# Large Function Algebras with Certain Topological Properties

Authors: Artur Bartoszewicz; Szymon Głąb
Journal: Journal of Function Spaces
 (2015)

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## Abstract

LetF be a family of continuous functions defined on a compact interval. We give a sufficient condition so that F∪{0} contains a dense c-generated free algebra; in other words, F is densely c-strongly algebrable. As an application we obtain dense c-strong algebrability of families of nowhere Hölder functions, Bruckner-Garg functions, functions with a dense set of local maxima and local minima, and nowhere monotonous functions differentiable at all but finitely many points. We also study the problem of the existence of large closed algebras within F∪{0} where F⊂RX or F⊂CX. We prove that the set of perfectly everywhere surjective functions together with the zero function contains a 2c-generated algebra closed in the topology of uniform convergence while it does not contain a nontrivial algebra closed in the pointwise convergence topology. We prove that an infinitely generated algebra which is closed in the pointwise convergence topology needs to contain two valued functions and infinitely valued functions. We give an example of such an algebra; namely, it was shown that there is a subalgebra of RR with 2c generators which is closed in the pointwise topology and, for any function f in this algebra, there is an open set U such that f-1(U) is a Bernstein set.


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## Body

## 1. Introduction

The algebraic properties of sets of functions have been considered in analysis for many years. One direction of such research is finding the so-called maximal (additive, multiplicative, and so on) classes for certain families of functions. For example, it was proved in [1] that the maximal additive class for Darboux real functions is the set of all constant functions. Recently, a new point of looking on the largeness of sets of functions has appeared. One can call a set A, contained in some algebraic structure of functions, a big one if A (or A∪{0}) contains a large, nice substructure inside. The first papers written in this direction were [2–4] and then [5–7]. In these papers, the notions contained in the following definition can be found.Definition 1.
Letκ be a cardinal number.(1)
LetL be a vector space and let A⊂L. We say that A is κ-lineable if A∪{0} contains a κ-dimensional vector space.(2)
LetL be a Banach space and A⊂L. We say that A is spaceable if A∪{0} contains an infinite dimensional closed vector space.(3)
LetL be a commutative algebra and let A⊂L. We say that A is κ-algebrable if A∪{0} contains a κ-generated algebra B (i.e., the minimal cardinality of the set generating B equals κ).Bartoszewicz and Głąb in [8] introduced the notion of strong algebrability.Definition 2.
Letκ be a cardinal number, let L be a commutative algebra, and let A⊂L. One says that A is strongly κ-algebrable if A∪{0} contains a κ-generated free algebra.Let us observe that the notion of spaceability is not a fully algebraic property but it has a topological ingredient (we ask about the existence ofclosed subspace of given Banach space). Ciesielski et al. in [9] asked about the existence of large linear subspaces, closed in the pointwise or uniform convergence topology in RR or CC. So, following this way, one can define spaceability in linear topological spaces.Some authors were interested in searching for a large substructure with some other topological property, namely, dense lineability (or algebrability) of some classes of functions. For example, Bayart and Quarta in [10] proved that the set NH of all nowhere Hölder functions is densely ω-algebrable in C[0,1]. In [11], Bastin et al. proved that the set of all nowhere Gevrey functions is densely c-algebrable in C∞[0,1].The aim of our paper is to formulate, prove, and apply some techniques of constructing densec-generated free algebras in the space of continuous functions on a compact interval and to consider the possibility of the existence of closed algebras in some sets of real or complex functions.


## 2. Dense Strongc-Algebrability in C[0,1]

It is a simple observation that the set{x↦exp⁡(rx):r∈R} is linearly independent in RR. Moreover, if X⊂R is linearly independent over Q, then {x↦exp⁡(rx):r∈X} is the set of free generators. In [12] the authors, using the composition of a function with some needed properties with such an exponential function, proved the c-algebrability of the set CMm([0,1]) of continuous functions with dense sets of local extrema. Recently, this idea has been further developed in [13, 14].Let us call, after [13], a function f:R→R exponential-like of rank m whenever f is given by the formula f(x)=∑i=1m‍aiexp⁡(βix) for some pairwise distinct nonzero numbers β1,…,βm and some nonzero numbers a1,…,am. We have the following.Theorem 3 (see [13]).
LetF⊆R[0,1] and assume that there exists a function F∈F such that f∘F∈F∖{0} for every exponential-like function f:R→R. Then F is strongly c-algebrable. More exactly, if H⊆R is a set of cardinality c and linearly independent over the rationals Q, then exp⁡∘(rF), r∈H, are free generators of an algebra contained in F∪{0}.Using Stone-Weierstrass theorem, it is not difficult to observe that the algebra described in Theorem3 is dense in C[0,1] if and only if the function F is continuous and strictly monotonic. This argument is described in the last section of [14]. To illustrate this, consider the following two examples. Let F stand for the set of all continuous functions which are differentiable n-1 times but not differentiable n times at any point of their domains. Let F be the (n-1)th antiderivative of a strictly positive nowhere differentiable function. Then by [14, Theorem 4.5], the family F is densely c-strongly algebrable. In turn, using [14, Theorem 4.9] and a similar argument, one can prove that the set of all functions from C1, whose derivative is not α-Hölder (for any α∈(0,1]) at all but finitely many points, is densely c-strongly algebrable.However, for many classes of functions, the monotonic representative does not exist. Here we propose some method of construction of a dense algebra even ifF does not contain any monotonic function.


