Real-Analytic Non-Integrable Functions on the Plane with Equal Iterated Integrals

Results Math (2022) 77:30 c 2021 The Author(s) 1422-6383/22/010001-12 published online December 16, 2021 https://doi.org/10.1007/s00025-021-01571-7 Results in Mathematics Real-Analytic Non-Integrable Functions on the Plane with Equal Iterated Integrals Luis Bernal-Gonz´alez, Mar´ıa del Carmen Calder´on-Moreno ,and Andreas Jung Abstract. In this note, a vector space of real-analytic functions on the plane is explicitly constructed such that all its nonzero functions are nonintegrable but yet their two iterated integrals exist as real numbers and coincide. Moreover, it is shown that this vector space is dense in the space of all real continuous functions on the plane endowed with the compactopen topology. Mathematics Subject Classification. (MSC 2020) 26E05, 26B15, 28A35, 46B87. Keywords. Fubini’s theorem, Real analytic functions, Iterated integrals. 1. Introduction Fubini’s theorem asserts that if a real function f, which is defined on a measure space X×Ythat is the product of two σ-finite product spaces, is integrable, then its two iterated integrals exist as real numbers and coincide, and in fact their common value is the integral of fon the product space; see e.g. [8, Chapter 17]. Throughout this paper we shall consider the real line Ras endowed with the Lebesgue measure dx, so that the plane R2=R×Ris equipped with the bidimensional Lebesgue measure dxdy, which is nothing but the product measure of dx with itself. This research has been supported by the Plan Andaluz de Investigaci´on de la Junta de Andaluc´ıa FQM-127 Grant P20-00637 and by MCINN Grant PGC2018-098474-B-C21. 30 Page 2 of 12 L. Bernal-Gonz´alez et al. Results Math Of course, Fubini’s theorem is only a sufficient criterion, so that it is not too difficult to provide examples of non-integrable functions R2→Rsuch that the iterated integrals exist as real numbers and share the same value. For instance, if Ik:= (k,k+1) (k∈N:= {1,2,...}), then the function f:R2→R given by f(x, y)=⎧ ⎨ ⎩ 1if(x, y)∈I2 1∪n∈N[(I2n×I2n+1)∪(I2n+1 ×I2n)] −1if(x, y)∈n∈N[(I2n×I2n−1)∪(I2n−1×I2n)] 0 otherwise, (1) is Lebesgue-measurable and satisfies RR f(x, y)dxdy =R 0dy =0=R 0dx =RR f(x, y)dydx, but it is not integrable on R2because, as |f|= 1 on a set of infinite area, we have R2|f(x, y)|dxdy =+∞. In the much more general setting of the product X×Yof two measurable spaces, the existence of measurable functions X×Y→Rsatisfying analogous properties –as well as the size of the set of them– is analyzed in [3]. A natural question is whether such “pseudo-Fubini functions” fcan be found to be continuous on R2. This is, again, not too difficult: for instance, just replace “1” and “−1” in (1) by, respectively, ϕm(x)ϕp(y)and−ϕm(x)ϕp(y) on each product Im×Ip(m, p ∈N), where ϕk(t):=t−kif t∈(k,k +1 2] k+1−tif t∈(k+1 2,k+1) (2) for every k∈N. Going one step further, we may wonder whether fcan even be chosen to live in the class C∞(R2)ofinfinitely differentiable functions R2→R.The answer is, again, affirmative: the construction of such an fis as in the preceding paragraph, with the sole exception that each function ϕk:(k,k +1) →R (k∈N)in(2) is this time given by ϕk(t)=e1 (t−k)(t−k−1) . Then the following questions arises: Are there real-analytic pseudo-Fubini functions on R2? Recall, if Ωis an open subset of R2, then a function f:Ω→ Ris said to be real-analytic on Ωwhenever, given (x0,y 0)∈Ω, there are an neighborhood Uof (x0,y 0) with U⊂Ωand a double sequence {ajk}j,k≥0⊂R such that f(x, y)=j,k≥0ajk(x−x0)j(y−y0)kfor every (x, y)∈U,the convergence being absolute. The set of all such functions will be denoted by Cω(Ω). Note that, due to the Analytic Continuation Principle, the piece-by- piece construction of the last three paragraphs is not possible in the analytic case. The reader is referred to [4, Chap. 4] for background on real or complex analytic functions of several variables. Vol. 77 (2022) Real-Analytic Non-Integrable Functions... Page 3 of 12 30 The aim of this short note is to show that not only such real-analytic functions exist but also that there is a plethora of them. Specifically, we shall construct a vector space of real-analytic functions in the plane all of whose nonzero members are non-integrable but satisfy that both iterated integrals exist as real numbers and share the same value. Moreover, this vector space has the dimension of the continuum and is dense in the space of all continuous functions in the plane when endowed with the topology of uniform convergence on compacta. The problem of replacing the class of analytic functions by the more stringent one of entire functions is also considered. 2. Construction of a Special Class of Real-Analytic Functions In our construction we shall consider the vector space C(Ω) of continuous functions Ω→R, where Ωis a nonempty open subset of R2. Note that Cω(Ω)⊂C∞(Ω)⊂C(Ω). Recall that C(Ω) becomes a separable complete metrizable topological vector space when it is endowed with the compact-open topology, that is, the topology of uniform convergence on compact subsets of Ω(see e.g. [6, Chapter 2]). For instance, a translation-invariant distance generating this topology is given by d(f,g):=sup n∈N min{1/n, sup Bn |f−g|}, where Bnis the Euclidean closed ball centered at the origin with radius n. It is also well-known (an easy proof follows, for instance, from the Baire category theorem) that dim C(Ω)=c, the cardinality of the continuum. Hence, cis the maximal value that one can expect for the dimension of a vector subspace of C(Ω). Our results concerning pseudo-Fubini analytic functions are collected in the next theorem. Note that it tells us that there is a plethora of such functions, in both topological and algebraic senses. Theorem 1. There is a family Aof functions R2→Renjoying the following properties: (a) A⊂Cω(R2). (b) For every f∈A both iterated Lebesgue integrals RRf(x, y)dxdy, RRf(x, y)dydx exist as real numbers and have the same value. (c) No nonzero function in Ais Lebesgue-integrable on R2. (d) Ais a c-dimensional vector space. (e) Ais dense in C(R2). 