[EN] In this short note we show that there exist uncountably generated alge- bras every non-zero element of which is a smooth function having uncount- ably many zeros. This result complements some recent ones by Enflo et al. [7, 9].The authors would like to thank the anonymous referee, whose thorough analysis and insightful remarks improved the text. The authors were supported by CNPq Grant 401735/2013-3 (PVE - Linha 2), MTM2012-34341, MEC Project MTM2013-47093-P, and Programa de Investigacion y Desarrollo de la UPV, Referencia SP2012070. The third author is also supported by a FPU grant of MEC Project MTM2010-14909.Conejero, JA.; Muñoz-Fernández, GA.; Murillo Arcila, M.; Seoane Sepúlveda, JB. (2015). Smooth functions with uncountably many zeros. Bulletin of the Bengian Mathematical Society. 22:1-5. http://hdl.handle.net/10251/64844152

Conejero, J. Alberto

Murillo Arcila, Marina

Muñoz-Fernández, Gustavo A.

Seoane Sepúlveda, Juan Benigno

Bulletin of the Belgian Mathematical Society - Simon Stevin

English

RiuNet

 Document downloaded from:  This paper must be cited as:                  The final publication is available at   Copyright  Additional Information  http://hdl.handle.net/10251/64844Belgian Mathematical SocietyConejero, JA.; Muñoz-Fernández, GA.; Murillo Arcila, M.; Seoane Sepúlveda, JB. (2015).Smooth functions with uncountably many zeros. Bulletin of the Bengian MathematicalSociety. 22:1-5. http://hdl.handle.net/10251/64844.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROSJ. A. CONEJERO, G. A. MUÑOZ-FERNÁNDEZ, M. MURILLO ARCILA,AND J.B. SEOANE-SEPÚLVEDAAbstract. In this short note we show that there exist uncountably generatedalgebras every non-zero element of which is a smooth function having uncount-ably many zeros. This result complements some recent ones by Enflo et al.[7, 9].As it nowadays is common terminology, a subset M of a topological vectorspace X is called lineable (respectively, spaceable) in X if there exists an infi-nite dimensional linear space (respectively, infinite dimensional closed linear space)Y ⊂ M ∪ {0}. Recently there have been several results regarding the linear struc-ture of certain subsets of real functions having a large set of zeros. For instance,in [9], Enflo et al. proved that, for every infinite dimensional closed subspace X ofC[0, 1], the set of functions in X having infinitely many zeros in [0, 1] is spaceablein X. Also, in [7], Conejero et al. constructed an algebra of functions A enjoyingthe following properties: (i) A is uncountably infinitely generated (that is, the car-dinality of a minimal system of generators of A is uncountable), (ii) every nonzeroelement of A is nowhere analytic, (iii) A ⊂ C∞(R), (iv) every element of A hasinfinitely many zeros in R, and (v) for every f ∈ A \ {0} and n ∈ N, f (n) (then-th derivative of f) enjoys the same properties as the elements in A \ {0}. Also,let us recall the notion of algebrability (see, e.g. [1–5, 10]). Given an algebra A, asubset B ⊂ A, and a cardinal number κ, we say that B is: (i) algebrable if thereis a subalgebra C of A so that C ⊂ B ∪ {0} and the cardinality of any system ofgenerators of C is infinite. (ii) κ-algebrable if there exists a κ-generated subalgebraC of A with C ⊂ B ∪ {0}. (iii) strongly κ-algebrable if there exists a κ-generatedfree algebra C contained in B ∪ {0}.On a totally different framework, and somehow related to the study of the setof zeros of functions on a given interval, Aron and Gurariy in 2003, asked whetherthere exists an infinite dimensional subspace of `∞ every non-zero element of whichhas a finite number of zeros. This question was recently answered, in the negative,in [6].Let us also recall that both of the results from [7, 9] share a common ground:The cardinality of the considered set of zeros was always countable. Of course, bymeans of a Baire category argument (as seen in [8]) one can show that almost everycontinuous function having zeros has, actually, an uncountable amount of them.2010 Mathematics Subject Classification. 