### 2.1. Nowhere Constant Continuous Functions

LetF:[a,b]→R be a continuous function. Then F is called left nondecreasing at x∈(a,b] if there is δ>0 such that F(y)≤F(x) for any y∈(x-δ,x). Analogously we define a left nonincreasing function at x∈(a,b] and right nondecreasing (nonincreasing) function at x∈[a,b). We say that x∈(a,b) is a point of local monotonicity, provided that F is left nondecreasing or left nonincreasing and F is right nondecreasing or right nonincreasing; see [15, 16]. Note that if x is a point of local minimum (local minimizer) or a point of local maximum (local maximizer) of F, then x is a point of local monotonicity. We say that F is nowhere constant, provided that its restriction to any open interval is not constant.Fix a functiong∈C[α,β] which is nowhere constant and such that α and β are points of (one-sided) monotonicity of g. For x∈[α,β], denote by H(x) the largest possible y∈[x,β] such that g(t) is between g(x) and g(y) for every t∈[x,y] (here by [x,x] we mean the singleton {x}). Such a number H(x) always exists by the continuity of g. Let x=H0(x) and inductively Hn+1(x)=H(Hn(x)) for x∈[α,β].Lemma 4.
Letn∈N. If Hn+1(α)<β, then(i)
Hn+1(α)>Hn(α);(ii)
Hn+1(α) is a point of local extremum of g;(iii)
Hn+1(α) is a local minimizer of g if and only if Hn(α) is a local maximizer of g.Proof.
Sinceα is a point of right local monotonicity of g, say g is right nondecreasing at α, then there is δ>0 such that g(α)≤g(y) for every y∈(α,α+δ). Let t∈(α,α+δ] such that g(t)=max⁡{g(y):y∈[α,α+δ]}. Since g is nowhere constant, then g(t)>g(α). Hence, H(α)≥t>α.
Now, we will show thatH(α) is a point of right local monotonicity of g. Suppose not, then by the definition of H, g(y)≥g(α) for y∈[α,H(α)]. Moreover g(H(α))>g(α). Let δ>0 such that gHα-gy<gHα-gα/2 for y∈(H(α),H(α)+δ]. Then g|[H(α),H(α)+δ] attains its maximum at some w∈[H(α),H(α)+δ]. Since g is not right nonincreasing at H(α), then g(H(α))<g(w). Moreover g(α)≤g(y)≤g(w) for any y∈[α,w]. This contradicts the definition of H.
Proceeding inductively we obtain thatHn+1(α)>Hn(α) or Hn+1(α)=β. Note that H1(α), H2(α),…,Hn(α) are local extrema of g. Moreover if Hi(α) is a local minimizer of g, then Hi+1(α) is a local maximizer of g and vice versa.Lemma 5.
There isn∈N such that Hn(α)=β.Proof.
Suppose thatHn(α)<β for every n∈N. By Lemma 4, the sequence (Hn(α))n∈N is strictly increasing. Let w=lim⁡n→∞Hn(α). If g is left nondecreasing at w, then there is δ>0 such that g(y)≤g(w) for every y∈[w-δ,w]. Let n be such that Hn(α) is a local minimizer of g with Hn(α)>w-δ. Then for any y∈(Hn(α),w) we have g(Hn(α))≤g(y)≤g(w) which contradicts the definition of Hn+1(α). In the same manner, we show that g is not left nonincreasing. Therefore w<β, since g is left monotonous at β.
Suppose now thatg is not right nondecreasing at w. Let v∈[w,β] be a minimizer of g on [w,β]. Then v>w and g(v)<g(w). Let δ>0 be such that gy-gw<(g(w)-g(v))/2 for every y∈(w-δ,w). Then fix n∈N such that Hn(α)∈(w-δ,w) is a local maximizer of g and g(Hk(α))≤g(Hn(α)) for k≥n. This is possible since lim⁡k→∞g(Hk(α))=g(w)<g(Hn(α)) and g is continuous. Therefore, g(Hn(α))≥g(y)≥g(v) for y∈(Hn(α),v), which contradicts the definition of Hn+1(α). Similarly one can prove that the assumption that g is not right nonincreasing at w also leads to contradiction. Hence, g is both right nondecreasing and right nonincreasing at w. This means that g is constant on [w,w+δ] for some positive δ, which contradicts the fact that g is nowhere constant. This shows that Hn(α)=β for some n∈N.Lemma 6.
LetF∈C[a,b] be nowhere constant and F(a)≤F(x)≤F(b) for any x∈[a,b]. Let ε>0. Then there is a partition a=x0<x1<x2<⋯<xn=b such that(i)
F(x) is between F(xk) and F(xk+1) for xk≤x≤xk+1 and k=0,1…,n-1;(ii)
the meshmax⁡{xi+1-xi:i=0,1,…,n-1} of the partition is smaller than ε.Proof.
Leta=t0<t1<t2<⋯<tm=b be any partition of [a,b] with the mesh smaller than ε. We will find a new partition a=v0<v1<v2<⋯<vk=b of [a,b] such that each interval [vi,vi+1) contains at most one tj and each vi is a point of local monotonicity of F. This new partition will also have a mesh smaller than ε. We construct it in the following way.
Ifti is a point of local monotonicity of F, then ti remains in the new partition. Otherwise, by the fact that F is nowhere constant the restriction F[ti-ε/3,ti] attains its minimum at some wL∈[ti-ε/3,ti] and maximum at some wL′∈[ti-ε/3,ti]. If one of the points wL,wL′ is in (ti-ε/3,ti), then it is a point of local monotonicity and we put it to the new partition. However, it may happen that {wL,wL′}={ti-ε/3,ti}; that is, wL and wL′ are the endpoints of the interval [ti-ε,ti]. We may assume that wL=ti-ε/3 and F(wL)<F(ti). Take any t∈(ti-ε/3,ti). If t is a point of local monotonicity of F, then we are done. Assume now that t is not a point of local monotonicity of F. This means that either t is not a point of left monotonicity of F or it is not a point of right monotonicity of F. We may assume that t is not a point of left monotonicity of F. Then, F attains its maximum on [wL,t] on some w∈(wL,t) and w is a both-sided monotonicity point of F; w is between ti-ε/3 and ti, and we put it to the new partition. Similarly one can find an appropriate both-sided monotonicity point in (ti,ti+ε/3) which we put into the new partition.
In the next step we will find a refinementa=x0<x1<x2<⋯<xn=b of a=v0<v1<v2<⋯<vk=b for which (i) holds true. To find such a refinement, for every i<k, we use Lemma 5 for the restriction g=F[vi,vi+1], α=vi, and β=vi+1.The assumption thatF is nowhere constant in Lemma 6 is essential. To see it, consider a function F given by
(1)F(x)=x,x∈1,2,xsinπ2x,x∈0,1,0,x∈-1,0,2x+2,x∈-2,-1.
Note that F(-2)≤F(x)≤F(2). For every partition -2=x0<x1<⋯<xn=2 with the mesh smaller than 1, there is the largest k with xk≤0. Then, F(xk)=0 and we may assume that F(xk+1)≥0. But there is x∈(0,xk+1) with F(x)<0, which means that the assertion of Lemma 6 does not hold for F. The problem is that F is constant on [-1,0].Lemma 7.
LetE⊂R be a finite set which is linearly independent over Q. Let F:[a,b]→R be a nowhere constant continuous function with F(a)≤F(x)≤F(b) for any x∈[a,b]. Then, for any ε>0, there are a=x0<x1<x2<⋯<xn=b and F0∈C[a,b] such that(i)
F0-idsup⁡<ε;(ii)
F0(x)=akF(x)+bk for xk≤x≤xk+1, k=0,1,…,n-1;(iii)
the set{a0,a1,…,an-1}∪E is linearly independent over Q.Proof.
By the previous lemma there area=x0<x1<x2<⋯<xn=b such that(i)
F(x) is between F(xk) and F(xk+1) for xk≤x≤xk+1 and k=0,1…,n-1;(ii)
max⁡{xi+1-xi:i=0,1,…,n-1}<ε/3.
We can find real numbersa0,b0 such that the set E∪{a0} is linearly independent over Q, a0F(x0)+b0=x0(=a), and |a0F(x1)+b0-x1|<ε/9. Let x∈[x0,x1]. Since F(x) is between F(x0) and F(x1), a0F(x)+b0 is between a0F(x0)+b0 and a0F(x1)+b0. We have
(2)a0Fx+b0-x≤a0Fx+b0-a0Fx0+b0+a0Fx0+b0-x0+x0-x≤a0Fx1+b0-a0Fx0+b0+0+ε3≤a0Fx1+b0-x1+x1-x0+x0-a0Fx0+b0+ε3<ε9+ε3+0+ε3≤ε.
In the second step, we can find real numbers a1,b1 such that the set E∪{a0,a1} is linearly independent over Q, a1F(x1)+b1=a0F(x1)+b0, and a1Fx2+b1-x2<ε/9. Let x∈[x1,x2]. Since F(x) is between F(x1) and F(x2), then a1F(x)+b1 is between a1F(x1)+b1 and a1F(x2)+b1. We have
(3)a1Fx+b1-x≤a1Fx+b1-a1Fx1+b1+a1Fx1+b1-x1+x1-x≤a1Fx2+b1-a1Fx1+b1+ε9+ε3≤a1Fx2+b1-x2+x2-x1+x1-a1Fx1+b1+ε9+ε3<ε9+ε3+ε9+ε3+ε9=ε.
After n steps, the construction is complete.


