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Frigyes Riesz

Riesz Frigyes · Frédéric Riesz · Győr 1880 – Budapest 1956

A course on the Hungarian schoolteacher turned professor who gave functional analysis its basic theorems: the completeness of \(L^2\), the representation of linear functionals, the \(L^p\) spaces and the theory of compact operators.

Born 22 January 1880, Győr Died 28 February 1956, Budapest Founded Bolyai Institute, Szeged, 1922 Level advanced undergraduate → graduate
Riesz’s rising sun lemma (1932). Light comes horizontally from the right. The points of the axis whose graph point lies in shadow form disjoint open intervals, and at the two ends of each interior interval the function takes the same value (dashed level lines).
Lesson 00

How to use this course

Open any textbook of functional analysis and Frigyes Riesz is on most pages. The Riesz–Fischer theorem makes \(L^2\) a Hilbert space. The Riesz representation theorems identify the dual of a Hilbert space and the dual of \(C[a,b]\). The \(L^p\) spaces and their duality are his, from 1910. Riesz’s lemma and the Riesz theory of compact operators, from 1918, are the first chapter of operator theory in Banach spaces. In complex analysis there are the F. and M. Riesz theorem, written with his younger brother Marcel, and the theory of subharmonic functions; in real analysis the rising sun lemma; in ergodic theory a short proof of the mean ergodic theorem; in order theory the Riesz spaces.

Riesz did much of his early work as a secondary-school teacher in a small town in what is now Slovakia. He then built, with Alfréd Haar, one of the best mathematical centres of interwar Europe in Szeged. His style was to find the simplest proof, and many of his arguments are still the ones taught today.

Prerequisites for Part II
  • Lebesgue measure and integration on an interval, including Fatou’s lemma and dominated convergence.
  • Normed spaces, inner products, and the idea of a bounded linear map.
  • For Lesson 08, basic complex analysis: harmonic functions, Poisson integral, Fourier series.
  • For Lesson 10, a first course in point-set topology helps.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, proofs or proof sketches, and exercises with folded solutions. Riesz worked with concrete spaces of functions on an interval; the statements here are given in the generality that his proofs actually support, with remarks where later authors extended them.

Reference

Timeline

1880Born 22 January in Győr, Hungary, eldest child of the physician Ignácz Riesz and Szidónia Nagel.
1886Birth of his brother Marcel Riesz, later a mathematician in Sweden.
1897Leaves school in Győr. Studies at the Polytechnic (ETH) in Zurich.
1899Moves to the University of Budapest.
1902Doctorate, with Gyula Vályi as advisor, on point configurations on a space quartic curve.
1903–04Winter semester at Göttingen, where he hears David Hilbert.
1904Teacher at the state school in Lőcse (now Levoča, Slovakia).
1907The Riesz–Fischer theorem; the representation of functionals on \(L^2\), at the same time as Maurice Fréchet.
1908Lecture on abstract sets and continuity at the International Congress in Rome. Moves to a Budapest gymnasium.
1909Representation of continuous linear functionals on \(C[a,b]\) by Stieltjes integrals.
1910The \(L^p\) spaces, their duality and weak convergence, in Mathematische Annalen.
1912Extraordinary professor at Kolozsvár (now Cluj-Napoca); full professor in 1914.
1913Les systèmes d’équations linéaires à une infinité d’inconnues.
1915Corresponding member of the Hungarian Academy of Sciences.
1916F. and M. Riesz theorem, with Marcel, at the Scandinavian Congress in Stockholm.
1918Über lineare Funktionalgleichungen, Acta Mathematica 41: compact operators.
1920The Hungarian university of Kolozsvár relocates to Szeged.
1922Founds the Bolyai Institute and the journal Acta Scientiarum Mathematicarum with Alfréd Haar.
1926First of two Acta Mathematica memoirs on subharmonic functions (the second in 1930).
1928Lecture at the Bologna International Congress on the decomposition of linear functionals.
1932Rising sun lemma.
1936Full member of the Hungarian Academy of Sciences.
1938Elementary proof of the mean ergodic theorem.
1944–45Rector of the University of Szeged.
1945Chair of mathematics at the University of Budapest.
1948Corresponding member of the Paris Academy of Sciences.
1949Kossuth Prize; again in 1953.
1952Leçons d’analyse fonctionnelle, with Béla Szőkefalvi-Nagy.
1956Dies 28 February in Budapest, aged 76.
Part I · Formation
Lesson 01

Győr, Zurich, Budapest, 1880–1902

Frigyes Riesz was born on 22 January 1880 in Győr, a market town in western Hungary on the road between Vienna and Budapest. His father, Ignácz Riesz, was a physician; his mother was Szidónia Nagel. The family was Jewish. Frigyes was the eldest of six children. His brother Marcel, born in 1886, also became a leading analyst. Their sister Margit was murdered at Auschwitz in 1944 with her husband and daughter.

Riesz attended the secondary school in Győr from 1889 and finished in 1897. He then went to the Polytechnic in Zurich. In 1899 he moved to the University of Budapest, where he took courses from Gyula Kőnig, József Kürschák and Manó Beke and attended the lectures of the physicist Loránd Eötvös.

His doctoral thesis of 1902, written in Hungarian with Gyula Vályi as advisor, was in the synthetic geometry of position: a treatment of configurations of points on a space curve of the fourth order. Nothing in it hints at analysis. In 1903 he qualified as a secondary teacher of mathematics and physics.

Lesson 02

A schoolteacher’s revolution, 1903–1911

From November 1903 to April 1904 Riesz studied at Göttingen. Hilbert was then beginning his work on integral equations, prompted by Ivar Fredholm’s theory, and the young Hungarian absorbed the new questions. Over the next few years he combined these questions with Henri Lebesgue’s new integral (1902) and Maurice Fréchet’s abstract spaces (1906).

In August 1904 he took a teaching post at the state school in Lőcse, a small town in the Szepes region, now Levoča in Slovakia. He stayed four years. In 1906 he joined the Reformed Church. From this provincial post he published, in 1906–1910, a sequence of papers that created much of functional analysis. In 1907 came the Riesz–Fischer theorem (Lesson 04) and the representation of functionals on \(L^2\) (Lesson 05). In April 1908 he lectured at the International Congress in Rome on abstract sets with a notion of accumulation point, an early axiomatic approach to topology (Lesson 10). In 1909 he characterized the continuous linear functionals on \(C[a,b]\), and in 1910 he introduced the spaces \(L^p\) (Lesson 06).

