{"id":22439,"date":"2026-01-23T11:20:48","date_gmt":"2026-01-23T10:20:48","guid":{"rendered":"https:\/\/djalil.chafai.net\/blog\/?p=22439"},"modified":"2026-01-25T18:44:07","modified_gmt":"2026-01-25T17:44:07","slug":"herglotz-and-nevanlinna-integral-representations","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2026\/01\/23\/herglotz-and-nevanlinna-integral-representations\/","title":{"rendered":"Herglotz and Nevanlinna integral representations"},"content":{"rendered":"<figure id=\"attachment_22440\" aria-describedby=\"caption-attachment-22440\" style=\"width: 206px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/en.wikipedia.org\/wiki\/Gustav_Herglotz\"><img loading=\"lazy\" src=\"http:\/\/djalil.chafai.net\/blog\/wp-content\/uploads\/2026\/01\/Herglotz-206x300.jpeg\" alt=\"Photo of Gustav Herglotz\" width=\"206\" height=\"300\" class=\"size-medium wp-image-22440\" srcset=\"https:\/\/djalil.chafai.net\/blog\/wp-content\/uploads\/2026\/01\/Herglotz-206x300.jpeg 206w, https:\/\/djalil.chafai.net\/blog\/wp-content\/uploads\/2026\/01\/Herglotz.jpeg 274w\" sizes=\"(max-width: 206px) 100vw, 206px\" \/><\/a><figcaption id=\"caption-attachment-22440\" class=\"wp-caption-text\">Gustav Herglotz (1881 - 1953), Bohemian mathematician and physicist, attended the lectures of Ludwig Boltzmann, developed pure and applied mathematics.<\/figcaption><\/figure>\n<p style=\"text-align:justify;\">The set of non-negative harmonic functions is a convex cone, raising the question of integral representation using extreme rays, \u00e0 la Krein-Milman\/Choquet.  This post is devoted to an interplay of this kind between complex analysis and convex geometry. More precisely, a theorem due to Gustav Herglotz (1881 - 1953) on the integral representation of holomorphic functions on the open unit disc with non-negative real part, and a theorem due to Rolf Herman Nevanlinna (1895 - 1980) on the integral representation of holomorphic functions on the open upper half-plane with non-negative imaginary part. In both cases, the set is a convex cone, and the integral representation involves extremal rays given by M\u00f6bius transforms. In what follows, we denote by<\/p>\n<p style=\"text-align:justify;\">\n<ul>\n<li>$\\mathbb{D}=\\{z\\in\\mathbb{C}:|z| < 1\\}$ the open unit disk <\/li>\n<li>$\\mathbb{T}=\\{z\\in\\mathbb{C}:|z|=1\\}$ the unit circle or one-dimensional   torus <\/li>\n<li>$\\mathbb{C}_+=\\{z\\in\\mathbb{C}:\\Im z > 0\\}$ the open upper half plane. <\/li>\n<\/ul>\n<p style=\"text-align:justify;\"><strong>Herglotz theorem.<\/strong> For all $F:\\mathbb{D}\\to\\mathbb{C}$, the following two properties are equivalent:<\/p>\n<p style=\"text-align:justify;\">\n<ul>\n<li>$F$ is holomorphic and $\\Re F\\geq 0$ <\/li>\n<li>$F$ admits an integral representation of the form   \\[     F(z)= \\mathrm{i}\\beta + \\int\\frac{\\zeta+z}{\\zeta-z}\\mathrm{d}\\nu(\\zeta),\\quad z\\in\\mathbb{D},   \\]   where $\\beta\\in\\mathbb{R}$ and $\\nu$ is a finite positive Borel measure on $\\mathbb{T}$   (possibly $=0$). <\/li>\n<\/ul>\n<p style=\"text-align:justify;\">Since $\\nu(\\mathbb{T})=\\Re F(0)$, we have $\\nu=0$ iff $\\Re F(0)=0$.<\/p>\n<p style=\"text-align:justify;\">The theorem gives an integral formula for the non-negative harmonic function $\\Re F$.