### 2.2. Main Theorem

LetF:[a,b]→R be a continuous function. We consider the following operation on F. Let a=x0<x1<⋯<xn=b be a partition of [a,b]. Let EF:[a,b]→R be such that LF(x)=fi(F(x)) for xi≤x≤xi+1, fi is exponential-like and LF is continuous. We say that LF is a continuous piecewise exponential-like transformation of F.We say that a familyF of continuous functions defined on compact intervals isflexible, provided(1)
F consists of nowhere constant functions;(2)
there isf∈F with f∈C[0,1] and f(0)≤f(x)≤f(1) for x∈[0,1];(3)
Ef∈F for every f∈F and for any of its continuous piecewise exponential-like transformation Ef.From now on we assume thatF is flexible.Theorem 8.
F
∩
C
[
0,1
] is densely c-strongly algebrable in C[0,1].Proof.
LetF∈F be such that F∈C[0,1] and F(0)≤F(x)≤F(1) for any x∈[0,1]. Using Lemma 7 for ε=1/2 and E=∅, we find a partition 0=x01<x11<⋯<xn11=1 of the unit interval and a continuous function F1 such that(i)
F1-idsup⁡<ε;(ii)
F1(x)=ak1F(x)+bk1 for xk1≤x≤xk+11, k=0,1,…,n1-1;(iii)
the set{a01,a11,…,an1-11} is linearly independent over Q.In the next step we use Lemma 7 for ε=1/4 and E={a01,a11,…,an1-11}, and we find a refinement 0=x02<x12<⋯<xn22=1 of the partition 0=x01<x11<⋯<xn11=1 and a continuous function F2 such that(i)
F2-idsup⁡<ε;(ii)
F2(x)=ak2F(x)+bk2 for xk2≤x≤xk+12, k=0,1,…,n2-1;(iii)
the set{a01,a11,…,an1-11}∪{a02,a12,…,an2-12} is linearly independent over Q and so forth.Inductively we define F1,F2,…. Let E=⋃k=1∞‍{a0k,a1k,…,ank-1k}. By the construction, E is linearly independent over Q. We extend E to a linearly independent set H over Q of cardinality c. We may assume that there is {hn:n∈N}⊂H∖E with hn→0. By the assumption, {exp⁡∘Fp:p∈N}∪{exp⁡∘(rF):r∈H∖E}⊆F. Let P be a polynomial in m variables without a constant term. Consider a function g=P(eF1,…,eFp,erp+1F,…,ernF). Then, g restricted to [xlp,xl+1p] is of the form
(4)∑i=1mciexp⁡Fxd1ki1+d2ki2+⋯+dpkipciexp⁡ciexp⁡cip+rp+1kip+1+⋯+rnkin,
where d1,…,dp∈E, rp+1,…,rn∈H∖E are pairwise distinct and the vectors of integers [ki1,ki2,…,kin] are pairwise distinct. Therefore, the numbers d1ki1+d2ki2+⋯+dpkip+rp+1kip+1+⋯+rnkin, i=1,…,m, are distinct as well. Thus, the mapping
(5)x⟼∑i=1mciexp⁡F(x)d1ki1+d2ki2+⋯+dpkipx⟼x⟼x⟼xuw+rp+1kip+1+⋯+rnkin
is a continuous exponential-like transformation of F on [xlp,xl+1p]. Since F is closed under continuous piecewise exponential-like transformations, g∈F.
This shows that the algebraA generated by {exp⁡∘Fp:p∈N}∪{exp⁡∘(rF):r∈H∖E} is a free algebra of c generators. To see that A is dense in C[0,1], note that the sequence exp⁡(F1),exp⁡(F2),… tends to x↦exp⁡(x), and therefore A separates the points of [0,1]. Moreover, note that lim⁡n→∞exp⁡(hnF)=1, which means that the closure of A contains all constant functions. Using Stone-Weierstrass theorem, we obtain the assertion.