Later in 1908 he moved to a gymnasium in Budapest. In 1911 the committee for a Budapest university chair ranked him first, but the post went to another candidate. Kolozsvár then invited him.

Lesson 03

Kolozsvár, 1911–1920

The Franz Joseph University in Kolozsvár (now Cluj-Napoca, Romania) was Hungary’s second university. The faculty invited Riesz in October 1911; he was appointed extraordinary professor on 6 April 1912 and full professor in 1914. Alfréd Haar, a doctoral student of Hilbert, joined the faculty in the same period, beginning a partnership that lasted until Haar’s death in 1933.

At Kolozsvár Riesz wrote his first book, Les systèmes d’équations linéaires à une infinité d’inconnues (1913), which put Hilbert’s infinite systems of equations in the setting of \(\ell^2\) and \(\ell^p\). In 1916 he and Marcel presented the F. and M. Riesz theorem (Lesson 08) at the Scandinavian Congress in Stockholm. In 1918 he published in Acta Mathematica his theory of completely continuous operators, now compact operators (Lesson 07), the paper in which he came closest to the axioms of a Banach space. The Hungarian Academy elected him a corresponding member in 1915.

The end of the war destroyed this world. Romanian troops entered Kolozsvár in December 1918, and in 1919 the Hungarian university was taken over by the Romanian state. Its Hungarian staff continued in Budapest for a time. After the Treaty of Trianon of 1920 the university was re-established in Szeged, in southern Hungary, and Riesz went with it.

Part II · The mathematics
Lesson 04

The Riesz–Fischer theorem

For the general reader. A Fourier series describes a function by a list of numbers, its coefficients. Which lists arise? If the sum of the squares of the numbers is finite, Riesz proved in 1907, then some function has exactly those coefficients. The catch is that the function may be wild, and only Lebesgue’s new integral can handle it. The theorem made Lebesgue’s integral indispensable and showed that functions and square-summable sequences are two descriptions of one space.

Goals
  • State the Riesz–Fischer theorem in Riesz’s form and in Fischer’s.
  • Prove that \(L^2[a,b]\) is complete.
  • Deduce that \(L^2\) and \(\ell^2\) are isometric.
Theorem (Riesz, 1907)

Let \((\varphi_n)\) be an orthonormal system in \(L^2[a,b]\) and \((c_n)\) a sequence of real numbers with \(\sum_nc_n^2<\infty\). Then there is \(f\in L^2[a,b]\) with \(\int_a^bf\varphi_n\,dx=c_n\) for every \(n\).

Theorem (Fischer, 1907)

\(L^2[a,b]\) is complete: if \(\|f_n-f_m\|_2\to0\) as \(n,m\to\infty\), there is \(f\in L^2\) with \(\|f_n-f\|_2\to0\).

Riesz announced his theorem in the Comptes rendus in March 1907. Ernst Fischer, who would later teach Emmy Noether, published his version, in terms of convergence in mean, in the same journal about two months later. The two statements are equivalent, and the combined result is known by both names.

Proof of completeness

Let \((f_n)\) be Cauchy in \(L^2\). Choose \(n_1<n_2<\cdots\) with \(\|f_{n_{k+1}}-f_{n_k}\|_2<2^{-k}\), and set \(g=\sum_k|f_{n_{k+1}}-f_{n_k}|\). By Minkowski’s inequality and monotone convergence, \(\|g\|_2\le\sum_k2^{-k}<\infty\), so \(g<\infty\) almost everywhere. Hence the telescoping series \(f_{n_1}+\sum_k(f_{n_{k+1}}-f_{n_k})\) converges absolutely almost everywhere to a function \(f\). By Fatou’s lemma, \(\|f-f_{n_k}\|_2\le\liminf_j\|f_{n_j}-f_{n_k}\|_2\), which is small for large \(k\); so \(f\in L^2\) and the subsequence converges to \(f\) in \(L^2\). A Cauchy sequence with a convergent subsequence converges.

Riesz’s form from Fischer’s. Put \(f_N=\sum_{n\le N}c_n\varphi_n\). By orthonormality \(\|f_N-f_M\|_2^2=\sum_{M<n\le N}c_n^2\to0\), so \(f_N\to f\) in \(L^2\), and \(\int f\varphi_n=\lim_N\int f_N\varphi_n=c_n\) by the Cauchy–Schwarz inequality.

Corollary

If \((\varphi_n)\) is a complete orthonormal system in \(L^2[a,b]\), the map \(f\mapsto(\int f\varphi_n)_n\) is a linear isometry of \(L^2[a,b]\) onto \(\ell^2\). In particular \(L^2[a,b]\) is a Hilbert space.

Isometry is Parseval’s identity, and surjectivity is the Riesz–Fischer theorem. This identification of a space of functions with Hilbert’s space of sequences was the step that later let John von Neumann show that Werner Heisenberg’s matrix mechanics and Erwin Schrödinger’s wave mechanics are the same theory.

Exercise 4.1Show that the theorem fails for the Riemann integral: give a Cauchy sequence in the \(L^2\) norm of Riemann-integrable functions on \([0,1]\) whose \(L^2\) limit is not almost everywhere equal to any Riemann-integrable function.

Show solution
Let \(K\subset[0,1]\) be a closed nowhere dense set of positive measure (a fat Cantor set), \(U=[0,1]\setminus K\), and let \(f_n\) be the indicator of the union of the first \(n\) open intervals removed in its construction. Each \(f_n\) is a step function, and \(f_n\to\mathbf 1_U\) in \(L^2\) by monotone convergence. Let \(g=\mathbf 1_U\) a.e. At a Lebesgue density point \(x\) of \(K\), every neighbourhood contains a subset of \(K\) of positive measure, where \(g=0\) a.e., and an open interval of \(U\), where \(g=1\) a.e. So \(g\) has oscillation at least 1 at \(x\). Almost every point of \(K\) is a density point, so \(g\) is discontinuous on a set of positive measure and is not Riemann integrable.

Exercise 4.2Using the trigonometric system on \([-\pi,\pi]\), show that \(\sum_{n\ge1}\frac{\sin nx}{n}\) converges in \(L^2\), identify its sum, and deduce \(\sum_{n\ge1}1/n^2=\pi^2/6\) from Parseval.

