the identity is subordinate to $f$ hence the first Taylor coefficients increase in absolute value from $z \to z$ to $z \to f(z)$ so $1 \le |a_1(f)|=|f'(0)|$ and we actually have strict ineqwaulity since $f(z) \ne \alpha z, |\alpha|=1$ from the hypothesis $\endgroup$ – Conrad Nov 5 '20 at 22:30
  • $\begingroup$ @Conrad, I am interested in your alternative proof. Can you explain it a bit, and maybe post it as an answer please? $\endgroup$ – Malkoun Nov 5 '20 at 22:32

If $f,g : 0 \in U \to \mathbb C, f(0)=g(0), g$ univalent, where $U$ is some region (open connected), we call $f$ subordinate to $g$ if $f(U) \subset g(U)$ ($0$ here is a convenience, any base point will do).

If in addition $U =\mathbb D$ the unit disc, $\omega =g^{-1} \circ f$ gives a map that satisfies the conditions of Schwarz Lemma, for which $f=g \circ \omega$ which is sometimes taken as the definition of subordination. This immediately implies that $|a_1(f)|=|f'(0)| \le |g'(0)| =|a_1(g)|$.

Less obvious results are the fact that $\int_0^{2\pi}|f(re^{i\theta})|^pd\theta \le \int_0^{2\pi}|g(re^{i\theta})|^pd\theta$ for all $0 <r<1, p>0$ and $\sum_{k=1}^n|a_k|^2 \le \sum_{k=1}^n|b_k|^2$ for all $n \ge 1$ where $f(z)=f(0)+\sum_{k \ge 1}a_kz^k, g(z)=f(0)+\sum_{k \ge 1}b_kz^k$ and in all cases equality happens only when $f$ is a rotation of $g$ so $f(z)=\alpha g(\beta z), |\alpha|=|\beta|=1$

In our case the fact that $\mathbb D \subset f(\mathbb D), f(0)=0$ implies that $z$ is subordinate to $f$ hence indeed $1 \le |f'(0)|$. Since by hypothesis we assume that the inclusion is strict we actually have strict inequality too.

(note also that for any $n \ge 2$, $z^n$ is subordinate to $z$, so in general, we cannot have simple inequalities of the type $|a_n| \le |b_n|$ for $n \ge 2$ and the ones above are close to the best we can do)


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