15A03, 26B05.Key words and phrases. lineability, spaceability, algebrability, zero set.The first author was supported by Programa de Investigación y Desarrollo de la UPV (ref.SP20120700) and by MTM2013-47093-P. The third author was supported by MTM2013-47093-Pand by a grant from the FPU program of MEC. The second and fourth authors were supportedby MTM2012-34341.12 CONEJERO, MUÑOZ, MURILLO, AND SEOANE−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.050.10.150.20.25Figure 1. Sketch of d(x) on [−1, 1].In this short note we complement the previously mentioned results by proving,constructively, the following:Theorem. The subset of smooth functions in R having a uncountable set ofzeros is strongly c−algebrable.Let us start by fixing Z ⊂ R with |Z| = c and a function 0 6= f ∈ C∞(R) suchthat f(z) = 0 for every z ∈ Z and f does not have horizontal asymptotes. Such afunction can be defined as follows. Let C be a copy of the Cantor set in the interval[−1, 1]. Observe that [−1, 1] \ C =⋃n In, where the In’s are pairwise disjoint openintervals. Now define the function d : [−1, 1]→ R asd(x) ={kan,bn · (x− an)(bn − x) if x /∈ C, and x ∈ In = (an, bn) for some n,0 if x ∈ C,where kan,bn is a positive constant depending on an and bn. Next, let g be thefunction, on [−1, 1], given by:g(x) ={0 if x ∈ C,e−1/d(x) if x /∈ C.The value of the constant kan,bn does not affect at all the smoothness of g. Forinstance, in Figures 1 and 2 we used kan,bn = 1/(bn − an)1.8. This constant playsthe role of a “scaling factor”.We leave as an exercise to the reader to check that g is smooth. Next, we candefine our function f : R→ R by extending g in a usual way by making it smoothon R and by making it not have horizontal asymptotes.Let us go back to our main construction now. Let H be a Hamel basis of R as aQ−vector space such that all elements in H are positive. Also, let (for r ∈ H andx ∈ R),fr(x) = erx sin(f(x)).Our aim is to show that the algebra generated by the fr’s, A = A(fr : r ∈ H), isuncountably generated and that every element in A has an uncountable set of zeros.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 3−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.0050.010.014Figure 2. Sketch of g(x) on [−1, 1].In order to do so, let k ∈ N, P ∈ R[z1, z2, . . . , zk] be any non-constant polynomialin k real variables, and r1, r2, . . . , rk ∈ H. Now we need to see that:(i.-) φ(z) := P (fr1 , fr2 , . . . , frk)(z) = 0 for every z ∈ Z.(ii.-) The algebra A is c−generated.First, notice that, since P can be written asP (z1, . . . , zk) =q∑j=1aj · zn1,j1 · . . . · znk,jk ,with q ∈ N, {ni,j : 1 ≤ i ≤ k, 1 ≤ j ≤ q} ⊂ N, and aj ∈ R \ {0} for everyj ∈ {1, . . . , q}, then φ can be expressed asφ(z) =q∑j=1aj · (sin f(z))∑ki=1 ni,j · e∑ki=1(rini,jz) =q∑j=1aj · (sin f(z))mj · ezsj ,where mj =∑ki=1 ni,j and sj =∑ki=1 rini,j for j ∈ {1, . . . , q}.Once we have that, it is straightforward to check that φ(z) = 0 for every z ∈ Z.Next, let us check some properties of the sj ’s that appear in the expression of φ.First of all, notice that sj 6= 0 for every j ∈ {1, . . . , q}. Indeed, suppose that forsome j ∈ {1, . . . , q} we have sj = 0. Then, it would ber1n1,j + r2n2,j + r3n3,j + · · ·+ rknk,j = 0,which contradicts the fact that H is a Hamel basis. Similarly it can be also shownthat si 6= sj if i 6= j. Thus, we can assume without loss of generality, that s1 <s2 < · · · < sq.Now, let us show that the set {fr : r ∈ H} is algebraically independent. Toachieve this, suppose that φ ≡ 0, we shall show that aj = 0 for every j ∈ {1, . . . , q}.This will amount to P ≡ 0, and we