Show solution
\(\sum1/n^2<\infty\), so the series converges in \(L^2\) by the theorem. Its sum is \(f(x)=(\pi-x)/2\) on \((0,\pi]\), extended oddly, since \(\frac1\pi\int_{-\pi}^{\pi}f(x)\sin nx\,dx=\frac1n\). Parseval: \(\frac1\pi\int_{-\pi}^\pi f^2=\sum b_n^2\). Here \(\frac1\pi\cdot2\int_0^\pi\frac{(\pi-x)^2}{4}dx=\frac{\pi^2}{6}\), so \(\sum1/n^2=\pi^2/6\).

Exercise 4.3Show that \(L^2\) convergence does not imply pointwise convergence anywhere, but that every \(L^2\)-convergent sequence has an almost everywhere convergent subsequence.

Show solution
The “typewriter” sequence of indicators of \([j2^{-k},(j+1)2^{-k}]\), \(0\le j<2^k\), enumerated by \(k\) then \(j\), has \(L^2\) norms \(2^{-k/2}\to0\), but at every \(x\) it takes the value 1 infinitely often. The second statement is exactly what the proof above shows: the fast subsequence converges a.e.
Lesson 05

Representing linear functionals

For the general reader. A linear functional assigns a number to each function, linearly and continuously: the average of the function, its value at a point, a weighted integral. Riesz asked for all of them. On \(L^2\) the answer is that every such functional is an inner product with a fixed function. On continuous functions every one is an integral against a generalized weight, a function of bounded variation. These two answers are the prototypes of duality, one of the central ideas of modern analysis.

Goals
  • Prove the representation theorem for Hilbert spaces.
  • State Riesz’s 1909 theorem for \(C[a,b]\) and the measure-theoretic form that followed.
  • Compute norms of functionals.
Theorem (Fréchet; Riesz, 1907)

Let \(H\) be a Hilbert space and \(A:H\to\mathbb C\) a bounded linear functional. There is a unique \(g\in H\) with \(A(f)=\langle f,g\rangle\) for all \(f\in H\), and \(\|A\|=\|g\|\).

For \(H=L^2[a,b]\) this was proved by Riesz and by Fréchet in notes published in the same volume of the Comptes rendus in 1907. Proof. If \(A=0\) take \(g=0\). Otherwise \(N=\ker A\) is a closed proper subspace. By the projection theorem there is a unit vector \(e\perp N\). For any \(f\), the vector \(A(f)e-A(e)f\) lies in \(N\), so it is orthogonal to \(e\): \(A(f)=A(e)\langle f,e\rangle\). Take \(g=\overline{A(e)}\,e\). Uniqueness: if \(\langle f,g-g'\rangle=0\) for all \(f\), take \(f=g-g'\). The norm follows from Cauchy–Schwarz with equality at \(f=g\).

Theorem (Riesz, 1909)

Every bounded linear functional \(A\) on \(C[a,b]\) with the supremum norm has the form

\[ A(f)=\int_a^bf(x)\,d\alpha(x), \]

a Riemann–Stieltjes integral, where \(\alpha\) is of bounded variation on \([a,b]\). If \(\alpha\) is normalized by \(\alpha(a)=0\) and right-continuity on \((a,b)\), it is unique, and \(\|A\|\) equals the total variation of \(\alpha\).

Jacques Hadamard had shown in 1903 that every such functional is a limit of ordinary integrals \(\int fk_n\). Riesz replaced the limit by a single Stieltjes integral. A short modern proof extends \(A\) with the same norm to bounded functions, by the Hahn–Banach theorem, and sets \(\alpha(x)=A(\mathbf 1_{(a,x]})\), followed by a normalization. In modern form, with Johann Radon’s 1913 integrals, Andrey Markov’s 1938 extension and Shizuo Kakutani’s 1941 version for compact Hausdorff spaces, the theorem reads: the dual of \(C(X)\) is the space of regular signed (or complex) Borel measures on \(X\), with the total variation norm. It is now called the Riesz–Markov–Kakutani theorem.

Exercise 5.1On \(C[0,1]\), find \(\alpha\) and \(\|A\|\) for \(A(f)=f(\tfrac12)-\int_0^1f(x)\,dx\).

Show solution
Point evaluation at \(\tfrac12\) corresponds to a unit jump at \(\tfrac12\); the integral to \(\alpha(x)=x\). So \(\alpha(x)=\mathbf 1_{[1/2,1]}(x)-x\). Its total variation is \(1\) (the jump) plus \(1\) (the continuous part), so \(\|A\|=2\). The norm is approached, not attained, by continuous \(f\) equal to \(1\) at \(\tfrac12\) and \(-1\) away from a tiny neighbourhood of it.

Exercise 5.2Use the Hilbert-space theorem to show that the operator \(T\mapsto T^*\) is well defined: for bounded \(T\) on \(H\) there is a unique bounded \(T^*\) with \(\langle Tf,h\rangle=\langle f,T^*h\rangle\), and \(\|T^*\|=\|T\|\).

Show solution
For fixed \(h\), \(f\mapsto\langle Tf,h\rangle\) is bounded with norm \(\le\|T\|\|h\|\). Representation gives a unique \(T^*h\) with \(\|T^*h\|\le\|T\|\|h\|\); linearity in \(h\) follows from uniqueness. So \(\|T^*\|\le\|T\|\), and since \(T^{**}=T\), equality holds.

Exercise 5.3Show that point evaluation \(\delta_0\) on \(C[0,1]\) is not of the form \(f\mapsto\int fg\,dx\) for any \(g\in L^1\). Why did Riesz need Stieltjes integrals?

Show solution
If \(f(0)=\int_0^1fg\) for all continuous \(f\), take \(f_n\) continuous with \(0\le f_n\le1\), \(f_n(0)=1\), supported in \([0,1/n]\). Then \(1=\int f_ng\le\int_0^{1/n}|g|\to0\) by absolute continuity of the integral, a contradiction. Point masses have no density, so the representing object must allow jumps: a function of bounded variation, or a measure.
Lesson 06

The spaces \(L^p\)

For the general reader. The space \(L^2\) measures a function by the square root of the integral of its square. Riesz saw in 1910 that every power \(p>1\) gives a space with its own geometry, and that the space for \(p\) is paired with the space for the dual exponent \(q\), where \(1/p+1/q=1\). This pairing, with the notion of weak convergence he introduced alongside it, is the model for all later duality theory.