30 Page 4 of 12 L. Bernal-Gonz´alez et al. Results Math Proof. Our first task is to locate an analytic pseudo-Fubini function on R2. We start with the function f:(x, y)∈(0,1)2−→ x−y (x+y)3∈R. Clearly, f∈Cω((0,1)2). Now, we practice a number of elementary calculations. If we fix y∈(0,1) then we have 1 0 f(x, y)dx =1 0 (x+y)−2y (x+y)3dx =1 0(x+y)−2−2y(x+y)−3dx =−(x+y)−1+y(x+y)−2x=1 x=0 =−1 1+y+1 y+y (1 + y)2−1 y=−1 (1 + y)2. Therefore 1 01 0 f(x, y)dxdy =1 0 −1 (1 + y)2dy =1 1+yy=1 y=0 =1 2−1 1=−1 2. Since f(x, y)=−f(y,x) for all (x, y)∈(0,1)2, by exchanging the roles of x and ywe obtain 1 01 0 f(x, y)dydx =−1 01 0 f(y,x)dydx =−−1 2=1 2. Next, we consider the function g:(0,1)2−→ Rdefined as g(x, y)=f(x, y)−f(1 −x, 1−y)=(x−y)·(x+y)−3+(2−x−y)−3. (3) Again, g∈Cω((0,1)2). Let Φ(y):=1 0f(x, y)dx for y∈(0,1). Now, observe that by applying successively the changes of variables x→ 1−xand y→ 1−y we get 1 01 0 f(1 −x, 1−y)dxdy =1 01 0 f(x, 1−y)dxdy =1 0 Φ(1 −y)dy =1 0 Φ(y)dy =1 01 0 f(x, y)dxdy. It follows from (3) that 1 01 0 g(x, y)dxdy = 0 (4) and, analogously, Vol. 77 (2022) Real-Analytic Non-Integrable Functions... Page 5 of 12 30 1 01 0 g(x, y)dydx =0.(5) Next, let us observe that the mappings x∈(0,1) −→ xα∈(0,1) (α≥1) and ϕ:s∈R−→ 1 2+1 π·arctan s∈(0,1) are analytic diffeomorphisms. Since analyticity is stable under composition, we get that each mapping ϕα:s∈R−→ (ϕ(s))α∈(0,1) (α≥1) is an analytic diffeomorphism. In order to use the theorem of change of variables for one-dimensional or multidimensional integrals (see, e.g., [5, Chapter 3]), we remark that ϕ α(s)=α·ϕα−1(s) π·(1+s2)(s∈R), and that the mapping Ψ:(s, t)∈R2−→ (ϕ(s),ϕ(t)) ∈(0,1) ×(0,1) is a diffeomorphism whose Jacobian determinant is det JΨ(s, t)= 1 π2(1+s2)(1+t2). For every fixed α∈[1,+∞), define the function fα:R2→Rby fα(s, t):=ϕα−1(s)ϕα−1(t)g(ϕα(s),ϕ α(t)) (1 + s2)(1 + t2).(6) We shall need for our purposes a subset of integrable real-analytic functions that is dense in C(R2). Consider the countable set D:= {P·G1:Pis a polynomial of two real variables with coefficients in Q}, where Qis the set of rational numbers and G1is the Gaussian function G1(x, y)=e−x2−y2, that is integrable on the plane (with integral equal to π) together with every function Gα(x, y):=e−α(x2+y2)with α>0. Plainly, D⊂Cω(R2). On the one hand, each function h:= P·G1∈D is integrable. Indeed, since Pis a polynomial, there are a constant M∈(0,+∞)andanN∈N such that |P(x, y)|≤M(1 + (x2+y2)N) for all (x, y)∈R2. And since limt→+∞M(1+tN) et/2= 0, we derive the existence of a constant γ∈(0,+∞) such that |P(x, y)|≤γ·ex2+y2 2on R2, and therefore |h(x, y)|≤γ·e−x2−y2 2 on R2. The comparison principle yields the integrability of h. In particular, Fubini’s theorem implies that RR h(x, y)dxdy = R2 h(x, y)dxdy =RR h(x, y)dydx (7) for all h∈D. On the other hand, it follows from the Stone–Weierstrass theorem in its version for completely regular spaces (see [7, Theorem 16.5.7]; note that, since R2is metrizable, it is completely regular), that the algebra generated by the functions (x, y)→ x, (x, y)→ yis dense in C(R2). But this algebra is the set of all polynomials of two real variables. Fix a function F∈C(R2), as well as an ε>0 and a compact set K⊂R2. Select a polynomial P0with |P0−F G1|<ε 2on K. Now, pick a polynomial Phaving rational coefficients such that |P−P0|<ε 2on K. This together with the triangle inequality and 30 Page 6 of 12 L. Bernal-Gonz´alez et al. Results Math the fact |G1|≤1 yields |h−F|<ε on K, where h:= P·G1∈D.Wehave proved the denseness of Din C(R2). Our next step is to construct the desired family Asatisfying properties (a) to (e). With this aim, put D={hn:n∈N}, which is possible because of the countability of D. Select any sequence {cn}n∈Nof pairwise different points of [1,2]. For each n∈N, the fact that C(R2) is a topological vector space implies the existence of a constant μn>0 such that d(μnfcn,0) <1/n, where fcnis defined by (6). Now, we define A:= span {hn+μnfcn:n∈N}∪{fα:α>2}.