### 2.3. Applications

(1) We say that a continuous functionF:[a,b]→R is nowhere Hölder, provided that for any x∈[a,b] and any α∈(0,1](6)limsup⁡y→xFx-Fyx-yα=∞.
Let us denote the set of all nowhere Hölder functions by NH. It was proved in [14] that f∘F∈NH for any nonconstant analytic function f:R→R and any F∈NH. It can be easily seen that if F:[a,b]→R and F′:[b,c]→R are nowhere Hölder with F(b)=F′(b), then F∪F′:[a,c]→R is also nowhere Hölder. Therefore, NH is closed under taking continuous piecewise exponential-like transformations. Clearly NH does not contain a function which is constant on some open interval.Now, we prove that condition (2) in definition of flexibility is fulfilled. LetF∈NH∩C[0,1]. We may assume that F(0)≤F(1) (otherwise, consider -F which is also nowhere Hölder). If F(0)≤F(x)≤F(1) for x∈[0,1], then we are done. Otherwise, find a maximizer x0∈(0,1) of F. Then, F(x)≤F(x1) for x∈[0,x0]. If F(0)≤F(x)≤F(x0) for x∈[0,x0], then an affine transformation t↦F(t/x0) of F|[0,x0] fulfills condition (2) in the definition of a flexible family. Otherwise, find a minimizer x1∈(0,x0) of F|[0,x0]. Then, F(x1)≤F(x)≤F(x0) for x∈[x1,x0]. Then, an affine transformation t↦F(t/(x0-x1)-x1/(x0-x1)) of F|[x1,x0] fulfills condition (2) in the definition of a flexible family. This argument will hold also for the next families.Finally, by Theorem8, the set of all nowhere Hölder functions in C[0,1] is densely c-strongly algebrable.(2) We say that a continuous functionf:[a,b]→R is Bruckner-Garg of rank k∈N (shortly f∈BGk), provided that there exists a countable set A⊆(min⁡f,max⁡f) with the property that for all x∈A the preimage f-1({x}) is a union of a Cantor set with at most k many isolated points and for all x∈(min⁡f,max⁡f)∖A the preimage f-1({x}) is a Cantor set. A function f is Bruckner-Garg (shortly f∈BGω), provided it is Bruckner-Garg of rank k for some k∈N. Bruckner-Garg functions of rank 1 were investigated in [17], where it was shown that BG1 is residual in C[0,1]. By [14, Theorem 4.13] we can easily conclude that BGω is flexible and hence it is densely c-strongly algebrable.(3) LetCMm([0,1]) be the set of all continuous functions such that both sets of their proper local minima and maxima are dense in [0,1]. Using a similar argument to that in [12], one can prove that the set of all functions from CMm([0,1]) is flexible and thereby it is densely c-strongly algebrable.(4) Denote byDNM the set of all functions in C[0,1] which are nowhere monotonic and differentiable in all but finitely many points; see [18]. It can be shown in a standard way that DNM is flexible; thus, it is densely c-strongly algebrable.