will be done.4 CONEJERO, MUÑOZ, MURILLO, AND SEOANEIf φ ≡ 0, then we would have thatφ(z)es1z= a1(sin z)m1 +q∑j=2aj · (sin z)mj · ez(sj−s1)is also 0 for every z ∈ R.Let, now, take the limit when z → −∞. Then, we have that0 = limz→−∞a1(sin f(z))m1 +q∑j=2aj · limz→−∞(sin f(z))mj · ez(sj−s1)= a1 · limz→−∞(sin f(z))m1 +q∑j=2aj · limz→−∞sin(f(z))mjez(sj−s1) == a1 · sin(limz→−∞f(z))mj+ 0,and we obtain that a1 = 0 (since f has no horizontal asymptotes). We can nowproceed similarly (dividing now the expression∑qj=2 aj · (sin z)mj · esjz by es2z andtaking again limits when z → −∞) and we would obtain that all the aj ’s are 0.Thus, P ≡ 0, the set {fr : r ∈ H} is algebraically independent, and we are done.Remark. Notice that this result is the best possible in terms of dimension, sincethe set of continuous functions has dimension c. Let us also recall that this con-struction can also be done using any other types of fractal sets with arbitrary fractaldimension. We chose the Cantor set for convenience.Acknowledgments. The authors would like to thank the anonymous referee,whose thorough analysis and insightful remarks improved the text. The authorswere supported by CNPq Grant 401735/2013-3 (PVE - Linha 2), MTM2012-34341,MEC Project MTM2013-47093-P, and Programa de Investigación y Desarrollo dela UPV, Referencia SP2012070. The third author is also supported by a FPU grantof MEC Project MTM2010-14909.References[1] R. M. Aron, D. Pérez-Garćıa, and J. B. Seoane-Sepúlveda, Algebrability of the set of non-convergent Fourier series, Studia Math. 175 (2006), no. 1, 83–90.[2] R. M. Aron and J. B. Seoane-Sepúlveda, Algebrability of the set of everywhere surjectivefunctions on C, Bull. Belg. Math. Soc. Simon Stevin 14 (2007), no. 1, 25–31.[3] A. Bartoszewicz and S. G la̧b, Strong algebrability of sets of sequences of functions, Proc.Amer. Math. Soc. 141 (2013), 827–835.[4] L. Bernal-González and M. Ordóñez Cabrera, Lineability criteria, with applications, J. Funct.Anal. 266 (2014), no. 6, 3997–4025, DOI 10.1016/j.jfa.2013.11.014.[5] L. Bernal-González, D. Pellegrino, and J. B. Seoane-Sepúlveda, Linear subsets of nonlinearsets in topological vector spaces, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 1, 71–130, DOI10.1090/S0273-0979-2013-01421-6.[6] D. Cariello and J. B. Seoane-Sepúlveda, Basic sequences and spaceability in `p spaces, J.Funct. Anal. 266 (2014), no. 6, 3797–3814, DOI 10.1016/j.jfa.2013.12.011.[7] J. A. Conejero, P. Jiménez-Rodŕıguez, G. A. Muñoz-Fernández, and J. B. Seoane-Sepúlveda,When the identity theorem “seems” to fail, Amer. Math. Monthly 121 (2014), no. 1, 60–68,DOI 10.4169/amer.math.monthly.121.01.060.[8] T. Domı́nguez Benavides, How many zeros does a continuous function have?, Amer. Math.Monthly 93 (1986), no. 6, 464–466, DOI 10.2307/2323473.[9] P. H. Enflo, V. I. Gurariy, and J. B. Seoane-Sepúlveda, Some results and open questions onspaceability in function spaces, Trans. Amer. Math. Soc. 366 (2014), no. 2, 611–625, DOI10.1090/S0002-9947-2013-05747-9.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 5[10] J. B. Seoane-Sepúlveda, Chaos and lineability of pathological phenomena in analysis, Pro-Quest LLC, Ann Arbor, MI, 2006. Thesis (Ph.D.)–Kent State University.Instituto Universitario de Matemática Pura y Aplicada,Universitat Politècnica de València,46022, València, Spain.E-mail address: aconejero@upv.esDepartamento de Análisis Matemático,Facultad de Ciencias Matemáticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: gustavo fernandez@mat.ucm.esInstituto Universitario de Matemática Pura y Aplicada,Universitat Politècnica de València,46022, València, Spain.E-mail address: mamuar1@posgrado.upv.esDepartamento de Análisis Matemático,Facultad de Ciencias Matemáticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: jseoane@mat.ucm.es