(8) Observe first that Ais a vector subspace of C(R2). In addition, each fα with α≥1 and each hnis analytic, which yields A⊂Cω(R2), that is (a). Moreover, since {hn}n∈Nis dense in C(R2)and 0≤d(hn,h n+μnfcn)=d(μnfcn,0) <1/n −→ 0(n→∞), one easily derives that the set {hn+μnfcn}n∈Nis also dense in C(R2). But A⊃{hn+μnfcn}n∈N, which shows that Ais dense in C(R2), that is (e). Now, our aim is to prove that any nonzero function F∈A is not integrable on R2.By(8), for such a function there are respective finite subsets F1⊂N,F 2⊂(2,+∞) with F1∪F2=∅as well as scalars βi,δ j∈R\{0} (i∈F1,j∈F2) satisfying F=i∈F1βi(hi+μifci)+j∈F2δjfj, where any of the sums i∈F1or j∈F2is understood as 0 if, respectively, F1=∅or F2=∅. After relabeling, we get F=h+ p  i=1 λifαi for certain p∈N,{λ1,...,λ p}⊂R\{0}and {α1<α 2<··· <α p}⊂ [1,+∞), where h∈D. Assume, by way of contradiction, that Fis integrable on R2. Since his integrable, this would yield the integrability of the function G:= p i=1 λifαi.Accordingto(6), we have G(s, t)= p  i=1 λi·ϕαi−1(s)ϕαi−1(t)g(ϕαi(s),ϕ αi(t)) (1 + s2)(1 + t2), where ϕγ(s)=ϕ(s)γ=1 2+1 π·arctan sγfor all γ≥0 and all s∈R.If we consider the diffeomorphism Ψ(s, t)=(ϕ(s),ϕ(t)), the change of variables theorem leads us to the integrability on (0,1)2of the function  G:(0,1)2→R given by  G(x, y)= p  i=1 λi·xαi−1yαi−1·g(xαi,yαi), Vol. 77 (2022) Real-Analytic Non-Integrable Functions... Page 7 of 12 30 which, according to (3), can be written as  G(x, y)= p  i=1 λi·xαi−1yαi−1 ·(xαi−yαi)·(xαi+yαi)−3+(2−xαi−yαi)−3. Trivially,  Gwould also be integrable on the square (0,1/2)2. But, since each function xαi−1yαi−1(xαi−yαi) (2−xαi−yαi)3is continuous on the closed square [0,1/2]2,itis integrable on (0,1/2)2, and this implies that the function H(x, y):= p  i=1 λi·xαi−1yαi−1(xαi−yαi) (xαi+yαi)3 is also integrable on (0,1/2)2. For every c∈(0,1/2) we define the triangular set Sc:= (x, y)∈R2:0<x<1 2and cx < y < 2cx. Observe that Scis open (so Lebesgue-measurable) and S⊂(0,1/2)2. Then His integrable on Sc. Letting ψα(x, y):=xα−1yα−1(xα−yα) (xα+yα)3we are allowed to write H= p  i=1 λi·ψαi. Moreover, for every (x, y)∈Scand every α≥1, it is easy to check that 0<1 8·cα−1(1 −2αcα)·x−2≤ψα(x, y)≤2α−1cα−1·x−2.(9) We assume that p>1 (the case p= 1 is easier to deal with). Let us define the functions μ, ν :(0,1/2) →Rby μ(c)=|λ1|·1 8·cα1−1(1 −2α1cα1)andν(c)= p  j=2 |λj|2αj+2cαj−α1 |λ1|(1 −2α1cα1).(10) Observe that μ(c)>0 for all c∈(0,1/2). In addition, it follows from 0 < α1<α j(j=2,...,p) that lim c→0+ν(c)=0. 