## 2.1. Nowhere Constant Continuous Functions

LetF:[a,b]→R be a continuous function. Then F is called left nondecreasing at x∈(a,b] if there is δ>0 such that F(y)≤F(x) for any y∈(x-δ,x). Analogously we define a left nonincreasing function at x∈(a,b] and right nondecreasing (nonincreasing) function at x∈[a,b). We say that x∈(a,b) is a point of local monotonicity, provided that F is left nondecreasing or left nonincreasing and F is right nondecreasing or right nonincreasing; see [15, 16]. Note that if x is a point of local minimum (local minimizer) or a point of local maximum (local maximizer) of F, then x is a point of local monotonicity. We say that F is nowhere constant, provided that its restriction to any open interval is not constant.Fix a functiong∈C[α,β] which is nowhere constant and such that α and β are points of (one-sided) monotonicity of g. For x∈[α,β], denote by H(x) the largest possible y∈[x,β] such that g(t) is between g(x) and g(y) for every t∈[x,y] (here by [x,x] we mean the singleton {x}). Such a number H(x) always exists by the continuity of g. Let x=H0(x) and inductively Hn+1(x)=H(Hn(x)) for x∈[α,β].Lemma 4.
Letn∈N. If Hn+1(α)<β, then(i)
Hn+1(α)>Hn(α);(ii)
Hn+1(α) is a point of local extremum of g;(iii)
Hn+1(α) is a local minimizer of g if and only if Hn(α) is a local maximizer of g.Proof.
Sinceα is a point of right local monotonicity of g, say g is right nondecreasing at α, then there is δ>0 such that g(α)≤g(y) for every y∈(α,α+δ). Let t∈(α,α+δ] such that g(t)=max⁡{g(y):y∈[α,α+δ]}. Since g is nowhere constant, then g(t)>g(α). Hence, H(α)≥t>α.
Now, we will show thatH(α) is a point of right local monotonicity of g. Suppose not, then by the definition of H, g(y)≥g(α) for y∈[α,H(α)]. Moreover g(H(α))>g(α). Let δ>0 such that gHα-gy<gHα-gα/2 for y∈(H(α),H(α)+δ]. Then g|[H(α),H(α)+δ] attains its maximum at some w∈[H(α),H(α)+δ]. Since g is not right nonincreasing at H(α), then g(H(α))<g(w). Moreover g(α)≤g(y)≤g(w) for any y∈[α,w]. This contradicts the definition of H.
Proceeding inductively we obtain thatHn+1(α)>Hn(α) or Hn+1(α)=β. Note that H1(α), H2(α),…,Hn(α) are local extrema of g. Moreover if Hi(α) is a local minimizer of g, then Hi+1(α) is a local maximizer of g and vice versa.Lemma 5.
There isn∈N such that Hn(α)=β.Proof.
Suppose thatHn(α)<β for every n∈N. By Lemma 4, the sequence (Hn(α))n∈N is strictly increasing. Let w=lim⁡n→∞Hn(α). If g is left nondecreasing at w, then there is δ>0 such that g(y)≤g(w) for every y∈[w-δ,w]. Let n be such that Hn(α) is a local minimizer of g with Hn(α)>w-δ. Then for any y∈(Hn(α),w) we have g(Hn(α))≤g(y)≤g(w) which contradicts the definition of Hn+1(α). In the same manner, we show that g is not left nonincreasing. Therefore w<β, since g is left monotonous at β.
Suppose now thatg is not right nondecreasing at w. Let v∈[w,β] be a minimizer of g on [w,β]. Then v>w and g(v)<g(w). Let δ>0 be such that gy-gw<(g(w)-g(v))/2 for every y∈(w-δ,w). Then fix n∈N such that Hn(α)∈(w-δ,w) is a local maximizer of g and g(Hk(α))≤g(Hn(α)) for k≥n. This is possible since lim⁡k→∞g(Hk(α))=g(w)<g(Hn(α)) and g is continuous. Therefore, g(Hn(α))≥g(y)≥g(v) for y∈(Hn(α),v), which contradicts the definition of Hn+1(α). Similarly one can prove that the assumption that g is not right nonincreasing at w also leads to contradiction. Hence, g is both right nondecreasing and right nonincreasing at w. This means that g is constant on [w,w+δ] for some positive δ, which contradicts the fact that g is nowhere constant. This shows that Hn(α)=β for some n∈N.Lemma 6.
LetF∈C[a,b] be nowhere constant and F(a)≤F(x)≤F(b) for any x∈[a,b]. Let ε>0. Then there is a partition a=x0<x1<x2<⋯<xn=b such that(i)
F(x) is between F(xk) and F(xk+1) for xk≤x≤xk+1 and k=0,1…,n-1;(ii)
the meshmax⁡{xi+1-xi:i=0,1,…,n-1} of the partition is smaller than ε.Proof.
Leta=t0<t1<t2<⋯<tm=b be any partition of [a,b] with the mesh smaller than ε. We will find a new partition a=v0<v1<v2<⋯<vk=b of [a,b] such that each interval [vi,vi+1) contains at most one tj and each vi is a point of local monotonicity of F. This new partition will also have a mesh smaller than ε. We construct it in the following way.
Ifti is a point of local monotonicity of F, then ti remains in the new partition. Otherwise, by the fact that F is nowhere constant the restriction F[ti-ε/3,ti] attains its minimum at some wL∈[ti-ε/3,ti] and maximum at some wL′∈[ti-ε/3,ti]. If one of the points wL,wL′ is in (ti-ε/3,ti), then it is a point of local monotonicity and we put it to the new partition. However, it may happen that {wL,wL′}={ti-ε/3,ti}; that is, wL and wL′ are the endpoints of the interval [ti-ε,ti]. We may assume that wL=ti-ε/3 and F(wL)<F(ti). Take any t∈(ti-ε/3,ti). If t is a point of local monotonicity of F, then we are done. Assume now that t is not a point of local monotonicity of F. This means that either t is not a point of left monotonicity of F or it is not a point of right monotonicity of F. We may assume that t is not a point of left monotonicity of F. Then, F attains its maximum on [wL,t] on some w∈(wL,t) and w is a both-sided monotonicity point of F; w is between ti-ε/3 and ti, and we put it to the new partition. Similarly one can find an appropriate both-sided monotonicity point in (ti,ti+ε/3) which we put into the new partition.
In the next step we will find a refinementa=x0<x1<x2<⋯<xn=b of a=v0<v1<v2<⋯<vk=b for which (i) holds true. To find such a refinement, for every i<k, we use Lemma 5 for the restriction g=F[vi,vi+1], α=vi, and β=vi+1.The assumption thatF is nowhere constant in Lemma 6 is essential. To see it, consider a function F given by
(1)F(x)=x,x∈1,2,xsinπ2x,x∈0,1,0,x∈-1,0,2x+2,x∈-2,-1.
Note that F(-2)≤F(x)≤F(2). For every partition -2=x0<x1<⋯<xn=2 with the mesh smaller than 1, there is the largest k with xk≤0. Then, F(xk)=0 and we may assume that F(xk+1)≥0. But there is x∈(0,xk+1) with F(x)<0, which means that the assertion of Lemma 6 does not hold for F. The problem is that F is constant on [-1,0].Lemma 7.
LetE⊂R be a finite set which is linearly independent over Q. Let F:[a,b]→R be a nowhere constant continuous function with F(a)≤F(x)≤F(b) for any x∈[a,b]. Then, for any ε>0, there are a=x0<x1<x2<⋯<xn=b and F0∈C[a,b] such that(i)
F0-idsup⁡<ε;(ii)
F0(x)=akF(x)+bk for xk≤x≤xk+1, k=0,1,…,n-1;(iii)
the set{a0,a1,…,an-1}∪E is linearly independent over Q.Proof.
By the previous lemma there area=x0<x1<x2<⋯<xn=b such that(i)
F(x) is between F(xk) and F(xk+1) for xk≤x≤xk+1 and k=0,1…,n-1;(ii)
max⁡{xi+1-xi:i=0,1,…,n-1}<ε/3.
We can find real numbersa0,b0 such that the set E∪{a0} is linearly independent over Q, a0F(x0)+b0=x0(=a), and |a0F(x1)+b0-x1|<ε/9. Let x∈[x0,x1]. Since F(x) is between F(x0) and F(x1), a0F(x)+b0 is between a0F(x0)+b0 and a0F(x1)+b0. We have
(2)a0Fx+b0-x≤a0Fx+b0-a0Fx0+b0+a0Fx0+b0-x0+x0-x≤a0Fx1+b0-a0Fx0+b0+0+ε3≤a0Fx1+b0-x1+x1-x0+x0-a0Fx0+b0+ε3<ε9+ε3+0+ε3≤ε.
In the second step, we can find real numbers a1,b1 such that the set E∪{a0,a1} is linearly independent over Q, a1F(x1)+b1=a0F(x1)+b0, and a1Fx2+b1-x2<ε/9. Let x∈[x1,x2]. Since F(x) is between F(x1) and F(x2), then a1F(x)+b1 is between a1F(x1)+b1 and a1F(x2)+b1. We have
(3)a1Fx+b1-x≤a1Fx+b1-a1Fx1+b1+a1Fx1+b1-x1+x1-x≤a1Fx2+b1-a1Fx1+b1+ε9+ε3≤a1Fx2+b1-x2+x2-x1+x1-a1Fx1+b1+ε9+ε3<ε9+ε3+ε9+ε3+ε9=ε.
After n steps, the construction is complete.