Smooth functions with uncountably many zeros

In this short note we show that there exist uncountably generated
algebras every non-zero element of which is a smooth function having uncountably
many zeros. This result complements some recent ones by En
o et al.The authors
were supported by CNPq Grant 401735/2013-3 (PVE - Linha 2), MTM2012-34341,
MEC Project MTM2013-47093-P, and Programa de Investigaci on y Desarrollo de
la UPV, Referencia SP2012070. The third author is also supported by a FPU grant
of MEC Project MTM2010-14909

Muñoz-Fernández, G. A.

murillo arcila, marina

Seoane Sepúlveda, Juan

Repositori Institucional de la Universitat Jaume I

SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROSJ. A. CONEJERO, G. A. MUN˜OZ-FERNA´NDEZ, M. MURILLO ARCILA,AND J.B. SEOANE-SEPU´LVEDAAbstract. In this short note we show that there exist uncountably generatedalgebras every non-zero element of which is a smooth function having uncount-ably many zeros. This result complements some recent ones by Enflo et al.[7, 9].As it nowadays is common terminology, a subset M of a topological vectorspace X is called lineable (respectively, spaceable) in X if there exists an infi-nite dimensional linear space (respectively, infinite dimensional closed linear space)Y ⊂ M ∪ {0}. Recently there have been several results regarding the linear struc-ture of certain subsets of real functions having a large set of zeros. For instance,in [9], Enflo et al. proved that, for every infinite dimensional closed subspace X ofC[0, 1], the set of functions in X having infinitely many zeros in [0, 1] is spaceablein X. Also, in [7], Conejero et al. constructed an algebra of functions A enjoyingthe following properties: (i) A is uncountably infinitely generated (that is, the car-dinality of a minimal system of generators of A is uncountable), (ii) every nonzeroelement of A is nowhere analytic, (iii) A ⊂ C∞(R), (iv) every element of A hasinfinitely many zeros in R, and (v) for every f ∈ A \ {0} and n ∈ N, f (n) (then-th derivative of f) enjoys the same properties as the elements in A \ {0}. Also,let us recall the notion of algebrability (see, e.g. [1–5, 10]). Given an algebra A, asubset B ⊂ A, and a cardinal number κ, we say that B is: (i) algebrable if thereis a subalgebra C of A so that C ⊂ B ∪ {0} and the cardinality of any system ofgenerators of C is infinite. (ii) κ-algebrable if there exists a κ-generated subalgebraC of A with C ⊂ B ∪ {0}. (iii) strongly κ-algebrable if there exists a κ-generatedfree algebra C contained in B ∪ {0}.On a totally different framework, and somehow related to the study of the setof zeros of functions on a given interval, Aron and Gurariy in 2003, asked whetherthere exists an infinite dimensional subspace of `∞ every non-zero element of whichhas a finite number of zeros. This question was recently answered, in the negative,in [6].Let us also recall that both of the results from [7, 9] share a common ground:The cardinality of the considered set of zeros was always countable. Of course, bymeans of a Baire category argument (as seen in [8]) one can show that almost everycontinuous function having zeros has, actually, an uncountable amount of them.2010 Mathematics Subject Classification. 15A03, 26B05.Key words and phrases. lineability, spaceability, algebrability, zero set.The first author was supported by Programa de Investigacio´n y Desarrollo de la UPV (ref.SP20120700) and by MTM2013-47093-P. The third author was supported by MTM2013-47093-Pand by a grant from the FPU program of MEC. The second and fourth authors were supportedby MTM2012-34341.12 CONEJERO, MUN˜OZ, MURILLO, AND SEOANE−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.050.10.150.20.25Figure 1. Sketch of d(x) on [−1, 1].In this short note we complement the previously mentioned results by proving,constructively, the following:Theorem. The subset of smooth functions in R having a uncountable set ofzeros is strongly c−algebrable.Let us start by fixing Z ⊂ R with |Z| = c and a function 0 6= f ∈ C∞(R) suchthat f(z) = 0 for every z ∈ Z and f does not have horizontal asymptotes. Such afunction can be defined as follows. Let C be a copy of the Cantor set in the interval[−1, 1]. Observe that [−1, 1] \ C = ⋃n In, where the In’s are pairwise disjoint openintervals. Now define the function d : [−1, 1]→ R asd(x) ={kan,bn · (x− an)(bn − x) if x /∈ C, and x ∈ In = (an, bn) for some n,0 if x ∈ C,where kan,bn is a positive constant depending on an and bn. Next, let g be thefunction, on [−1, 1], given by:g(x) ={0 if x ∈ C,e−1/d(x) if x /∈ C.The value of the constant kan,bn does not affect at all the smoothness of g. Forinstance, in Figures 1 and 2 we used kan,bn = 1/(bn − an)1.8. This constant playsthe role of a “scaling factor”.We leave as an exercise to the reader to check that g is smooth. Next, we candefine our function f : R→ R by extending g in a usual way by making it smoothon R and by making it not have horizontal asymptotes.Let us go back to our main construction now. Let H be a Hamel basis of R as aQ−vector space such that all elements in H are positive. Also, let (for r ∈ H andx ∈ R),fr(x) = erx sin(f(x)).Our aim is to show that the algebra generated by the fr’s, A = A(fr : r ∈ H), isuncountably generated and that every element in A has an uncountable set of zeros.