Univalent function

A univalent function is a holomorphic injective function.

For an injective holomorphic function $f(z)={\sum }_{n\ge 0}{c}_n{z}^n$ on the unit disc, its real Jacobian determinant is $|f^{\prime }(z){|}^2$. Change of variables gives image area ${\int }_{|z|<r}|f^{\prime }(z){|}^{2}dA$. Angular orthogonality removes cross terms of the differentiated power series; integrating $2\pi n^2|c_n{|}^2{\rho }^{2n-1}$ radially gives the title formula. For $r<1$, uniform convergence on a slightly larger closed disk justifies termwise operations. Injectivity is essential for this to be image area rather than multiplicity of a parametrized surface integral.

For a univalent exterior map $G(\zeta )=\zeta +b_0+{\sum }_{n\ge 1}{b}_n{\zeta }^{-n}$, the image of $|\zeta |=R>1$ is a positively oriented Jordan curve. Green's theorem gives its enclosed area as $\pi (R^2-{\sum }_{n\ge 1}n|b_n{|}^2{R}^{-2n})$. Nonnegativity and $R\downarrow 1$ give the displayed bound. Laurent-series differentiation and integration on a circle of radius greater than one are justified by uniform convergence there.

A normalized univalent function is a holomorphic function on the unit disc which is injective and has the displayed normalization. Its Taylor series begins $z+a_2{z}^2+\cdots$. Normalization makes coefficient bounds and image-size estimates independent of translation and dilation; the Koebe quarter theorem is the sharp universal image bound.

For $f(z)=z+{\sum }_{n\ge 2}{a}_n{z}^n$ in the normalized univalent function class, the integral mean bound for an odd square-root transform gives $(2\pi {)}^{-1}\int |f(r{e}^{i\theta })|d\theta \le r/(1-r)$. The Cauchy integral formula for Taylor coefficients therefore yields $|a_n|\le r^{1-n}/(1-r)$. Choosing $r=(n-1)/n$ and applying the logarithm tangent bound gives the displayed bound for every $n\ge 2$.

The odd square-root transform of a normalized univalent function has reciprocal exterior expansion $1/h(1/\zeta )=\zeta -(a_2/2){\zeta }^{-1}+\cdots$. Its first negative Laurent coefficient has modulus at most one by the area theorem for univalent functions, proving the bound. If $w$ is omitted by $f$, the normalized map $wf/(w-f)$ has second coefficient $a_2+1/w$. Applying the bound twice gives $|1/w|\le 4$, the Koebe quarter theorem.

Because $f(z)/z$ is holomorphic and nowhere zero on the unit disc, its square root with value one at zero exists. Defining $h(z)=z\sqrt{f(z^2)/z^2}$ gives an odd normalized univalent function. If $h(z)=h(w)$, squaring gives $z^2=w^2$; the possibility $z=-w$ and oddness force both to be zero. Applying the area theorem for univalent functions to $1/h(1/\zeta )$ yields $|a_2|\le 2$.

Suppose $h(z{)}^2=f(z^2)$ is the odd square-root transform of a normalized univalent function. The Koebe growth theorem gives $|h(z)|\le |z|/(1-|z{|}^2)$, so its radius-$r$ image has area at most $\pi r^2/(1-r^2{)}^2$. Write $h=\sum c_n{z}^n$ and $M(r)=\sum |c_n{|}^2{r}^{2n}$, the mean-square boundary integral by the Parseval identity. The area formula for a univalent power series gives $M^{\prime }(r)=2\mathrm{Area}(h(r\mathbb{D}))/(\pi r)\le 2r/(1-r^2{)}^2$. Integrating from zero proves the bound. The relation $|h(\sqrt{r}{e}^{i\theta }){|}^2=|f(r{e}^{2i\theta })|$ then gives the corresponding first absolute mean bound $r/(1-r)$ for $f$.

For an injective holomorphic function $f$ on an open subset of $\mathbb{C}$, $f^{\prime }(z)$ is nowhere zero. If $f(z)-f(z_0)=(z-z_0{)}^{m}h(z)$ with $h(z_0)\ne 0$, choose a small circle where both $h$ and $mh+(z-z_0)h^{\prime }$ have no zeros. The argument principle counts $m$ zeros of $f(z)-f(z_0)-w$ inside that circle for sufficiently small $w\ne 0$: deforming $w$ to zero cannot change the count. All these zeros are simple, since $f^{\prime }$ only vanishes at $z_0$ inside the circle and $z_0$ is not a zero when $w\ne 0$. Injectivity forces $m=1$, and hence $f^{\prime }(z_0)\ne 0$.

A locally uniform convergence limit of univalent functions on a domain is either constant or univalent. For a nonconstant limit $f$, suppose distinct $x,y$ satisfy $f(x)=f(y)$. Choose a small closed disk around $x$ excluding $y$ with $f-f(y)$ nonzero on its boundary. The Rouche theorem then forces $f_j-f_j(y)$ to have a zero in that disk for large $j$, contradicting injectivity of $f_j$. This supplies the injectivity step in an extremal proof of the Riemann mapping theorem without assuming it as a special lemma.

The Koebe function $k(z)=z/(1-z{)}^2$ is a univalent function from the unit disc onto $\mathbb{C}\setminus (-\infty ,-1/4]$. To see the image, $(1+z)/(1-z)$ maps the disc onto the right half-plane and $k(z)=(((1+z)/(1-z){)}^2-1)/4$. Its normalization $k(0)=0$, $k^{\prime }(0)=1$ proves sharpness of the Koebe quarter theorem.

If $\varphi$ maps the unit disc conformally to a proper simply connected domain, the nonvanishing holomorphic quotient $F(w)=(\varphi (w)-\varphi (0))/w$ has harmonic log modulus. Its radial mean equals $\log |{\varphi }^{\prime }(0)|$. Boundary values are understood as radial limits: the standard integral-mean bound ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, together with the Koebe distortion theorem lower bound $|F(w)|\ge |{\varphi }^{\prime }(0)|/4$, gives uniform integrability of these logarithms. Hence