Goals
  • Define \(L^p\), prove Hölder’s and Minkowski’s inequalities.
  • State and prove the duality \((L^p)^*=L^q\) for \(1<p<\infty\).
  • Understand weak convergence and Riesz’s criterion for it.

For \(1\le p<\infty\), \(L^p[a,b]\) consists of measurable \(f\) with \(\|f\|_p=\big(\int_a^b|f|^p\big)^{1/p}<\infty\), modulo functions vanishing almost everywhere. Riesz’s 1910 paper in Mathematische Annalen, Untersuchungen über Systeme integrierbarer Funktionen, studied these spaces for \(p>1\), proved their completeness, and solved systems of linear equations in them.

Hölder and Minkowski inequalities

If \(1<p<\infty\) and \(1/p+1/q=1\), then \(\int|fg|\le\|f\|_p\|g\|_q\), and \(\|f+g\|_p\le\|f\|_p+\|g\|_p\).

Theorem (Riesz, 1910)

Let \(1<p<\infty\) and \(1/p+1/q=1\). For every bounded linear functional \(A\) on \(L^p[a,b]\) there is a unique \(g\in L^q[a,b]\) with \(A(f)=\int_a^bfg\) for all \(f\in L^p\), and \(\|A\|=\|g\|_q\).

Proof sketch. Set \(G(x)=A(\mathbf 1_{[a,x]})\). Using that \(|A(\mathbf 1_E)|\le\|A\||E|^{1/p}\), one shows \(G\) is absolutely continuous, so \(G(x)=\int_a^xg\) with \(g=G'\in L^1\). Then \(A(f)=\int fg\) for step functions. Testing \(A\) on truncations of \(f=|g|^{q-1}\operatorname{sign}g\), for which \(\int fg=\|f\|_p^p=\|g\|_q^q\), gives \(\|g\|_q\le\|A\|\). Step functions are dense in \(L^p\), so the formula extends. Hölder gives \(\|A\|\le\|g\|_q\).

Weak convergence

Riesz called a sequence \(f_n\) in \(L^p\) weakly convergent to \(f\) if \(\int f_ng\to\int fg\) for every \(g\in L^q\). He proved that this happens if and only if \(\sup_n\|f_n\|_p<\infty\) and \(\int_a^xf_n\to\int_a^xf\) for every \(x\in[a,b]\). Bounded sequences in \(L^p\), \(1<p<\infty\), have weakly convergent subsequences. This weak compactness, already used by Hilbert in \(\ell^2\), was made general by Banach and Leonidas Alaoglu.

Exercise 6.1Show that \(f_n(x)=\sin(nx)\) converges weakly to 0 in \(L^2[0,2\pi]\) but not in norm.

Show solution
\(\|f_n\|_2=\sqrt\pi\) for all \(n\), so there is no norm convergence to 0. By Riesz’s criterion it suffices that \(\int_0^x\sin nt\,dt=(1-\cos nx)/n\to0\), which holds. Equivalently, the Riemann–Lebesgue lemma gives \(\int\sin(nx)g\to0\) for every \(g\in L^2\).

Exercise 6.2Prove Hölder’s inequality from Young’s inequality \(ab\le\frac{a^p}{p}+\frac{b^q}{q}\) for \(a,b\ge0\).

Show solution
Assume \(\|f\|_p=\|g\|_q=1\) by scaling (the cases with norm 0 are trivial). Apply Young pointwise with \(a=|f(x)|\), \(b=|g(x)|\) and integrate: \(\int|fg|\le\frac1p+\frac1q=1\). Young’s inequality itself follows from concavity of \(\log\): \(\log(\frac{a^p}{p}+\frac{b^q}{q})\ge\frac1p\log a^p+\frac1q\log b^q=\log ab\).

Exercise 6.3Show that \(f_n=n\,\mathbf 1_{[0,1/n]}\) is bounded in \(L^1[0,1]\) but has no weakly convergent subsequence in \(L^1\). Why does this not contradict Riesz’s compactness result?

Show solution
\(\|f_n\|_1=1\). For continuous \(g\), \(\int f_ng\to g(0)\), so a weak limit \(f\in L^1\) would satisfy \(\int fg=g(0)\) for all continuous \(g\), impossible by Exercise 5.3. Riesz’s result needs \(1<p<\infty\): \(L^1\) is not reflexive, and mass can concentrate into a point measure that lies outside the space.
Lesson 07

Compact operators

For the general reader. Fredholm proved that integral equations behave like finite systems of linear equations, using determinants that only make sense for integral kernels. In 1918 Riesz found the reason behind Fredholm’s results without determinants or kernels. All that matters is one property of the operator, compactness, and one geometric fact about infinite-dimensional spaces, now called Riesz’s lemma. His paper is the first in which the methods of Banach-space theory appear in full.

Goals
  • Prove Riesz’s lemma and the non-compactness of the unit ball in infinite dimensions.
  • Prove the Fredholm alternative for \(I-T\) with \(T\) compact.
  • Describe the spectrum of a compact operator.

Riesz worked in the space \(C[a,b]\) with the maximum norm, but his proofs used only the norm and completeness, and he remarked on this. In modern terms: a linear operator \(T\) on a Banach space \(X\) is compact (Riesz said completely continuous) if it maps bounded sets to sets with compact closure. Integral operators with continuous kernels on \(C[a,b]\) are compact by the Arzelà–Ascoli theorem.

Riesz’s lemma (1918)

Let \(Y\) be a closed proper subspace of a normed space \(X\), and \(0<\theta<1\). There is \(x\in X\) with \(\|x\|=1\) and \(\|x-y\|\ge\theta\) for every \(y\in Y\). Consequently, the closed unit ball of \(X\) is compact if and only if \(X\) is finite-dimensional.

Proof. Pick \(z\notin Y\); then \(d=\operatorname{dist}(z,Y)>0\) since \(Y\) is closed. Choose \(y_0\in Y\) with \(\|z-y_0\|\le d/\theta\) and set \(x=(z-y_0)/\|z-y_0\|\). For \(y\in Y\), \(\|x-y\|=\|z-(y_0+\|z-y_0\|y)\|/\|z-y_0\|\ge d/\|z-y_0\|\ge\theta\). For the consequence, if \(X\) is infinite-dimensional, apply the lemma to the spans of \(x_1,\dots,x_n\) repeatedly to get unit vectors with \(\|x_n-x_m\|\ge\tfrac12\), a sequence with no convergent subsequence.