## 2.2. Main Theorem

LetF:[a,b]→R be a continuous function. We consider the following operation on F. Let a=x0<x1<⋯<xn=b be a partition of [a,b]. Let EF:[a,b]→R be such that LF(x)=fi(F(x)) for xi≤x≤xi+1, fi is exponential-like and LF is continuous. We say that LF is a continuous piecewise exponential-like transformation of F.We say that a familyF of continuous functions defined on compact intervals isflexible, provided(1)
F consists of nowhere constant functions;(2)
there isf∈F with f∈C[0,1] and f(0)≤f(x)≤f(1) for x∈[0,1];(3)
Ef∈F for every f∈F and for any of its continuous piecewise exponential-like transformation Ef.From now on we assume thatF is flexible.Theorem 8.
F
∩
C
[
0,1
] is densely c-strongly algebrable in C[0,1].Proof.
LetF∈F be such that F∈C[0,1] and F(0)≤F(x)≤F(1) for any x∈[0,1]. Using Lemma 7 for ε=1/2 and E=∅, we find a partition 0=x01<x11<⋯<xn11=1 of the unit interval and a continuous function F1 such that(i)
F1-idsup⁡<ε;(ii)
F1(x)=ak1F(x)+bk1 for xk1≤x≤xk+11, k=0,1,…,n1-1;(iii)
the set{a01,a11,…,an1-11} is linearly independent over Q.In the next step we use Lemma 7 for ε=1/4 and E={a01,a11,…,an1-11}, and we find a refinement 0=x02<x12<⋯<xn22=1 of the partition 0=x01<x11<⋯<xn11=1 and a continuous function F2 such that(i)
F2-idsup⁡<ε;(ii)
F2(x)=ak2F(x)+bk2 for xk2≤x≤xk+12, k=0,1,…,n2-1;(iii)
the set{a01,a11,…,an1-11}∪{a02,a12,…,an2-12} is linearly independent over Q and so forth.Inductively we define F1,F2,…. Let E=⋃k=1∞‍{a0k,a1k,…,ank-1k}. By the construction, E is linearly independent over Q. We extend E to a linearly independent set H over Q of cardinality c. We may assume that there is {hn:n∈N}⊂H∖E with hn→0. By the assumption, {exp⁡∘Fp:p∈N}∪{exp⁡∘(rF):r∈H∖E}⊆F. Let P be a polynomial in m variables without a constant term. Consider a function g=P(eF1,…,eFp,erp+1F,…,ernF). Then, g restricted to [xlp,xl+1p] is of the form
(4)∑i=1mciexp⁡Fxd1ki1+d2ki2+⋯+dpkipciexp⁡ciexp⁡cip+rp+1kip+1+⋯+rnkin,
where d1,…,dp∈E, rp+1,…,rn∈H∖E are pairwise distinct and the vectors of integers [ki1,ki2,…,kin] are pairwise distinct. Therefore, the numbers d1ki1+d2ki2+⋯+dpkip+rp+1kip+1+⋯+rnkin, i=1,…,m, are distinct as well. Thus, the mapping
(5)x⟼∑i=1mciexp⁡F(x)d1ki1+d2ki2+⋯+dpkipx⟼x⟼x⟼xuw+rp+1kip+1+⋯+rnkin
is a continuous exponential-like transformation of F on [xlp,xl+1p]. Since F is closed under continuous piecewise exponential-like transformations, g∈F.
This shows that the algebraA generated by {exp⁡∘Fp:p∈N}∪{exp⁡∘(rF):r∈H∖E} is a free algebra of c generators. To see that A is dense in C[0,1], note that the sequence exp⁡(F1),exp⁡(F2),… tends to x↦exp⁡(x), and therefore A separates the points of [0,1]. Moreover, note that lim⁡n→∞exp⁡(hnF)=1, which means that the closure of A contains all constant functions. Using Stone-Weierstrass theorem, we obtain the assertion.


## 2.3. Applications

(1) We say that a continuous functionF:[a,b]→R is nowhere Hölder, provided that for any x∈[a,b] and any α∈(0,1](6)limsup⁡y→xFx-Fyx-yα=∞.
Let us denote the set of all nowhere Hölder functions by NH. It was proved in [14] that f∘F∈NH for any nonconstant analytic function f:R→R and any F∈NH. It can be easily seen that if F:[a,b]→R and F′:[b,c]→R are nowhere Hölder with F(b)=F′(b), then F∪F′:[a,c]→R is also nowhere Hölder. Therefore, NH is closed under taking continuous piecewise exponential-like transformations. Clearly NH does not contain a function which is constant on some open interval.Now, we prove that condition (2) in definition of flexibility is fulfilled. LetF∈NH∩C[0,1]. We may assume that F(0)≤F(1) (otherwise, consider -F which is also nowhere Hölder). If F(0)≤F(x)≤F(1) for x∈[0,1], then we are done. Otherwise, find a maximizer x0∈(0,1) of F. Then, F(x)≤F(x1) for x∈[0,x0]. If F(0)≤F(x)≤F(x0) for x∈[0,x0], then an affine transformation t↦F(t/x0) of F|[0,x0] fulfills condition (2) in the definition of a flexible family. Otherwise, find a minimizer x1∈(0,x0) of F|[0,x0]. Then, F(x1)≤F(x)≤F(x0) for x∈[x1,x0]. Then, an affine transformation t↦F(t/(x0-x1)-x1/(x0-x1)) of F|[x1,x0] fulfills condition (2) in the definition of a flexible family. This argument will hold also for the next families.Finally, by Theorem8, the set of all nowhere Hölder functions in C[0,1] is densely c-strongly algebrable.(2) We say that a continuous functionf:[a,b]→R is Bruckner-Garg of rank k∈N (shortly f∈BGk), provided that there exists a countable set A⊆(min⁡f,max⁡f) with the property that for all x∈A the preimage f-1({x}) is a union of a Cantor set with at most k many isolated points and for all x∈(min⁡f,max⁡f)∖A the preimage f-1({x}) is a Cantor set. A function f is Bruckner-Garg (shortly f∈BGω), provided it is Bruckner-Garg of rank k for some k∈N. Bruckner-Garg functions of rank 1 were investigated in [17], where it was shown that BG1 is residual in C[0,1]. By [14, Theorem 4.13] we can easily conclude that BGω is flexible and hence it is densely c-strongly algebrable.(3) LetCMm([0,1]) be the set of all continuous functions such that both sets of their proper local minima and maxima are dense in [0,1]. Using a similar argument to that in [12], one can prove that the set of all functions from CMm([0,1]) is flexible and thereby it is densely c-strongly algebrable.(4) Denote byDNM the set of all functions in C[0,1] which are nowhere monotonic and differentiable in all but finitely many points; see [18]. It can be shown in a standard way that DNM is flexible; thus, it is densely c-strongly algebrable.