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 3−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.0050.010.014Figure 2. Sketch of g(x) on [−1, 1].In order to do so, let k ∈ N, P ∈ R[z1, z2, . . . , zk] be any non-constant polynomialin k real variables, and r1, r2, . . . , rk ∈ H. Now we need to see that:(i.-) φ(z) := P (fr1 , fr2 , . . . , frk)(z) = 0 for every z ∈ Z.(ii.-) The algebra A is c−generated.First, notice that, since P can be written asP (z1, . . . , zk) =q∑j=1aj · zn1,j1 · . . . · znk,jk ,with q ∈ N, {ni,j : 1 ≤ i ≤ k, 1 ≤ j ≤ q} ⊂ N, and aj ∈ R \ {0} for everyj ∈ {1, . . . , q}, then φ can be expressed asφ(z) =q∑j=1aj · (sin f(z))∑ki=1 ni,j · e∑ki=1(rini,jz) =q∑j=1aj · (sin f(z))mj · ezsj ,where mj =∑ki=1 ni,j and sj =∑ki=1 rini,j for j ∈ {1, . . . , q}.Once we have that, it is straightforward to check that φ(z) = 0 for every z ∈ Z.Next, let us check some properties of the sj ’s that appear in the expression of φ.First of all, notice that sj 6= 0 for every j ∈ {1, . . . , q}. Indeed, suppose that forsome j ∈ {1, . . . , q} we have sj = 0. Then, it would ber1n1,j + r2n2,j + r3n3,j + · · ·+ rknk,j = 0,which contradicts the fact that H is a Hamel basis. Similarly it can be also shownthat si 6= sj if i 6= j. Thus, we can assume without loss of generality, that s1 <s2 < · · · < sq.Now, let us show that the set {fr : r ∈ H} is algebraically independent. Toachieve this, suppose that φ ≡ 0, we shall show that aj = 0 for every j ∈ {1, . . . , q}.This will amount to P ≡ 0, and we will be done.4 CONEJERO, MUN˜OZ, MURILLO, AND SEOANEIf φ ≡ 0, then we would have thatφ(z)es1z= a1(sin z)m1 +q∑j=2aj · (sin z)mj · ez(sj−s1)is also 0 for every z ∈ R.Let, now, take the limit when z → −∞. Then, we have that0 = limz→−∞ a1(sin f(z))m1 +q∑j=2aj · limz→−∞(sin f(z))mj · ez(sj−s1)= a1 · limz→−∞(sin f(z))m1 +q∑j=2aj · limz→−∞ sin(f(z))mjez(sj−s1) == a1 · sin(limz→−∞ f(z))mj+ 0,and we obtain that a1 = 0 (since f has no horizontal asymptotes). We can nowproceed similarly (dividing now the expression∑qj=2 aj · (sin z)mj · esjz by es2z andtaking again limits when z → −∞) and we would obtain that all the aj ’s are 0.Thus, P ≡ 0, the set {fr : r ∈ H} is algebraically independent, and we are done.Remark. Notice that this result is the best possible in terms of dimension, sincethe set of continuous functions has dimension c. Let us also recall that this con-struction can also be done using any other types of fractal sets with arbitrary fractaldimension. We chose the Cantor set for convenience.Acknowledgments. The authors would like to thank the anonymous referee,whose thorough analysis and insightful remarks improved the text. The authorswere supported by CNPq Grant 401735/2013-3 (PVE - Linha 2), MTM2012-34341,MEC Project MTM2013-47093-P, and Programa de Investigacio´n y Desarrollo dela UPV, Referencia SP2012070. The third author is also supported by a FPU grantof MEC Project MTM2010-14909.References[1] R. M. Aron, D. Pe´rez-Garc´ıa, and J. B. Seoane-Sepu´lveda, Algebrability of the set of non-convergent Fourier series, Studia Math. 175 (2006), no. 1, 83–90.[2] R. M. Aron and J. B. Seoane-Sepu´lveda, Algebrability of the set of everywhere surjectivefunctions on C, Bull. Belg. Math. Soc. Simon Stevin 14 (2007), no. 1, 25–31.[3] A. Bartoszewicz and S. G la¸b, Strong algebrability of sets of sequences of functions, Proc.Amer. Math. Soc. 141 (2013), 827–835.[4] L. Bernal-Gonza´lez and M. Ordo´n˜ez Cabrera, Lineability criteria, with applications, J. Funct.Anal. 266 (2014), no. 6, 3997–4025, DOI 10.1016/j.jfa.2013.11.014.[5] L. Bernal-Gonza´lez, D. Pellegrino, and J. B. Seoane-Sepu´lveda, Linear subsets of nonlinearsets in topological vector spaces, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 1, 71–130, DOI10.1090/S0273-0979-2013-01421-6.[6] D. Cariello and J. B. Seoane-Sepu´lveda, Basic sequences and spaceability in `p spaces, J.Funct. Anal. 266 (2014), no. 6, 3797–3814, DOI 10.1016/j.jfa.2013.12.011.[7] J. A. Conejero, P. Jime´nez-Rodr´ıguez, G. A. Mun˜oz-Ferna´ndez, and J. B. Seoane-Sepu´lveda,When the identity theorem “seems” to fail, Amer. Math. Monthly 121 (2014), no. 1, 60–68,DOI 10.4169/amer.math.monthly.121.01.060.[8] T. Domı´nguez Benavides, How many zeros does a continuous function have?, Amer. Math.Monthly 93 (1986), no. 6, 464–466, DOI 10.2307/2323473.[9] P. H. Enflo, V. I. Gurariy, and J. B. Seoane-Sepu´lveda, Some results and open questions onspaceability in function spaces, Trans. Amer. Math. Soc. 366 (2014), no. 2, 611–625, DOI10.1090/S0002-9947-2013-05747-9.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 5[10] J. B. Seoane-Sepu´lveda, Chaos and lineability of pathological phenomena in analysis, Pro-Quest LLC, Ann Arbor, MI, 2006. Thesis (Ph.D.)–Kent State University.Instituto Universitario de Matema´tica Pura y Aplicada,Universitat Polite`cnica de Vale`ncia,46022, Vale`ncia, Spain.E-mail address: aconejero@upv.esDepartamento de Ana´lisis Matema´tico,Facultad de Ciencias Matema´ticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: gustavo fernandez@mat.ucm.esInstituto Universitario de Matema´tica Pura y Aplicada,Universitat Polite`cnica de Vale`ncia,46022, Vale`ncia, Spain.E-mail address: mamuar1@posgrado.upv.esDepartamento de Ana´lisis Matema´tico,Facultad de Ciencias Matema´ticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: jseoane@mat.ucm.es