Theorem (Riesz, 1918)

Let \(T\) be compact on a Banach space \(X\) and \(S=I-T\).

  1. \(\ker S\) is finite-dimensional and \(\operatorname{ran}S\) is closed.
  2. There is a smallest integer \(r\ge0\) with \(\ker S^r=\ker S^{r+1}\); for the same \(r\), \(\operatorname{ran}S^r=\operatorname{ran}S^{r+1}\), and \(X=\ker S^r\oplus\operatorname{ran}S^r\).
  3. \(S\) is injective if and only if it is surjective.
  4. The spectrum of \(T\) is countable, every nonzero point of it is an eigenvalue of finite multiplicity, and \(0\) is its only possible accumulation point.

Ideas of proof. (1) On \(\ker S\), \(T\) is the identity; since \(T\) is compact, the unit ball of \(\ker S\) is compact, so the kernel is finite-dimensional by Riesz’s lemma. (2) If the kernels \(\ker S^n\) increased strictly forever, Riesz’s lemma would give unit vectors \(x_n\in\ker S^n\) at distance \(\ge\tfrac12\) from \(\ker S^{n-1}\); then \(\|Tx_n-Tx_m\|\ge\tfrac12\) for \(m<n\), contradicting compactness. The same argument works for the decreasing ranges. (3) follows from (2): if \(\ker S=0\) then \(r=0\) and \(\operatorname{ran}S=X\). Juliusz Schauder added in 1930 the adjoint operator, which gives Fredholm’s solvability conditions, and the combined result is called the Riesz–Schauder theory.

Exercise 7.1Show that the diagonal operator \(T(x_n)=(x_n/n)\) on \(\ell^2\) is compact, find its spectrum, and check part 4.

Show solution
\(T\) is the norm limit of the finite-rank truncations \(T_N\), since \(\|T-T_N\|=1/(N+1)\), and norm limits of finite-rank operators are compact. The eigenvalues are \(1/n\), each with multiplicity 1. \(T\) is injective but not onto (the sequence \((1/n)\) is not in the range), so \(0\) is in the spectrum, as the only accumulation point.

Exercise 7.2Why does part 3 fail for operators that are not of the form \(I-T\)? Give an injective bounded operator on \(\ell^2\) that is not surjective, and an injective compact operator that is not surjective.

Show solution
The right shift \(S(x_0,x_1,\dots)=(0,x_0,x_1,\dots)\) is injective (an isometry) but its range misses \((1,0,0,\dots)\). The operator of Exercise 7.1 is compact, injective and not surjective. Neither is of the form \(I-T\) with \(T\) compact, so Riesz’s theorem does not apply. The shift has a closed range of codimension 1; the diagonal operator has a dense range that is not closed.

Exercise 7.3For the Volterra operator \((Vf)(x)=\int_0^xf\) on \(C[0,1]\), use part 3 to show that \(I-\mu V\) is invertible for every complex \(\mu\).

Show solution
\(\mu V\) is compact. If \((I-\mu V)f=0\), then \(f\) is \(C^1\), \(f(0)=0\) and \(f'=\mu f\), so \(f=0\). Injective implies surjective by part 3, and the inverse is bounded by the open mapping theorem (or directly from the Neumann series of Vito Volterra).
Lesson 08

Boundary values and subharmonic functions

For the general reader. An analytic function on a disc is controlled by its values on the boundary circle. In 1916 the Riesz brothers proved a striking rigidity: a measure on the circle whose Fourier series has only nonnegative frequencies cannot be singular; it must have a density. In the 1920s Frigyes Riesz built a general theory of subharmonic functions, functions that lie below their harmonic interpolations, and showed that each one is a potential of a positive mass distribution plus a harmonic function.

Goals
  • State the F. and M. Riesz theorem and its geometric consequence.
  • Define subharmonic functions and state the Riesz decomposition.
  • Compute Riesz measures and derive Jensen’s formula.
Theorem (F. and M. Riesz, 1916)

Let \(\mu\) be a complex Borel measure on the unit circle with \(\hat\mu(n)=\int e^{-in\theta}d\mu(\theta)=0\) for all \(n<0\). Then \(\mu\) is absolutely continuous with respect to arc length.

Their motivation was conformal mapping. If a Jordan region is bounded by a rectifiable curve, and \(F\) maps the disc conformally onto it, then \(F'\) belongs to the Hardy space \(H^1\), and the theorem shows that sets of zero length on the boundary curve correspond to sets of zero length on the circle and conversely. Riesz introduced the Hardy spaces \(H^p\) by name in 1923, after G. H. Hardy’s 1915 convexity theorem for their means.

Subharmonic functions

An upper semicontinuous function \(u:D\to[-\infty,\infty)\) on a domain \(D\subset\mathbb R^2\), not identically \(-\infty\), is subharmonic if for every closed disc \(\bar B(z,r)\subset D\), \(u(z)\le\frac{1}{2\pi}\int_0^{2\pi}u(z+re^{i\theta})\,d\theta\). For smooth \(u\) this is equivalent to \(\Delta u\ge0\). In general \(\Delta u\), taken in the sense of distributions, is a positive measure \(\mu_u\), the Riesz measure of \(u\).

Riesz decomposition (1926, 1930)

Let \(u\) be subharmonic on \(D\subset\mathbb R^2\) and \(G\) a relatively compact open subset of \(D\). Then on \(G\)

\[ u(z)=\frac{1}{2\pi}\int_G\log|z-w|\,d\mu_u(w)+h(z), \]

where \(h\) is harmonic on \(G\). The analogous statement holds in \(\mathbb R^n\), \(n\ge3\), with the Newtonian kernel \(-|z-w|^{2-n}\) and its normalizing constant.

The proof subtracts the logarithmic potential of \(\mu_u|_G\), whose Laplacian is \(\mu_u\) because \(\Delta\log|z|=2\pi\delta_0\), and applies Weyl’s lemma to the difference, which has zero Laplacian. Subharmonic functions became the basic tool of potential theory and of complex analysis in several variables, through the plurisubharmonic functions of Pierre Lelong and Kiyoshi Oka.

