I am trying to prove a very simple theorem that uses the general idea in complex analysis that if $f:\mathbb{D}\to\mathbb{C}$ is holomorphic, then the quantity $|f'(0)|$ is somehow responsible for how 'wide' the image $f(\mathbb{D})$ is.
In particular, I've got this:
Claim: Let $f,\,g:\mathbb{D}\to\mathbb{C}$ be injective holomorphic functions such that $g(0)=0=f(0)$ and $g(\mathbb{D})$ is a strict subset of $f(\mathbb{D})$. Then $|g'(0)| < |f'(0)|$.
I have written the following proof:
Since $g(\mathbb{D})\subset f(\mathbb{D})$ and both images contain the origin, it follows by Cauchy's integral formula that \begin{align*} |g'(0)| &= \frac{1}{2\pi}\left|\int_{\partial \mathbb{D}} \frac{g(z)}{z^2}\,\text{d}z \right|\\ &< \frac{1}{2\pi}\left|\int_{\partial \mathbb{D}} \frac{f(z)}{z^2}\,\text{d}z \right|\\ &= |f'(0)|. \end{align*}
I am trying to use the fact that since $f$ and $g$ are conformal (injective and holomorphic), the curve $f(\partial \mathbb{D})$ is going to enclose $g(\partial \mathbb{D})$ around the origin. I think that this is correct, and I wouldn't have any serious doubts if it weren't for the fact that Duren (in Univalent functions) calls this a corollary of Schwarz's lemma, which I haven't used.
I think that I haven't had to use Schwarz's lemma because I've simplified the statement quite a bit to suit the application I need it for (Duren doesn't say that $f$ and $g$ both fix the origin, and only requires $f$ to be conformal).
But it would set my mind at ease if it could be confirmed that I haven't overlooked something.
Hint: Consider the holomorphic function $h=f^{-1}\circ g.$ This is well defined since $g(\mathbb D)\subset f(\mathbb D).$ We have $h:\mathbb D\to \mathbb D$ and $h(0)=0.$ That should ring some bells.