As Grunsky showed, these inequalities hold if and only if the holomorphic function is univalent.
The Grunsky matrix and its associated inequalities were originally formulated in a more general setting of univalent functions between a region bounded by finitely many sufficiently smooth Jordan curves and its complement: the results of Grunsky, Goluzin and Milin generalize to that case.
Historically the inequalities for the disk were used in proving special cases of the Bieberbach conjecture up to the sixth coefficient; the exponentiated inequalities of Milin were used by de Branges in the final solution.
The Grunsky operators and their Fredholm determinants are also related to spectral properties of bounded domains in the complex plane.
The Grunsky coefficients of a normalized univalent function in |z| > 1 are polynomials in the coefficients bi which can be computed recursively in terms of the Faber polynomials Φn, a monic polynomial of degree n depending on g. Taking the derivative in z of the defining relation of the Grunsky coefficients and multiplying by z gives The Faber polynomials are defined by the relation Dividing this relation by z and integrating between z and ∞ gives This gives the recurrence relations for n > 0 with Thus so that for n ≥ 1 The latter property uniquely determines the Faber polynomial of g. Let g(z) be a univalent function on |z| > 1 normalized so that and let f(z) be a non-constant holomorphic function on C. If is the Laurent expansion on z > 1, then If Ω is a bounded open region with smooth boundary ∂Ω and h is a differentiable function on Ω extending to a continuous function on the closure, then, by Stokes' theorem applied to the differential 1-form
If then Applying Milin's area theorem, (Equality holds here if and only if the complement of the image of g has Lebesgue measure zero.)
So a fortiori Hence the symmetric matrix regarded as an operator on CN with its standard inner product, satisfies So by the Cauchy–Schwarz inequality With this gives the Grunsky inequality: Let g(z) be a holomorphic function on z > 1 with Then g is univalent if and only if the Grunsky coefficients of g satisfy the Grunsky inequalities for all N. In fact the conditions have already been shown to be necessary.
To see sufficiency, note that makes sense when |z| and |ζ| are large and hence the coefficients cmn are defined.
If the Grunsky inequalities are satisfied then it is easy to see that the |cmn| are uniformly bounded and hence the expansion on the left hand side converges for |z| > 1 and |ζ| > 1.
In 1972 the American mathematician James Hummel extended the Grunsky inequalities to this matrix, proving that for any sequence of complex numbers λ±1, ..., λ±N The proof proceeds by computing the area of the image of the complement of the images of |z| < r < 1 under F and |ζ| > R > 1 under g under a suitable Laurent polynomial h(w).
Letting r increase to 1 and R decrease to 1, it follows that with equality if and only if the complement of the images has Lebesgue measure zero.
So, roughly speaking, in the case of one function the image is a slit region in the complex plane; and in the case of two functions the two regions are separated by a closed Jordan curve.
If g(z) is a normalized univalent function in |z| > 1, z1, ..., zN are distinct points with |zn| > 1 and α1, ..., αN are complex numbers, the Goluzin inequalities, proved in 1947 by the Russian mathematician Gennadi Mikhailovich Goluzin (1906-1953), state that To deduce them from the Grunsky inequalities, let for k > 0.
Bergman & Schiffer (1951) gave another derivation of the Grunsky inequalities using reproducing kernels and singular integral operators in geometric function theory; a more recent related approach can be found in Baranov & Hedenmalm (2008).
They were also used by Schiffer and Charzynski in 1960 to give a completely elementary proof of the Bieberbach conjecture for the fourth coefficient; a far more complicated proof had previously been found by Schiffer and Garabedian in 1955.
In 1968 Pedersen and Ozawa independently used the Grunsky inequalities to prove the conjecture for the sixth coefficient.
Combining Gronwall's area theorem for f with the Grunsky inequalities for the first 2 x 2 minor of the Grunsky matrix of g leads to a bound for |a4| in terms of a simple function of a2 and a free complex parameter.
and hence, reversing the steps, In particular defining bn(w) by the identity the following inequality must hold for |w| < 1 The Beurling transform (also called the Beurling-Ahlfors transform and the Hilbert transform in the complex plane) provides one of the most direct methods of proving the Grunsky inequalities, following Bergman & Schiffer (1951) and Baranov & Hedenmalm (2008).
If f is a holomorphic univalent map from the unit disk D onto Ω then the Bergman space of Ω and its conjugate can be identified with that of D and TΩ becomes the singular integral operator with kernel It defines a contraction.
Its realization on D using a univalent function f mapping D onto Ω and the fact that TD = 0 shows that it is given by restriction of the kernel and is therefore a Hilbert–Schmidt operator.
By the spectral theorem for compact self-adjoint operators, there is an orthonormal basis un of H consisting of eigenvectors of A: where μn is non-negative by the positivity of A.
by antilinearity of T.) The non-zero λn (or sometimes their reciprocals) are called the Fredholm eigenvalues of Ω: If Ω is a bounded domain that is not a disk, Ahlfors showed that The Fredholm determinant for the domain Ω is defined by[6][7] Note that this makes sense because A = T2 is a trace class operator.
A similar formula applies in the case of a pair of univalent functions (see below).
Let Ω be a bounded simply connected domain in C with smooth boundary C = ∂Ω.