Exercise 8.1Let \(f\) be analytic on a neighbourhood of the closed unit disc, with zeros \(a_1,\dots,a_n\) inside (with multiplicity) and none on the circle, and \(f(0)\ne0\). Find the Riesz measure of \(u=\log|f|\), and use the decomposition to prove Jensen’s formula \(\log|f(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(e^{i\theta})|\,d\theta-\sum_k\log\frac{1}{|a_k|}\).

Show solution
Near a zero of order \(m\), \(\log|f|=m\log|z-a|+\)harmonic, so \(\mu_u=2\pi\sum_k\delta_{a_k}\). Then \(h=\log|f|-\sum_k\log|z-a_k|\) is harmonic on the closed disc, and the mean value property gives \(h(0)=\frac1{2\pi}\int h(e^{i\theta})d\theta\). Now \(h(0)=\log|f(0)|-\sum\log|a_k|\), and \(\frac1{2\pi}\int_0^{2\pi}\log|e^{i\theta}-a|\,d\theta=0\) for \(|a|<1\). So \(\log|f(0)|-\sum_k\log|a_k|=\frac1{2\pi}\int\log|f(e^{i\theta})|d\theta\), which rearranges to Jensen’s formula.

Exercise 8.2Show that \(u(z)=|z|^2\) is subharmonic, find its Riesz measure, and verify the sub-mean-value inequality directly.

Show solution
\(\Delta|z|^2=\Delta(x^2+y^2)=4>0\), so \(\mu_u=4\,dA\). Directly, \(\frac1{2\pi}\int|z+re^{i\theta}|^2d\theta=|z|^2+r^2\ge|z|^2\), since the cross term averages to zero.

Exercise 8.3Compute \(\hat\mu(n)\) for \(\mu=d\theta\), \(\mu=\delta_1\) (the point mass at \(\theta=0\)) and \(\mu=(1-\cos\theta)\,d\theta\). Which satisfy the hypothesis of the F. and M. Riesz theorem, and what does the last example show?

Show solution
\(d\theta\): \(\hat\mu(0)=2\pi\), other coefficients 0, so the hypothesis holds; it is absolutely continuous. \(\delta_1\): \(\hat\mu(n)=1\) for all \(n\), so the hypothesis fails, consistent with the theorem (a point mass is singular). \((1-\cos\theta)d\theta\): \(\hat\mu(0)=2\pi\), \(\hat\mu(\pm1)=-\pi\), so \(\hat\mu(-1)\ne0\) and the hypothesis fails, although the measure is absolutely continuous. The theorem is one-directional.
Lesson 09

The rising sun and the ergodic theorem

For the general reader. Riesz had a gift for replacing long proofs with a picture or a few lines. Imagine the graph of a function as a mountain range at dawn, with the sun rising far to the right. Which points along the range lie in shadow? The answer, a union of intervals with a simple property, gives in two lines the key inequality of Hardy and Littlewood on maximal functions. In 1938 he gave a similarly short proof of von Neumann’s mean ergodic theorem.

Goals
  • Prove the rising sun lemma.
  • Derive the one-sided Hardy–Littlewood maximal inequality.
  • Prove the mean ergodic theorem for contractions on Hilbert space.
Rising sun lemma (Riesz, 1932)

Let \(g\) be continuous on \([a,b]\), and let \(E\) be the set of \(x\in(a,b)\) for which some \(y>x\) in \([a,b]\) has \(g(y)>g(x)\). Then \(E\) is open, hence a countable disjoint union of open intervals \((a_k,b_k)\), and \(g(a_k)\le g(b_k)\) for each \(k\), with equality whenever \(a_k\ne a\).

Proof. \(E\) is open by continuity. Let \((a_k,b_k)\) be a component. Since \(b_k\notin E\), \(g(y)\le g(b_k)\) for all \(y>b_k\). Suppose \(g(x_0)>g(b_k)\) for some \(x_0\in(a_k,b_k)\), and let \(x_1\) be a point where \(g\) attains its maximum on \([x_0,b_k]\). Then \(g(x_1)>g(b_k)\), so \(x_1\in(a_k,b_k)\subset E\); yet no \(y>x_1\) has \(g(y)>g(x_1)\), a contradiction. Hence \(g\le g(b_k)\) on \((a_k,b_k)\), and by continuity \(g(a_k)\le g(b_k)\). If \(a_k\ne a\), then \(a_k\notin E\), so \(g(a_k)\ge g(b_k)\).

Corollary (one-sided maximal inequality)

For \(f\in L^1[a,b]\) let \(M^+f(x)=\sup_{x<y\le b}\frac{1}{y-x}\int_x^y|f|\). For every \(t>0\), \(\big|\{x:M^+f(x)>t\}\big|\le\frac1t\int_a^b|f|\).

Proof. Apply the lemma to \(g(x)=\int_a^x|f|-tx\). Then \(M^+f(x)>t\) exactly when some \(y>x\) has \(g(y)>g(x)\). On each component, \(g(a_k)\le g(b_k)\) means \(t(b_k-a_k)\le\int_{a_k}^{b_k}|f|\). Summing over \(k\) gives the inequality, with the best constant. Riesz used the lemma to give a short proof of Lebesgue’s theorem that monotone functions are differentiable almost everywhere.

The mean ergodic theorem

Theorem (von Neumann 1932; Riesz’s proof 1938)

Let \(T\) be a linear contraction (\(\|T\|\le1\)) on a Hilbert space \(H\), and \(P\) the orthogonal projection onto \(\{x:Tx=x\}\). Then for every \(x\in H\), \(\frac1n\sum_{k=0}^{n-1}T^kx\to Px\) in norm.

Proof. First, for a contraction \(Tx=x\) if and only if \(T^*x=x\): if \(T^*x=x\) then \(\langle Tx,x\rangle=\|x\|^2\), so \(\|Tx-x\|^2=\|Tx\|^2-2\|x\|^2+\|x\|^2\le0\). Hence \(\ker(I-T)=\ker(I-T^*)=\operatorname{ran}(I-T)^\perp\), and \(H=\ker(I-T)\oplus\overline{\operatorname{ran}(I-T)}\). On the first summand the averages are constant. For \(y=z-Tz\) they telescope: \(\frac1n\sum_{k<n}T^ky=\frac1n(z-T^nz)\), of norm at most \(2\|z\|/n\). The averages have norm at most 1, so convergence to 0 extends to the closure of the range. Riesz expanded this circle of ideas in his 1944 Geneva lectures, published in 1945 as Sur la théorie ergodique.