<\/p>\n<p style=\"text-align:justify;\">Geometrically, the set of holomorphic $F:\\mathbb{D}\\to\\mathbb{C}$ such that $\\Re F\\geq0$ is stable by linear combinations with non-negative coefficients (a convex cone). The Herglotz theorem shows that it contains the one-dimensional linear space $f=\\mathrm{i}\\beta$, $\\beta\\in\\mathbb{R}$. For this reason, it has no extremal points since $f=f-\\mathrm{i}\\beta+\\mathrm{i}\\beta$. However, for $\\beta=0$, the extreme rays are the M\u00f6bius transforms $z\\mapsto C\\frac{c+z}{c-z}$, $c\\in\\mathbb{T}$, $C > 0$, which correspond to $\\nu=C\\delta_c$.<\/p>\n<p style=\"text-align:justify;\"><strong>Proof.<\/strong> $\\Uparrow$. If $F$ admits an integral representation $(\\beta,\\nu)$, then for all $z\\in\\mathbb{D}$, \\[   \\Re F(z)=\\int\\frac{1-|z|^2}{|\\zeta-z|^2}\\mathrm{d}\\nu(\\zeta)\\geq0. \\] Since the function under the integral is $ > 0$, this also shows that if $\\Re F(z)=0$ for some $z\\in\\mathbb{D}$, then $\\nu=0$ and therefore $\\Re F(z)=0$ for all $z\\in\\mathbb{D}$. This observation suggests an alternative \"strict\" version of the theorem where $\\Re F > 0$ and $\\nu\\neq0$.<\/p>\n<p style=\"text-align:justify;\">Next, $|\\zeta|=1$ and $|z| < 1$, hence $|\\zeta-z|\\geq 1-|z|$, and  $\\frac{|\\zeta+z|}{|\\zeta-z|}\\leq\\frac{1+|z|}{1-|z|}=C_z < \\infty$, thus $\\int\\frac{\\zeta+z}{\\zeta-z}\\mathrm{d}\\nu(\\zeta)$ is absolutely convergent, and $F$ is holomorphic by dominated convergence.<\/p>\n<p style=\"text-align:justify;\">$\\Downarrow$. Let $F:\\mathbb{D}\\to\\mathbb{C}$ be holomorphic with $\\Re F\\geq0$. The first step is to extract the boundary measure $\\nu$. Let $u=\\Re F$. Since $u\\geq0$ is continuous, for all $0 < r&lt;1$, \\[   \\mathrm{d}\\nu_r(\\mathrm{e}^{\\mathrm{i}\\theta})   := u(r\\mathrm{e}^{\\mathrm{i}\\theta})\\frac{\\mathrm{d}\\theta}{2\\pi} \\] is a positive Borel measure on $\\mathbb{T}$. By the mean-value property for the harmonic function $u$, \\[ \\nu_r(\\mathbb{T})=\\int_{0}^{2\\pi}u(r\\mathrm{e}^{\\mathrm{i}\\theta})\\frac{\\mathrm{d}\\theta}{2\\pi}=u(0)=\\Re F(0). \\] If $\\Re F(0)=0$ then $\\nu_r=0\\to0$ as $r\\to1$, and we set $\\nu=0$. If $\\Re F(0) > 0$, then since $\\mathbb{T}$ is compact, ${(\\nu_r)}_{0 < r&lt;1}$ is tight, and by the Prohorov theorem, there exists a sequence $r_n\\to1$ in $(0,1)$ and a positive Borel measure $\\nu$ on $\\mathbb{T}$ such that $\\nu(\\mathbb{T})=\\Re F(0)$ and $\\nu_{r_n}\\to\\nu$ narrowly.<\/p>\n<p style=\"text-align:justify;\">Fix $z\\in\\mathbb D$. For $n$ large enough, we have $|z| < r_n$ and the Poisson representation of the harmonic function $u$ on $r_n\\mathbb{D}$ yields \\[   u(z)=\\int P_{\\frac{z}{r_n}}(\\zeta)\\mathrm{d}\\nu_{r_n}(\\zeta)   \\quad\\text{where}\\quad   P_w(\\zeta)=\\frac{1-|w|^2}{|\\zeta-w|^2},\\quad \\zeta\\in\\mathbb{T}. \\] Since $P_w$ is continuous on the compact set $\\mathbb{T}$, it is bounded. Since $P_{z\/r_n}\\to P_z$ uniformly and since $\\nu_{r_n}\\to\\nu$ narrowly, it follows that \\[ u(z)=\\int P_z(\\zeta)\\mathrm{d}\\nu(\\zeta). \\] Now, a natural holomorphic $G:\\mathbb{D}\\to\\mathbb{C}$ such that $\\Re G=u$ is \\[ G(z):=\\int\\frac{\\zeta+z}{\\zeta-z}\\mathrm{d}\\nu(\\zeta). \\] Indeed, as we have already seen, the integrand is bounded on $\\mathbb{T}$, $G$ is well defined, its holomorphicity follows from the one of the integrand and dominated convergence on compact subsets of $\\mathbb{D}$, and $\\Re G=u$. It remains to compare $F$ and $G$. Since $F-G$ is holomorphic on $\\mathbb{D}$ and $\\Re(F-G)=0$ on $\\mathbb{D}$, by the Cauchy-Riemann equations, $\\nabla\\Im(F-G)=0$ on $\\mathbb{D}$, hence $\\Im(F-G)$ is constant, in other words $F-G=\\mathrm{i}\\beta$ for some $\\beta\\in\\mathbb{R}$.