## 3. Closed Algebrability

Aron et al. posed the following problem [19, Problem 4.1]:Characterize when there exists a closed infinite dimensional algebra of functions with a particular “strange” property. Among the classes considered by the authors, there was the family of everywhere surjective functions f:C→C. In the space CX or RX, X≠∅, we consider two natural topologies, namely, the topology τp of pointwise convergence—the weakest topology in which each projection is continuous—and the topology τu of uniform convergence. We will show that the τp-closure of any nontrivial algebra contains a two-valued function (some characteristic function). Moreover, we will give a sufficient condition for the existence of a closed algebra inside F∪{0} of 2c generators.The following proposition shows that ifA is a τp-closed nontrivial algebra, then A contains a two-valued function.Proposition 9.
LetA be a subalgebra of CX or RX. Then for any f∈A the characteristic function χS of S∶={x∈X:f(x)≠0} is in cl⁡τp(A).Proof.
Letf∈A⊂RX. Let g=χS be the characteristic function of S. Take any x1,…,xn∈X and ε>0. Let V={h∈RX:hxi-gxi<ε for i=1,…,n}. We need to show that A∩V≠∅. Let Y={f(xi):f(xi)≠0,i=1,…,n}={y1,…,yk}. Put
(7)P(y)=∑j=1k‍yyj∏i≠j‍y-yiyj-yi.
Then, P is a polynomial without a constant term such that P(yj)=1 for any j=1,…,k. If f(xi)=0, then P(f)(xi)=0. Since xi∉S, f(xi)=g(xi). If f(xi)≠0, then yj=f(xi) for some j and P(f)(xi)=P(yj)=1=g(xi). This shows that P(f)∈A∩V.ByES(C), we denote the family of all everywhere surjective functions f:C→C, that is, functions which map any nonempty open subset of C onto C. This family appeared at first in terms of algebrability in [7]. By PES(C), we denote the family of all perfectly everywhere surjective functions f:C→C, that is, functions which map any perfect subset of C onto C. It was proved in [20] that PES(C) is 2c-strongly algebrable. Since PES(C)⊂ES(C), ES(C) is 2c-strongly algebrable too. Let D stand for the family of all nonconstant Darboux functions. Since any nonconstant Darboux function attains c many values, we obtain the following.Corollary 10.
D
∪
{
0
} does not contain a nontrivial closed algebra. In particular, the set ES(C) of all everywhere surjective functions is not 1-τp-closed-algebrable.Proposition9 says that any τp-closed algebra contains two-valued functions. The next step is searching for large τp-closed algebras in those consisting of functions with a finite range. Note that {f∈RR:f has a finite range} is an algebra of cardinality 2c. However, the following shows that it does not contain a large τp-closed (even τu-closed) algebra.Theorem 11.
LetA be an algebra consisting of functions with finite ranges. Then(i)
ifA is finitely generated, then A is τp-closed;(ii)
ifA is not finitely generated, then A is not τu-closed (in particular, it is not τp-closed).Proof.
(i) Assume thatA is generated by f1,…,fn. Since each fi has a finite range, we can write
(8)fi=∑j=1ki‍cijχAij,
where ci1,…,ciki are distinct and Ai1,…,Aiki is a partition of R. Let B stand for all finite Boolean combinations of {Aij:i=1,…,n,j=1,…,ki}. Clearly, any member of A is B-measurable. Let A∈B be a nonempty atom of the algebra B. Then, there are j1,…,jn such that A=A1j1∩⋯∩Anjn. For any i=1,…,n, there is a polynomial Pi such that Pi(ciji)=1 and Pi(cij)≤0 for j≠ji. Then,
(9)P1(f1(x))+⋯+Pn(fn(x))>0⟺x∈A.
Since P1(f1)+⋯+Pn(fn) is constant on A and has finitely many values, there is a polynomial P such that P(P1(f1)+⋯+Pn(fn)) is a characteristic function of A. Therefore, any B-measurable function is in A. Since B is a σ-algebra of sets, the family of all B-measurable functions is τp-closed (a pointwise limit of B-measurable functions is B-measurable).
(ii) Assume now thatA is not finitely generated. There are f1,f2,…∈A which are algebraically independent. As before, fi=∑j=1ki‍cijχAij and let B stand for the set of all finite Boolean combinations of {Aij:i∈N,j=1,…,ki}. Suppose that B is finite. Again, any characteristic function of an atom in B is an algebraic combination of finitely many fi’s. Therefore, there is n∈N such that any B-measurable function f is an algebraic combination of f1,…,fn. This yields a contradiction. Therefore, B is infinite. Hence, we can find pairwise disjoint sets A1,A2,…∈B. Define fn=∑k=1n‍(1/k)χAk. Since χAi∈A, each fn is in A. Clearly, fn tends uniformly to f=∑k=1∞‍(1/k)χAk∉A.ByEDF, denote the family of all functions f:R→R which are everywhere discontinuous and f(R) is finite. It was proved in [21] that EDF is 2c-algebrable. Immediately we obtain the following.Corollary 12.
E
D
F
∪
{
0
