Schwarz lemma with the equality cases

If $f\equiv 0$ then the two inequalities and the equality characterization are immediate, so assume $f$ is not identically zero and define $g(z)=f(z)\mathrm{/}z$ for $z\mathrm{\ne }0$ on the punctured disc.

Since $f^{\mathrm{\prime }}(0)={\lim }_{z\to 0}f(z)\mathrm{/}z$, the function $g$ has finite limit $f^{\mathrm{\prime }}(0)$ at $0$; by [F2] it extends holomorphically to $\mathbb{D}$, still denoted $g$, with $g(0)=f^{\mathrm{\prime }}(0)$.

Fix $0<r<1$. On $\mathrm{\mid }z\mathrm{\mid }=r$ one has $\mathrm{\mid }g(z)\mathrm{\mid }=\mathrm{\mid }f(z)\mathrm{\mid }\mathrm{/}r\le 1\mathrm{/}r$ because $f(\mathbb{D})\subseteq \mathbb{D}$; applying [F3] on the radius-$r$ disc gives $\mathrm{\mid }g(z)\mathrm{\mid }\le 1\mathrm{/}r$ whenever $\mathrm{\mid }z\mathrm{\mid }\le r$.

For any $w\in \mathbb{D}$, step 3.1 holds for every $r$ with $\mathrm{\mid }w\mathrm{\mid }<r<1$, so letting $r\to 1^{-}$ gives $\mathrm{\mid }g(w)\mathrm{\mid }\le 1$; therefore $\mathrm{\mid }f(w)\mathrm{\mid }=\mathrm{\mid }w\mathrm{\mid }\mathrm{\mid }g(w)\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$ for all $w\in \mathbb{D}$, and at $w=0$ this also gives $\mathrm{\mid }{f}^{\mathrm{\prime }}(0)\mathrm{\mid }=\mathrm{\mid }g(0)\mathrm{\mid }\le 1$.

If $\mathrm{\mid }f(z_0)\mathrm{\mid }=\mathrm{\mid }{z}_0\mathrm{\mid }$ for some $z_0\mathrm{\ne }0$, then step 4.1 gives $\mathrm{\mid }g(z_0)\mathrm{\mid }=1$, an interior maximum for $\mathrm{\mid }g\mathrm{\mid }$, so [F4] makes $g$ constant of modulus $1$; if instead $\mathrm{\mid }{f}^{\mathrm{\prime }}(0)\mathrm{\mid }=1$, then $\mathrm{\mid }g(0)\mathrm{\mid }=1$ and the same argument applies. Thus in either equality case $g(z)\equiv e^{i\theta }$ for some real $\theta$, so $f(z)=e^{i\theta }z$ on $\mathbb{D}$.

Conversely, for $f(z)=e^{i\theta }z$ one has $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ for all $z$ and $\mathrm{\mid }{f}^{\mathrm{\prime }}(0)\mathrm{\mid }=\mathrm{\mid }{e}^{i\theta }\mathrm{\mid }=1$.