Harnack theorem


It is frequent to make use of Cauchy integral formulaPlanetmathPlanetmath to represent analytically some functionsMathworldPlanetmath that are useful in mathematical physics applications. However, it must be noted that such representation is not unique, so that the same function can be represented by different integralsDlmfPlanetmath of Cauchy’s type. An important case has to do with the equality of the two Cauchy integrals


for all values of z in the interior of C. In general no conclusion can be drawn concerning the equality of the density functions ψ1(ζ) and ψ2(ζ). We shall see, however, that if some additional restriction are imposed on the density functions and on the contour C, then the equality will occur. That is the matter of Harnack theorem. In considering the applications of the theory of functions of a complex variable to problems in continuum mechanics, for instance, we shall most frequently deal with the region bounded by the unit circle, i.e. the compact disc |z|1 that we shall draw in the z-plane, its boundary will be denoted by γ and the points of γ by ζ=eiθ. All density functions of the argumentMathworldPlanetmath θ will be assumed to be 2π-periodic.

Theorem 1.
11A less restrictive form of Harnack theorem is discussed in [1].

Let f(θ) and g(θ) be continuousMathworldPlanetmath real functions of the argument θ defined on the boundary γ; if

12πiγf(θ)dζζ-z=12πiγg(θ)dζζ-z (1)


f(θ)g(θ)  if  |z|<1
f(θ)=g(θ)+const.  if  |z|>1.
  1. 1.

    |z|<1. From equality (1) we obtain


    where h(θ)f(θ)-g(θ). We shall prove that h(θ)0. Since |z|<1 we can write,



    12πiγh(θ)ζ-z𝑑ζ=12πiγn=0h(θ)ζn+1zndζ=12πin=0(an-ibn)zn, (2)

    the complex form of Fourier series 22The restrictions imposed upon the expanded function are known as the Dirichlet conditions, but it is sufficient to demand that it be a function of bounded variation.where the coefficients are given by (we are using the Euler’s formula)


    But (2) vanishes for all values of z, therefore an=bn=0,n and a reference to Fourier complex expansion


    shows that all Fourier coefficients of the function h(θ) vanish, and hence h(θ)0.

  2. 2.

    |z|>1. By analytic continuation, we have


    so that

    12πiγh(θ)ζ-z𝑑ζ=-12πiγn=1ζn-1h(θ)zndζ=-n=1an+ibnzn, (3)



    Since (3) vanishes for all values of |z|>1, an=bn=0,n\{0}. Thus, all Fourier coefficients of h(θ), with the possible exception of a0, vanish and hence


Moreover, from this theorem if |z|>1 and in addition to (1) we have the equality


then f(θ)=g(θ).

Corollary 1.

Given the continuous real functions f1,f2,g1,g2 and the following simultaneous equalities for all values of z.






By adding and substracting those equalities, this corollary follows from Harnack theorem.


  • 1 N. I. Muskhelishvili’s, Singular Integral Equations, p.64, 1953.
  • 2 E.C. Titchmarsh, The Theory of Functions, Oxford University Press, New York, 2d ed., pp. 64-101, 399-428.
  • 3 W.F. Osgood, Lehrbuch der Funktionentheorie, Teubner Verlagsgesellschaft, Leipzig, vol. 1.
  • 4 É. Goursat, Course d’analyse, Gauthiers-Villars & Cie, Paris, vol. 2.
  • 5 E. Picard, Leçons sur quelques types simples d’équations aux dérivées partielles, Gauthiers-Villars & Cie, Paris.
Title Harnack theorem
Canonical name HarnackTheorem
Date of creation 2013-03-22 16:02:55
Last modified on 2013-03-22 16:02:55
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 8
Author perucho (2192)
Entry type Theorem
Classification msc 30D10