<\/p>\n<p style=\"text-align:justify;\"><strong>Uniqueness.<\/strong> We have $\\beta=\\Im F(0)$. Let us show that $\\nu$ is unique by using Fourier coefficients. Namely, suppose that $(\\beta,\\nu_1)$ and $(\\beta,\\nu_2)$ represent $F$. From the proof above, \\[   0=\\Re\\int\\frac{\\zeta+z}{\\zeta-z}\\mathrm{d}(\\nu_1-\\nu_2)(\\zeta)   =\\int P_z(\\zeta)\\mathrm{d}(\\nu_1-\\nu_2)(\\zeta). \\] By writing $z=r\\mathrm{e}^{\\mathrm{i}\\theta}$, $0 < r&lt;1$, $\\zeta=\\mathrm{e}^{\\mathrm{i}t}$, and using $P_z(\\zeta)=P_r(\\mathrm{e}^{-\\mathrm{i}\\theta}\\zeta)=P_r(\\mathrm{e}^{\\mathrm{i}(t-\\theta)})$ and $P_r(\\mathrm{e}^{\\mathrm{i}s})=\\frac{1-r^2}{1-2r\\cos(s)+r^2}=\\sum_{n\\in\\mathbb{Z}}r^{|n|}\\mathrm{e}^{\\mathrm{i}ns}$, we get $\\widehat{\\nu_1-\\nu_2}=0$ for all $n\\in\\mathbb{Z}$, hence $\\nu_1=\\nu_2$.<\/p>\n<p style=\"text-align:justify;\"><strong>Nevanlinna theorem.<\/strong> For all $f:\\mathbb{C}_+\\to\\mathbb{C}$, the following properties are equivalent:<\/p>\n<p style=\"text-align:justify;\">\n<ul>\n<li>$f$ is holomorphic and $\\Im f\\geq0$. <\/li>\n<li>$f$ admits an integral representation of the form   \\[     f(z)=az+b+\\int\\Bigl(\\frac{1}{\\lambda-z}-\\frac{\\lambda}{1+\\lambda^2}\\Bigr)\\mathrm{d}\\mu(\\lambda)     =az+b+\\int\\frac{1+\\lambda z}{\\lambda-z}\\frac{\\mathrm{d}\\mu(\\lambda)}{1+\\lambda^2},     \\quad z\\in\\mathbb{C}_+,   \\]   where $a,b\\in\\mathbb{R}$, $a\\geq0$, and $\\mu$ is a positive Borel   measure on $\\mathbb{R}$ satisfying   \\[     \\int_{\\mathbb{R}}\\frac{\\mathrm{d}\\mu(\\lambda)}{1+\\lambda^2} < \\infty.   \\] <\/li>\n<\/ul>\n<p style=\"text-align:justify;\">The theorem gives an integral formula for the non-negative harmonic function $\\Im f$.<\/p>\n<p style=\"text-align:justify;\">Geometrically, the set of holomorphic $f:\\mathbb{C}_+\\to\\mathbb{C}$ such that $\\Im f\\geq0$ is stable by linear combinations with non-negative coefficients (convex cone). The Nevanlinna theorem shows that this cone contains the line of constant real functions, which corresponds to $b\\in\\mathbb{R}$ in the integral representation. For this reason it has no extremal points since $f=f-b+b$.  However, if we fix $b=0$, the extreme rays are given by $f(z)=az$, $a > 0$, for which $\\mu=0$, and the M\u00f6bius transforms $f(z)=\\frac{C}{1+c^2}\\frac{1+cz}{c-z}$, for which $a=0$ and $\\mu=C\\delta_c$.