Murillo-Arcila, Marina

Repositori UJI

In this short note we show that there exist uncountably generated algebras every non-zero element of which is a smooth function having uncountably many zeros. This result complements some recent ones by Enflo et al. [7, 9].CNPqMEC ProjectPrograma de Investigacion y Desarrollo de la UPVFPU grant of MEC ProjectDepto. de Análisis Matemático y Matemática AplicadaFac. de Ciencias MatemáticasTRUEpu

Conejero, Jose A.

Arcila,, M.

Seoane-Sepúlveda, Juan B.

Docta Complutense

SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROSJ. A. CONEJERO, G. A. MUÑOZ-FERNÁNDEZ, M. MURILLO ARCILA,AND J.B. SEOANE-SEPÚLVEDAAbstract. In this short note we show that there exist uncountably generatedalgebras every non-zero element of which is a smooth function having uncount-ably many zeros. This result complements some recent ones by Enflo et al.[7, 9].As it nowadays is common terminology, a subset M of a topological vectorspace X is called lineable (respectively, spaceable) in X if there exists an infi-nite dimensional linear space (respectively, infinite dimensional closed linear space)Y ⊂ M ∪ {0}. Recently there have been several results regarding the linear struc-ture of certain subsets of real functions having a large set of zeros. For instance,in [9], Enflo et al. proved that, for every infinite dimensional closed subspace X ofC[0, 1], the set of functions in X having infinitely many zeros in [0, 1] is spaceablein X. Also, in [7], Conejero et al. constructed an algebra of functions A enjoyingthe following properties: (i) A is uncountably infinitely generated (that is, the car-dinality of a minimal system of generators of A is uncountable), (ii) every nonzeroelement of A is nowhere analytic, (iii) A ⊂ C∞(R), (iv) every element of A hasinfinitely many zeros in R, and (v) for every f ∈ A \ {0} and n ∈ N, f (n) (then-th derivative of f) enjoys the same properties as the elements in A \ {0}. Also,let us recall the notion of algebrability (see, e.g. [1–5, 10]). Given an algebra A, asubset B ⊂ A, and a cardinal number κ, we say that B is: (i) algebrable if thereis a subalgebra C of A so that C ⊂ B ∪ {0} and the cardinality of any system ofgenerators of C is infinite. (ii) κ-algebrable if there exists a κ-generated subalgebraC of A with C ⊂ B ∪ {0}. (iii) strongly κ-algebrable if there exists a κ-generatedfree algebra C contained in B ∪ {0}.On a totally different framework, and somehow related to the study of the setof zeros of functions on a given interval, Aron and Gurariy in 2003, asked whetherthere exists an infinite dimensional subspace of `∞ every non-zero element of whichhas a finite number of zeros. This question was recently answered, in the negative,in [6].Let us also recall that both of the results from [7, 9] share a common ground:The cardinality of the considered set of zeros was always countable. Of course, bymeans of a Baire category argument (as seen in [8]) one can show that almost everycontinuous function having zeros has, actually, an uncountable amount of them.2010 Mathematics Subject Classification. 