} does not contain an infinitely generated τu-closed algebra.By Proposition9 and Theorem 11, any infinitely generated τp-closed algebra contains finite valued and countably valued functions. It turns out that there are large τp-closed algebras of countably valued functions. Such construction, using the existence of large σ-independent family, will be used in the next theorem.A family{Aα:α<κ} of subsets of Y is called σ-independent, if for every countable set X⊂κ and every ε:X→{0,1}(10)⋂α∈X‍Aαε(α)≠∅,
where A0=A and A1=Y∖A. By the Tarski theorem [22] there exists a σ-independent family on c of cardinality 2c.Theorem 13.
There is a linear algebraA⊂RR of 2c generators such that for any function f∈cl⁡τp(A)∖{0} there is open set U such that f-1(U) is a Bernstein set. In particular, if F is the family of all nonmeasurable functions (having no Baire property, nonmeasurable in the sense of Marczewski), then F∪{0} contains a τp-closed algebra of 2c generators.Proof.
We use the method of independent Bernstein sets which was introduced in [21]. Let {Bα:α<c} be a partition of R into c many pairwise disjoint Bernstein sets. Let {Aξ:ξ<2c} be a σ-independent family on c. For any ξ<2c, put Cξ=⋃{Bα:α∈Aξ}. Let B be the σ-algebra generated by {Cξ:ξ<2c}.
LetA be the linear algebra generated by {χCξ:ξ<2c}. Then each function in A is a simple function of the form ∑k=12n‍ckχDk, where Dk are Boolean combinations of Cξ1,…,Cξn for some distinct ξ1,…,ξn<2c. If f∈clτp(A)∖{0}, then there are fn∈A which tend pointwisely to f. Let X⊂2c be the smallest set such that each fn is measurable with respect to σ-algebra BX generated by {Cξ:ξ∈X}. Clearly X is countable. There is α<c which does not belong to any Aξ, ξ∈X. Consequently, Bα⊂⋂ξ∈X‍R∖Cξ. Therefore, fn|Bα=0 and f|Bα=0. Since f is not the zero function, f(x)≠0 for some x∈R. There is δ>0 such that f-1(f(x)-δ,f(x)+δ) is disjoint with f-1(0). Since f is BX-measurable, f-1(f(x)-δ,f(x)+δ) contains a Bernstein set of the form ⋂ξ∈X‍Cξε(ξ) for some ε:X→{0,1}. Finally, a set which contains a Bernstein set and is disjoint with some other Bernstein sets is also a Bernstein set.Letf∈RX (or f∈CX). Fix the partition {Bξ:ξ<κ} of R (or C). By V(f) we define the set
(11)⋃ξ<κ‍tξfBξ:t∈Rκ,⋃ξ<κ‍tξfBξ:t∈Cκ,resp..
Let g1,…,gn∈V(f) and let P(y1,…,yn) be a polynomial in n variables. Let ti(ξ) be such that gi|Bξ=ti(ξ)f|Bξ. Then,
(12)Pg1,…,gnBξ=Pg1Bξ,…,gnBξ=Pt1ξfBξ,…,tnξfBξ=P′fBξ,
where P′(y)=Pt1ξy,…,tnξy. Therefore, the algebra A(f) generated by V(f) is of form
(13)Af=⋃ξ<κ‍fξ:fξ∈Aξ,
where Aξ is a subalgebra of CBξ generated by fBξ.Theorem 14.
Assume thatfBξ is unbounded for every ξ<κ. Then, A(f) is τu-closed.Proof.
Note thatτu is metrizable by the metric d(g,h)=min⁡{1,sup⁡{gx-hx:x∈C}}. To prove that A(f) is τu-closed, take a sequence (gn) in A(f) tending with respect to d to some function g. Fix ξ<κ. If g is zero on Bξ, then obviously g∈Aξ. Otherwise, gBξ is nonzero. Then, the sequence (gn|Bξ)n∈N eventually consists of nonzero functions. Note that gnBξ=Pn(f)Bξ for some nonzero polynomials Pn in one variable. By the assumption f(Bξ) is unbounded. Note that the sequence Pn:f(Bξ)→C is a Cauchy sequence with respect to d(g,h)=min⁡{1,sup⁡{gy-hy:y∈f(Bξ)}} for g,h∈Cf(Bξ). Since f(Bξ) is unbounded, then, for distinct polynomials in one variable P,Q:f(Bξ)→C without constant term, we have sup⁡{Py-Qy:y∈f(Bξ)}=∞. Therefore, the sequence (Pn) is eventually constant and equal to some polynomial P. Thus, gBξ=P(f)Bξ∈Aξ.Corollary 15.
There exists aτu-closed algebra A of cardinality 2c and hence 2c-generated, such that A∖{0} consists of perfectly everywhere surjective functions.Proof.
Let{Bξ:ξ<c} be a decomposition of C into c many Bernstein sets. For any ξ<c, let fξ:Bξ→C be a free generator such that algebra generated by fξ consists of perfectly everywhere surjective functions; the existence of such a function was proved in [20]. Put f=⋃ξ<c‍fξ:C→C. Then, A(f) is the desired algebra.For a sequencex∈l∞, put LIM(x)={y∈R:x(nk)→y for some increasing (nk)k∈N}. It was proved in [8] that the set of x∈l∞ for which LIM(x) is homeomorphic to the Cantor set is strongly c-algebrable and comeager. We complete this result with the following.Theorem 16.
The set of thosex∈l∞, for which LIM(x) is homeomorphic to the Cantor set, does not contain any nontrivial closed algebra.Proof.
LetA be an algebra such that for any x∈A∖{0} the set of limit points LIM(x) is homeomorphic to the Cantor set. Fix nonzero x∈A and let C=LIM(x). There is a continuous function f:[min⁡C,max⁡C]→[0,1] such that f(C)=[0,1]. Let (Pn) be a sequence of polynomials, tending uniformly to f. It is evident that Pn(x) tends in l∞ to some y with LIM(y)=[0,1]. Since [0,1] is not homeomorphic to C, the algebra A cannot be closed.



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