30 Page 8 of 12 L. Bernal-Gonz´alez et al. Results Math In particular, we can select a value c∈(0,1/2) such that ν(c)<1 2.From(9), (10) and the triangle inequality we obtain that for every (x, y)∈Sc(with c chosen as before) the following holds: |H(x, y)|= p  i=1 λi·ψαi(x, y) ≥|λ1|·ψα1(x, y)·1− p  i=2 |λi|·ψαi(x, y) |λ1|·ψα1(x, y) ≥μ(c)·x−2·1−ν(c)≥μ(c) 2·x−2. Recall that the conclusion of Fubini’s theorem holds for nonnegative measurable functions. Therefore Sc |H(x, y)|dxdy ≥ Sc μ(c) 2·x−2dxdy =μ(c) 2·1/2 02cx cx x−2dydx =μ(c) 2·1/2 0 x−2(2cx −cx)dx =cμ(c) 2·1/2 0 x−1dx =+∞, and so His not integrable on Sc, which is absurd. This contradiction shows that Fis not integrable on R2, which is (c). Note that the same argument (taking h=0 and αi∈(2,+∞)) yields the linear independence of the functions fα(α>2). But these functions belong to A. Consequently, dim (A)=c, that is (d). Concerning the property (b), consider a function F∈A. Again, F has the shape F=h+p i=1 λifαifor certain p∈N,{λ1,...,λ p}⊂R, {α1,α 2,...,α p}⊂[1,+∞), and h∈D.The λi’s that are zero make no influence in the calculations, so that we can assume λi= 0 for all i=1,...,p. Let us denote L(y):=1 0 g(x, y)dx for y∈(0,1). It follows from (4), (5), (6), the linearity of the integral and the change of variable formula for integrals on R(twice applied) that both iterated integrals for Fexist as real numbers and RR F(s, t)dsdt =RR h(s, t)dsdt + p  i=1 λiRR fαi(s, t)dsdt =RR h(s, t)dsdt + p  i=1 λiπ2 α2 iR ϕ αi(t)R ϕ αi(s)g(ϕαi(s),ϕ αi(t)) dsdt Vol. 77 (2022) Real-Analytic Non-Integrable Functions... Page 9 of 12 30 =RR h(s, t)dsdt + p  i=1 λiπ2 α2 iR ϕ αi(t)1 0 g(x, ϕαi(t)) dxdt =RR h(s, t)dsdt + p  i=1 λiπ2 α2 iR ϕ αi(t)L(ϕαi(t)) dt =RR h(s, t)dsdt + p  i=1 λiπ2 α2 i1 0 L(y)dy =RR h(s, t)dsdt + p  i=1 λiπ2 α2 i1 01 0 g(x, y)dxdy =RR h(s, t)dsdt + p  i=1 λiπ2 α2 i ·0=RR h(s, t)dsdt, and analogously, RR F(s, t)dtds =RR h(s, t)dtds. An application of (7) gives RRF(s, t)dsdt =RRF(s, t)dtds,which is (b). The theorem is proved.  3. Final Remarks 1. Under the terminology of the modern theory of lineability (see [2] for background), what has been proved in Theorem 1is that the set of non-integrable real analytic functions on R2having finite equal iterated integrals is maximal dense-lineable in C(R2). A slight shortening of the proof could have been done by using [2, Theorem 7.3.1], but our approach has the virtue of being totally constructive. 2. The existence of real entire functions –that is, of functions f:R2→Rfor which an absolutely convergent expansion f(x, y)=j,k≥0ajkxjykis valid on the whole plane– that are Lebesgue-integrable is well known. The simplest instance of such functions is maybe f(x, y)=e−x2−y2, which, incidentally, has been used in the proof of Theorem 1. Note that every entire function is in Cω(R2). This raises in a natural way the problem –which we pose here– of existence of non-integrable real entire functions whose iterated Lebesgue integrals exist as real numbers and coincide. If the answer were affirmative, the study of the size of the set of such functions would also be interesting. 3. As a first step towards the solution of the problem posed in the preceding remark, we have been able to find a non-Lebesgue-integrable entire function f:R2→Rsuch that both iterated integrals do exist in the Riemann sense (but not in the Lebesgue one) and coincide. Indeed, consider the function f(x, y):=sin(x2+y2), which is entire because both functions t∈R→ sin t∈R and (x, y)∈R2→ x2+y2∈Rare. Then fis not Lebesgue-integrable