15A03, 26B05.Key words and phrases. lineability, spaceability, algebrability, zero set.The first author was supported by Programa de Investigación y Desarrollo de la UPV (ref.SP20120700) and by MTM2013-47093-P. The third author was supported by MTM2013-47093-Pand by a grant from the FPU program of MEC. The second and fourth authors were supportedby MTM2012-34341.12 CONEJERO, MUÑOZ, MURILLO, AND SEOANE−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.050.10.150.20.25Figure 1. Sketch of d(x) on [−1, 1].In this short note we complement the previously mentioned results by proving,constructively, the following:Theorem. The subset of smooth functions in R having a uncountable set ofzeros is strongly c−algebrable.Let us start by fixing Z ⊂ R with |Z| = c and a function 0 6= f ∈ C∞(R) suchthat f(z) = 0 for every z ∈ Z and f does not have horizontal asymptotes. Such afunction can be defined as follows. Let C be a copy of the Cantor set in the interval[−1, 1]. Observe that [−1, 1] \ C =⋃n In, where the In’s are pairwise disjoint openintervals. Now define the function d : [−1, 1]→ R asd(x) ={kan,bn · (x− an)(bn − x) if x /∈ C, and x ∈ In = (an, bn) for some n,0 if x ∈ C,where kan,bn is a positive constant depending on an and bn. Next, let g be thefunction, on [−1, 1], given by:g(x) ={0 if x ∈ C,e−1/d(x) if x /∈ C.The value of the constant kan,bn does not affect at all the smoothness of g. Forinstance, in Figures 1 and 2 we used kan,bn = 1/(bn − an)1.8. This constant playsthe role of a “scaling factor”.We leave as an exercise to the reader to check that g is smooth. Next, we candefine our function f : R→ R by extending g in a usual way by making it smoothon R and by making it not have horizontal asymptotes.Let us go back to our main construction now. Let H be a Hamel basis of R as aQ−vector space such that all elements in H are positive. Also, let (for r ∈ H andx ∈ R),fr(x) = erx sin(f(x)).Our aim is to show that the algebra generated by the fr’s, A = A(fr : r ∈ H), isuncountably generated and that every element in A has an uncountable set of zeros.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 3−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 100.0050.010.014Figure 2. Sketch of g(x) on [−1, 1].In order to do so, let k ∈ N, P ∈ R[z1, z2, . . . , zk] be any non-constant polynomialin k real variables, and r1, r2, . . . , rk ∈ H. Now we need to see that:(i.-) φ(z) := P (fr1 , fr2 , . . . , frk)(z) = 0 for every z ∈ Z.(ii.-) The algebra A is c−generated.First, notice that, since P can be written asP (z1, . . . , zk) =q∑j=1aj · zn1,j1 · . . . · znk,jk ,with q ∈ N, {ni,j : 1 ≤ i ≤ k, 1 ≤ j ≤ q} ⊂ N, and aj ∈ R \ {0} for everyj ∈ {1, . . . , q}, then φ can be expressed asφ(z) =q∑j=1aj · (sin f(z))∑ki=1 ni,j · e∑ki=1(rini,jz) =q∑j=1aj · (sin f(z))mj · ezsj ,where mj =∑ki=1 ni,j and sj =∑ki=1 rini,j for j ∈ {1, . . . , q}.Once we have that, it is straightforward to check that φ(z) = 0 for every z ∈ Z.Next, let us check some properties of the sj ’s that appear in the expression of φ.First of all, notice that sj 6= 0 for every j ∈ {1, . . . , q}. Indeed, suppose that forsome j ∈ {1, . . . , q} we have sj = 0. Then, it would ber1n1,j + r2n2,j + r3n3,j + · · ·+ rknk,j = 0,which contradicts the fact that H is a Hamel basis. Similarly it can be also shownthat si 6= sj if i 6= j. Thus, we can assume without loss of generality, that s1 <s2 < · · · < sq.Now, let us show that the set {fr : r ∈ H} is algebraically independent. Toachieve this, suppose that φ ≡ 0, we shall show that aj = 0 for every j ∈ {1, . . . , q}.This will amount to P ≡ 0, and we will be done.4 CONEJERO, MUÑOZ, MURILLO, AND SEOANEIf φ ≡ 0, then we would have thatφ(z)es1z= a1(sin z)m1 +q∑j=2aj · (sin z)mj · ez(sj−s1)is also 0 for every z ∈ R.Let, now, take the limit when z → −∞. Then, we have that0 = limz→−∞a1(sin f(z))m1 +q∑j=2aj · limz→−∞(sin f(z))mj · ez(sj−s1)= a1 · limz→−∞(sin f(z))m1 +q∑j=2aj · limz→−∞sin(f(z))mjez(sj−s1) == a1 · sin(limz→−∞f(z))mj+ 0,and we obtain that a1 = 0 (since f has no horizontal asymptotes). We can nowproceed similarly (dividing now the expression∑qj=2 aj · (sin z)mj · esjz by es2z andtaking again limits when z → −∞) and we would obtain that all the aj ’s are 0.Thus, P ≡ 0, the set {fr : r ∈ H} is algebraically independent, and we are done.Remark. Notice that this result is the best possible in terms of dimension, sincethe set of continuous functions has dimension c. Let us also recall that this con-struction can also be done using any other types of fractal sets with arbitrary fractaldimension. We chose the Cantor set for convenience.Acknowledgments. The authors would like to thank the anonymous referee,whose thorough analysis and insightful remarks improved the text. The authorswere supported by CNPq Grant 401735/2013-3 (PVE - Linha 2), MTM2012-34341,MEC Project MTM2013-47093-P, and Programa de Investigación y Desarrollo dela UPV, Referencia SP2012070. The third author is also supported by a FPU grantof MEC Project MTM2010-14909.References[1] R. M. Aron, D. Pérez-Garćıa, and J. B. Seoane-Sepúlveda, Algebrability of the set of non-convergent Fourier series, Studia Math. 175 (2006), no. 1, 83–90.[2] R. M. Aron and J. B. Seoane-Sepúlveda, Algebrability of the set of everywhere surjectivefunctions on C, Bull. Belg. Math. Soc. Simon Stevin 14 (2007), no. 1, 25–31.[3] A. Bartoszewicz and S. G la̧b, Strong algebrability of sets of sequences of functions, Proc.Amer. Math. Soc. 141 (2013), 827–835.[4] L. Bernal-González and M. Ordóñez Cabrera, Lineability criteria, with applications, J. Funct.Anal. 266 (2014), no. 6, 3997–4025, DOI 10.1016/j.jfa.2013.11.014.[5] L. Bernal-González, D. Pellegrino, and J. B. Seoane-Sepúlveda, Linear subsets of nonlinearsets in topological vector spaces, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 1, 71–130, DOI10.1090/S0273-0979-2013-01421-6.[6] D. Cariello and J. B. Seoane-Sepúlveda, Basic sequences and spaceability in `p spaces, J.Funct. Anal. 266 (2014), no. 6, 3797–3814, DOI 10.1016/j.jfa.2013.12.011.[7] J. A. Conejero, P. Jiménez-Rodŕıguez, G. A. Muñoz-Fernández, and J. B. Seoane-Sepúlveda,When the identity theorem “seems” to fail, Amer. Math. Monthly 121 (2014), no. 1, 60–68,DOI 10.4169/amer.math.monthly.121.01.060.[8] T. Domı́nguez Benavides, How many zeros does a continuous function have?, Amer. Math.Monthly 93 (1986), no. 6, 464–466, DOI 10.2307/2323473.[9] P. H. Enflo, V. I. Gurariy, and J. B. Seoane-Sepúlveda, Some results and open questions onspaceability in function spaces, Trans. Amer. Math. Soc. 366 (2014), no. 2, 611–625, DOI10.1090/S0002-9947-2013-05747-9.SMOOTH FUNCTIONS WITH UNCOUNTABLY MANY ZEROS 5[10] J. B. Seoane-Sepúlveda, Chaos and lineability of pathological phenomena in analysis, Pro-Quest LLC, Ann Arbor, MI, 2006. Thesis (Ph.D.)–Kent State University.Instituto Universitario de Matemática Pura y Aplicada,Universitat Politècnica de València,46022, València, Spain.E-mail address: aconejero@upv.esDepartamento de Análisis Matemático,Facultad de Ciencias Matemáticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: gustavo fernandez@mat.ucm.esInstituto Universitario de Matemática Pura y Aplicada,Universitat Politècnica de València,46022, València, Spain.E-mail address: mamuar1@posgrado.upv.esDepartamento de Análisis Matemático,Facultad de Ciencias Matemáticas,Plaza de Ciencias 3,Universidad Complutense de Madrid,Madrid, 28040, Spain.E-mail address: jseoane@mat.ucm.es

https://riunet.upv.es/bitstream/10251/64844/3/CMMS_22_June_2014_REVISED.pdf

Smooth functions with uncountably many zeros

Abstract

Similar works

Full text

Available Versions

RiuNet

Repositori Institucional de la Universitat Jaume I

Repositori UJI

Docta Complutense