Exercise 9.1Apply the rising sun lemma to \(g(x)=\sin x\) on \([0,4\pi]\). Find \(E\) and check the endpoint condition.

Show solution
The value 1 is attained last at \(5\pi/2\), so every \(x<5\pi/2\) with \(\sin x<1\) is in \(E\), while \(\pi/2\) is not. On \([5\pi/2,3\pi]\), \(\sin x\ge0\) and every later value is smaller or at most 0, so these points are not in \(E\). On \((3\pi,4\pi)\), \(\sin x<0=\sin4\pi\), so these points are in \(E\). Hence \(E=(0,\pi/2)\cup(\pi/2,5\pi/2)\cup(3\pi,4\pi)\). Endpoint values: \(g(0)=0\le g(\pi/2)=1\) (left end \(a\), inequality allowed), \(g(\pi/2)=g(5\pi/2)=1\) and \(g(3\pi)=g(4\pi)=0\) (equalities).

Exercise 9.2Let \(T\) be rotation of the circle by an irrational angle \(2\pi\alpha\), acting on \(L^2\) by \((Tf)(\theta)=f(\theta+2\pi\alpha)\). Identify \(P\) and interpret the mean ergodic theorem.

Show solution
\(T\) is unitary, \(Te^{in\theta}=e^{2\pi in\alpha}e^{in\theta}\). Since \(\alpha\) is irrational, \(e^{2\pi in\alpha}=1\) only for \(n=0\), so the fixed space is the constants and \(Pf=\frac{1}{2\pi}\int f\). Time averages along the rotation converge in \(L^2\) to the space average.

Exercise 9.3Show that the conclusion can fail for a non-contraction: find a \(2\times2\) matrix \(T\) with \(\sup_n\|T^n\|=\infty\) whose averages diverge.

Show solution
\(T=\begin{pmatrix}1&1\\0&1\end{pmatrix}\), \(T^k=\begin{pmatrix}1&k\\0&1\end{pmatrix}\). The averages are \(\begin{pmatrix}1&(n-1)/2\\0&1\end{pmatrix}\), unbounded. The key step \(\ker(I-T)=\ker(I-T^*)\) fails: \((1,0)\) is fixed by \(T\) but not by \(T^*\).
Lesson 10

Topology and order

For the general reader. Two of Riesz’s ideas were about structure rather than particular theorems. In 1908 he proposed describing a space by declaring which points are limits of which sets, without any distances, one of the earliest axiomatic approaches to topology. From 1928 he studied spaces of functions as ordered spaces, where one can compare two functions and take their maximum, and showed that linear functionals split into positive and negative parts. Ordered vector spaces of this kind are now called Riesz spaces.

Goals
  • Understand Riesz’s 1908 approach through accumulation points.
  • Define vector lattices and the Riesz decomposition of order-bounded functionals.
  • Compute positive and negative parts of functionals.

Abstract continuity, 1908

In his lecture Stetigkeitsbegriff und abstrakte Mengenlehre at the Rome Congress of 1908, Riesz considered an abstract set in which certain points are declared accumulation points of certain subsets, subject to axioms: a finite set has no accumulation points; an accumulation point of a set is one of every larger set; an accumulation point of \(A\cup B\) is one of \(A\) or of \(B\); and distinct points are distinguished by the sets of which they are accumulation points. From this he defined closed sets and continuity. Felix Hausdorff’s neighbourhood axioms of 1914 became the standard, but Riesz’s formulation is close in spirit to Kazimierz Kuratowski’s later closure axioms. His related notion of linked sets anticipated the proximity spaces of Vadim Efremovich.

Riesz spaces

A vector lattice or Riesz space is a real vector space \(E\) with a partial order compatible with addition and positive scaling, in which every pair \(f,g\) has a least upper bound \(f\vee g\). Then \(f^+=f\vee0\), \(f^-=(-f)\vee0\), \(f=f^+-f^-\) and \(|f|=f^++f^-\). Function spaces such as \(C[a,b]\) and \(L^p\) with the pointwise order are examples.

Theorem (Riesz, 1928 and 1940)

Let \(E\) be a Riesz space and \(A\) a linear functional that is order bounded (bounded on each order interval \([-u,u]\)). Then

\[ A^+(u)=\sup\{A(v):0\le v\le u\},\qquad u\ge0, \]

extends to a positive linear functional, \(A^-=A^+-A\) is positive, and \(A=A^+-A^-\). The order-bounded functionals form a Riesz space, in fact a Dedekind complete one, with \(A\vee0=A^+\).

Riesz presented the decomposition for functionals on \(C[a,b]\) at the Bologna Congress of 1928, without using the integral representation, and developed the general theory in Sur quelques notions fondamentales dans la théorie générale des opérations linéaires in the Annals of Mathematics (1940). For \(C[a,b]\) the decomposition is the Jordan decomposition of the representing function of bounded variation into increasing parts. The theory of Riesz spaces was developed further by Hidegorô Nakano, Leonid Kantorovich and their schools.

Exercise 10.1On \(C[0,1]\) let \(A(f)=f(0)-2f(1)+\int_0^1f\). Compute \(A^+\) and \(A^-\).

Show solution
For \(u\ge0\), \(A(v)\) over \(0\le v\le u\) is largest when \(v=u\) except near 1, where \(v\) should drop to 0 in a shrinking neighbourhood: the supremum is \(u(0)+\int_0^1u\). So \(A^+(f)=f(0)+\int_0^1f\) and \(A^-(f)=2f(1)\). In measure language \(\mu=\delta_0-2\delta_1+dx\), \(\mu^+=\delta_0+dx\), \(\mu^-=2\delta_1\).

Exercise 10.2In any Riesz space prove \(f\vee g+f\wedge g=f+g\), where \(f\wedge g=-((-f)\vee(-g))\).

Show solution
Translation is an order isomorphism, so \((f\vee g)-f-g=(-g)\vee(-f)=-(f\wedge g)\). Rearranging gives the identity. With \(g=0\): \(f^+-f^-=f\).

Exercise 10.3Using Riesz’s 1908 axioms, show that if every accumulation point of \(A\) belongs to \(A\) and every accumulation point of \(B\) belongs to \(B\), then the same holds for \(A\cup B\). Which axiom is essential here?