<\/p>\n<p style=\"text-align:justify;\">The term $\\int\\frac{\\mathrm{d}\\mu(\\lambda)}{\\lambda-z}$ is the Cauchy-Stieltjes transform of $\\mu$ at point $z$. The term $\\int\\frac{\\lambda\\mathrm{d}\\mu(\\lambda)}{1+\\lambda^2}$ is a correction to get a convergent integral, in other words to remove the singularity due to the possible lack of integrability of $1\/\\lambda$ for $\\mu$ at infinity. More precisely, as $\\lambda\\to\\infty$, \\[   \\frac{1}{\\lambda-z}-\\frac{\\lambda}{1+\\lambda^2}   =\\Bigl(\\frac{1}{\\lambda}+\\frac{z}{\\lambda^2}+O\\Bigl(\\frac{1}{\\lambda^3}\\Bigr)\\Bigr)   -\\Bigl(\\frac{1}{\\lambda}-\\frac{1}{\\lambda^3}+O\\Bigl(\\frac{1}{\\lambda^5}\\Bigr)\\Bigr)   =\\frac{z}{\\lambda^2}+O\\Bigl(\\frac{1}{\\lambda^3}\\Bigr). \\] When $\\mu$ is finite : $\\mu(\\mathbb{R}) < \\infty$, then the bound $1\/|\\lambda -z|\\leq1\/|\\Im z|$ implies that the Cauchy-Stieltjes transform of $\\mu$ is finite, and the correction is useless.<\/p>\n<p style=\"text-align:justify;\"><strong>Proof. $\\Uparrow$.<\/strong> Let $f:\\mathbb{C}_+\\to\\mathbb{C}$ with integral representation $(a,b,\\mu)$. Then for all $z\\in\\mathbb{C}_+$, \\[   \\Im f(z)   =a\\Im z+\\Im z\\int\\frac{\\mathrm{d}\\mu(\\lambda)}{|\\lambda -z|^2}\\geq0. \\] Moreover, for all $z\\in\\mathbb{C}_+$, the function $t\\in\\mathbb{R}\\mapsto\\bigl|\\frac{1+t z}{(t-z)}\\bigr|$ is continuous and tends to $|z|$ as $|t|\\to\\infty$, hence it is bounded by some constant $C_z$, and the integral with respect to $\\mathrm{d}\\mu(t)$ converges absolutely as soon as $1\/(1+t^2)\\in L^1(\\mu)$. The holomorphicity with respect to $z$ of this integral follows from dominated convergence on compact subsets of $\\mathbb C_+$.<\/p>\n<p style=\"text-align:justify;\"><strong>$\\Downarrow$.<\/strong> Let $f:\\mathbb{C}_+\\to\\mathbb{C}$ be holomorphic with $\\Im f\\geq0$. We proceed by reducing to the disk case (Herglotz theorem). We introduce the Cayley transform $\\phi:\\mathbb D\\to\\mathbb C_+$ and its inverse \\[ \\phi(w)= \\mathrm{i}\\frac{1+w}{1-w},\\quad  \\phi^{-1}(z)=\\frac{z-\\mathrm{i}}{z+\\mathrm{i}}. \\] The function $F:= -\\mathrm{i}(f\\circ\\phi)$ is holomorphic on $\\mathbb{D}$ and $\\Re F=\\Im(f\\circ\\phi)\\geq 0$. By the Herglotz theorem, there exist $\\beta\\in\\mathbb{R}$ and a finite positive Borel measure $\\nu$ on $\\mathbb{T}$ (possibly $\\nu=0$) such that \\[ F(w)=\\mathrm{i}\\beta +\\int\\frac{\\zeta+w}{\\zeta-w}\\mathrm{d}\\nu(\\zeta),\\quad w\\in\\mathbb{D}. \\] By composing with $\\phi^{-1}$ we get, for all $z\\in\\mathbb{C}_+$, \\[   f(z)=\\mathrm{i}F(\\phi^{-1}(z))=-\\beta +\\mathrm{i}\\int\\frac{\\zeta+w}{\\zeta-w}\\mathrm{d}\\nu(\\zeta),   \\quad w=\\frac{z-i}{z+i}. \\] At this step, we decompose $\\nu$ into its atom at $\\zeta=1$ and the remainder, namely \\[   \\nu = a\\delta_1+\\nu_0,\\quad a:=\\nu(\\{1\\})\\geq 0,\\quad \\nu_0(\\{1\\})=0. \\] Since $z=\\mathrm{i}\\frac{1+w}{1-w}$, we get $\\frac{1+w}{1-w} = -\\mathrm{i}z$, hence $\\mathrm{i}a\\frac{1+w}{1-w}=az$, and therefore \\[   f(z)   = a z -\\beta + \\mathrm{i}\\int\\frac{\\zeta+w}{\\zeta-w}\\mathrm{d}\\nu_0(\\zeta). \\] Let $\\widetilde{\\nu}$ be the image of $\\nu_0$ by $\\zeta\\in\\mathbb{T}\\setminus\\{1\\}\\mapsto t=\\mathrm{i}\\frac{1+\\zeta}{1-\\zeta}\\in\\mathbb{R}$. We have $\\mathrm{i}\\frac{\\zeta+w}{\\zeta-w} = \\frac{1+tz}{t-z}$, and thus \\[   f(z)= a z -\\beta +   \\int\\frac{1+tz}{t-z}\\mathrm{d}\\widetilde{\\nu}(t). \\] Note that  $\\widetilde{\\nu}(\\mathbb{R})=\\nu_0(\\mathbb{T})=\\nu(\\mathbb{T})-a < \\infty$. It remains to take $\\mathrm{d}\\mu(t):=(1+t^2)\\mathrm{d}\\widetilde\\nu(t)$ since \\[   \\frac{1+t z}{t-z}=(1+t^2)\\Bigl(\\frac{1}{t-z}-\\frac{t}{1+t^2}\\Bigr). \\]<\/p>\n<p style=\"text-align:justify;\"><strong>Uniqueness.