# A question on one-to-one holomorphic functions on the unit disk

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...

• 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 – Conrad Nov 5 '20 at 22:30
• @Conrad, I am interested in your alternative proof. Can you explain it a bit, and maybe post it as an answer please? – 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 $$0 0$$ 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)