Show solution
By the third axiom an accumulation point of \(A\cup B\) is one of \(A\) or of \(B\), hence lies in \(A\) or \(B\). Without that axiom, \(A\cup B\) could acquire new accumulation points not belonging to either set, and finite unions of closed sets would fail to be closed.
Part III · Life and legacy
Lesson 11

Szeged and the Bolyai Institute, 1920–1945

Szeged in 1920 was a provincial city with a university that had just arrived from somewhere else. Riesz and Haar made it a centre of analysis. In 1922 they founded the János Bolyai Mathematical Institute, named after the Hungarian discoverer of non-Euclidean geometry, and the journal Acta Scientiarum Mathematicarum, which Riesz edited. The journal soon attracted contributions from abroad. Riesz headed the faculty of natural sciences in 1925–26, and his 1925 address in that office was titled Elementary Methods in Higher Mathematics, a fair description of his own style.

His students and younger colleagues at Szeged included Tibor Radó, who took his doctorate in 1922 and later gave one of the first solutions of the Plateau problem; Béla Szőkefalvi-Nagy, doctorate 1936; and Alfréd Rényi, doctorate 1945. Visitors such as the American Edgar Lorch, in 1934, came to work with him. Haar died in 1933, and Riesz carried the institute on alone.

Honours accumulated: the Marczibányi Prize (1927), full membership of the Hungarian Academy (1936), and the regard of the international community, which saw him as one of the founders of functional analysis. His research shifted toward the elementary and the definitive: subharmonic functions, the rising sun lemma, the ergodic theorem, ordered spaces.

The war years were dangerous for a man of Jewish birth in Hungary, despite his conversion. Hungary’s anti-Jewish laws and the German occupation of 1944 destroyed much of his family; his sister Margit, her husband and their daughter were deported and murdered at Auschwitz. In 1944–45, as the front passed through Szeged, Riesz served as rector of the university.

Lesson 12

Budapest, the Leçons and legacy, 1945–1956

Late in 1945 Riesz took the chair of mathematics at the University of Budapest, the post that had eluded him in 1911. He received the Grand Prize of the Hungarian Academy in 1946 and honorary doctorates from Szeged (1946), Budapest (1950) and the Sorbonne (1954). From December 1947 to March 1948 he lectured at universities in Sweden and Denmark on a tour organised by Marcel. He became a corresponding member of the Paris Academy of Sciences in 1948 and an external member of the Bavarian Academy in 1954. The Hungarian state gave him the Kossuth Prize in 1949 and 1953. Among his last students in Budapest were Ákos Császár and János Aczél.

His final work was a book. Leçons d’analyse fonctionnelle, written with Béla Szőkefalvi-Nagy and published in 1952, gathered forty-five years of his methods: the Lebesgue integral built from step functions, Stieltjes integrals, integral equations, compact and self-adjoint operators, spectral theory, and his own proofs throughout. Translated into English as Functional Analysis in 1955 and into many other languages, it taught the subject to a generation and remains in print.

Riesz spent his last six months in a sanatorium in Budapest and died there on 28 February 1956, aged 76. He was buried in the Kerepesi Cemetery after a small family funeral in the Reformed Church.

Legacy

  • Function spaces: \(L^p\), its completeness and duality, and weak convergence. The completeness of \(L^2\) made the Lebesgue integral the standard integral of analysis.
  • Duality: the representation theorems for Hilbert spaces and for \(C(X)\) are the bridges between functional analysis and measure theory.
  • Operator theory: Riesz’s lemma and the Riesz theory of compact operators, the Riesz projection in spectral theory, and the methods behind Stefan Banach’s axiomatic theory.
  • Complex and harmonic analysis: Hardy spaces, the F. and M. Riesz theorem, subharmonic functions and potential theory.
  • Institutions: the Bolyai Institute in Szeged and Acta Scientiarum Mathematicarum, both still active.

DiscussionRiesz published several of his most important papers while teaching in secondary schools, far from any research centre. What did this setting give him and what did it cost? Compare his position with that of Fredholm in Stockholm, who worked part-time as an actuary.

Final projectRiesz’s theorem on compact operators (Lesson 07) and Fredholm’s determinant theory prove the same alternative by very different means. Write a short essay that proves the alternative for the kernel \(K(x,y)=\min(x,y)(1-\max(x,y))\) on \([0,1]\) in both ways, and explain what each proof reveals that the other hides.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Frigyes Riesz, MacTutor History of Mathematics.
  2. Frigyes Riesz, Wikipedia.
  3. Frigyes (Frédéric) Riesz, Mathematics Genealogy Project.
  4. B. Szőkefalvi-Nagy, Riesz Frigyes, a memoir of his life and work, in F. Riesz, Œuvres complètes, ed. Á. Császár, Akadémiai Kiadó, Budapest, 1960.
  5. Riesz Frédéric (1880–1956), Encyclopædia Universalis.

Mathematics

  1. F. Riesz, Sur les systèmes orthogonaux de fonctions, Comptes rendus 144 (1907).
  2. E. Fischer, Sur la convergence en moyenne, Comptes rendus 144 (1907).
  3. F. Riesz, Sur les opérations fonctionnelles linéaires, Comptes rendus 149 (1909).
  4. F. Riesz, Untersuchungen über Systeme integrierbarer Funktionen, Mathematische Annalen 69 (1910).
  5. F. Riesz, Über lineare Funktionalgleichungen, Acta Mathematica 41 (1918).
  6. F. Riesz and M. Riesz, Über die Randwerte einer analytischen Funktion, Quatrième Congrès des Mathématiciens Scandinaves, Stockholm, 1916.
  7. F. Riesz, Sur les fonctions subharmoniques et leur rapport à la théorie du potentiel, Acta Mathematica 48 (1926) and 54 (1930).
  8. F. Riesz, Sur un théorème de maximum de MM. Hardy et Littlewood, Journal of the London Mathematical Society 7 (1932).
  9. F. Riesz, Some mean ergodic theorems, Journal of the London Mathematical Society 13 (1938).
  10. F. Riesz, Sur quelques notions fondamentales dans la théorie générale des opérations linéaires, Annals of Mathematics 41 (1940).
  11. F. Riesz and B. Szőkefalvi-Nagy, Leçons d’analyse fonctionnelle, Akadémiai Kiadó, 1952; English translation Functional Analysis, Ungar, 1955; Dover reprint.
  12. J. Dieudonné, History of Functional Analysis, North-Holland, 1981.
  13. Riesz–Markov–Kakutani representation theorem and F. and M. Riesz theorem, Wikipedia.