<\/strong> Moreover the triple $(a,b,\\mu)$ is unique since  \\[   a=\\lim_{y\\to\\infty}\\frac{f(\\mathrm{i}y)}{\\mathrm{i}y},\\quad    b=\\Re f(\\mathrm{i}), \\] and we can recover $\\mu$ from the behavior of $\\Im f$ near $\\mathbb{R}$ by Poisson-Stieltjes inversion on \\[ \\Im f(x+\\mathrm{i}y)=ay+\\int\\frac{y}{(t-x)^2+y^2}\\mathrm{d}\\mu(t). \\] Alternatively, we could use the uniqueness of $\\nu$ in the Herglotz theorem.<\/p>\n<p style=\"text-align:justify;\"><strong>Note.<\/strong> Herglotz\/Nevanlinna functions were actually also studied by Thomas-Joannes Stieltjes (1856 - 1894), Georg Alexander Pick (1859 - 1942), Constantin Carath\u00e9odory (1873 - 1950), and many others. There are important ramifications and contemporary resonances in operator algebras, free probability theory, and random matrix theory.<\/p>\n<p style=\"text-align:justify;\"><strong>Further reading.<\/strong><\/p>\n<p style=\"text-align:justify;\">\n<ul>\n<li>Jim Agler, John E. McCarthy, Nicholas J. Young<br \/>   <strong>Operator Analysis: Hilbert Space Methods in Complex Analysis<\/strong><br \/>   Cambridge University Press (2020) <\/li>\n<\/ul>\n<figure id=\"attachment_22441\" aria-describedby=\"caption-attachment-22441\" style=\"width: 250px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/en.wikipedia.org\/wiki\/Rolf_Nevanlinna\"><img loading=\"lazy\" src=\"http:\/\/djalil.chafai.net\/blog\/wp-content\/uploads\/2026\/01\/Nevanlinna.jpg\" alt=\"Photo of Rolf Nevanlinna\" width=\"250\" height=\"300\" class=\"size-full wp-image-22441\" \/><\/a><figcaption id=\"caption-attachment-22441\" class=\"wp-caption-text\">Rolf Herman Nevanlinna (1895 - 1980) Finnish mathematician, known for his work in complex analysis, and also for the development of the teaching of computer science. His former student Lars Valerian Ahlfors (1907 - 1996) was one of the first two mathematicians to receive the Fields Medal in 1936.<\/figcaption><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The set of non-negative harmonic functions is a convex cone, raising the question of integral representation using extreme rays, &agrave; la Krein-Milman\/Choquet. This post is&#8230;<\/p>\n<div class=\"more-link-wrapper\"><a class=\"more-link\" href=\"https:\/\/djalil.chafai.net\/blog\/2026\/01\/23\/herglotz-and-nevanlinna-integral-representations\/\">Continue reading<span class=\"screen-reader-text\">Herglotz and Nevanlinna integral representations<\/span><\/a><\/div>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":171},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/22439"}],"collection":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/comments?post=22439"}],"version-history":[{"count":6,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/22439\/revisions"}],"predecessor-version":[{"id":22452,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/posts\/22439\/revisions\/22452"}],"wp:attachment":[{"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/media?parent=22439"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/categories?post=22439"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/djalil.chafai.net\/blog\/wp-json\/wp\/v2\/tags?post=22439"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}