# Quantitative Estimates for Positive Linear Operators in terms of the Usual Second Modulus

Authors: José A. Adell; A. Lekuona
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We give accurate estimates of the constantsCn(A(I),x) appearing in direct inequalities of the form |Lnf(x)-f(x)|≤Cn(A(I),x)ω2  f;σ(x)/n, f∈A(I), x∈I,  and  n=1,2,…, where Ln is a positive linear operator reproducing linear functions and acting on real functions f defined on the interval I, A(I) is a certain subset of such functions, ω2(f;·) is the usual second modulus of f, and σ(x) is an appropriate weight function. We show that the size of the constants Cn(A(I),x) mainly depends on the degree of smoothness of the functions in the set A(I) and on the distance from the point x to the boundary of I. We give a closed form expression for the best constant when A(I) is a certain set of continuous piecewise linear functions. As illustrative examples, the Szàsz-Mirakyan operators and the Bernstein polynomials are discussed.


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## Body

## 1. Introduction

LetI be a closed real interval with nonempty interior set I∘. The usual second modulus of smoothness of a function f:I→R is defined as (1)ω2f;τ=sup⁡⁡Δh2fx:0≤h≤τ,x-h,x+h⊆I,τ≥0,where (2)Δh2fx=fx+h-2fx+fx-h.Denote by M(I) the set of measurable functions f:I→R such that ω2(f;τ)<∞, τ≥0. Many sequences (Ln,n=1,2,…) of positive linear operators acting on M(I) allow for a probabilistic representation of the form (cf. [1]) (3)Lnfx=EfYnx,f∈MI,x∈I,n=1,2,…,where E stands for mathematical expectation and Yn(x) is an I-valued random variable whose mean and standard deviation are given, respectively, by (4)EYnx=x,EYnx-x2=σxn,x∈I,n=1,2,…,for some nonnegative function σ:I→R. The condition EYn(x)=x is equivalent to say that Ln reproduces linear functions.It is well known (see, for instance, [2–6] and the references therein) that such operators satisfy pointwise inequalities of the form(5)Lnfx-fx≤Cnxω2f;σxn,x∈I,n=1,2,…,which measure the rate of convergence from Lnf(x) to f(x) according to the degree of smoothness of f. In (5), Cn(x) is a positive constant only depending upon n and x. It is also interesting to consider in (5) the uniform constant(6)C=sup⁡⁡Cnx:x∈I,n=1,2,….Several authors have obtained estimates of this uniform constant. For instance, Adell and Sangüesa [7] gave C=1.385 for the Weierstrass operator. Păltănea [5, Corollary  4.1.2, pp. 93-94] obtained C=11/8 for the Bernstein polynomials, and Gonska and Păltănea [8] showed that C≃3/2 for a certain class of Bernstein-Durrmeyer operators. More generally, in Păltănea’s book [5, Corollary  2.2.1, p. 31] it is shown that C≃3/2 for a large class of positive linear operators reproducing linear functions.The aim of this paper is to give a general method to provide accurate estimates of the constantsCn(A(I),x) satisfying the inequalities(7)Lnfx-fx≤CnAI,xω2f;σxn,f∈AI,x∈I,n=1,2,…,where A(I) is a certain subset of M(I). Such a problem is meaningful, because in specific examples the estimates of the constants in (6) and (7) may be quite different, mainly depending on two facts: the degree of smoothness of the functions in the set A(I) and the distance from the point x to the boundary of I. In this way, we complete the general results shown by Păltănea [5].The method is based on the approximation of any functionf∈M(I) by a quasi interpolating piecewise linear function having an appropriate set of nodes. In doing this, special attention must be paid to the nodes near the endpoints of I, if any. The main results are Theorems 6 and 7 stated in Section 3. In particular, Theorem 6 provides inequalities of form (7), where the upper bound consists of various terms involving ω2(f;·) evaluated at different lengths. Theorem 7 gives a closed form expression for the best constant in (7) when A(I) is a certain set of continuous piecewise linear functions.As illustrative examples, we consider the Szàsz-Mirakyan operator (Section4) and the Bernstein polynomials (Section 5). Although the kind of estimates is similar in both examples, the results take on a simpler form in the first case, because the interval of definition I=[0,∞) has only one endpoint. In any case, both examples show that the size of the constants in front of ω2(f;·) heavily depends on the set of functions A(I) under consideration and on the distance from point x to boundary of I.We believe that the methods proposed in this paper could be applied to a wide class of positive linear operators, such as Baskakov operators, Stancu operators, and theirq-analogues, among others (see [9, 10] and the references therein). To obtain accurate estimates of the constants involved in each case, we essentially need to compute second moments (see Theorem 8 in Section 3) and tail probabilities of the underlying random variables defining the operators under consideration (see Lemmas 9 and 11 in Sections 4 and 5, resp.).


## 2. Continuous Piecewise Linear Functions

Throughout this paper,I is a closed real interval of positive length and I° is the interior set of I. If I=[a,b], we denote by N a finite ordered set of nodes a=x-(m+1)<x-m<⋯<x-1<x0<x1<⋯<xk<xk+1=b, for some m,k=0,1,…. If I is an infinite interval, N could also be infinite. In such a case, the finite endpoint of I, if any, is always in N. We denote by L(I) the set of continuous piecewise linear functions g:I→R whose set of nodes is N. Unless otherwise specified, we assume from now on that I=[a,b]. Given a sequence (ci,i∈Z), we denote by δci=ci+1-ci, i∈Z. We set y+=max⁡(0,y), y-=max⁡(0,-y) and denote by 1A the indicator function of the set A.Lemma 1.
For anyg∈L(I), one has the representations(8)gy-gx0=c1y-x0+-c0y-x0-+∑i=1kδciy-xi++∑i=-m-1δciy-xi-=c0+c12y-x0+δc02y-x0+∑i=1kδciy-xi++∑i=-m-1δciy-xi-,y∈I,where(9)ci=gxi-gxi-1xi-xi-1,i=-m,…,k+1.Proof.
The first equality in (8) follows from the fact that the two functions involved have the same Radon-Nikodym derivative in (xi-1,xi),  i=-m,…,k+1, given by the constant ci defined in (9). The second equality in (8) follows from the first one and the equalities(10)y+=12y+y,y-=12y-y,y∈R.The proof is complete.The following auxiliary result is taken from [5, Lemma  2.5.7] (see also [11]). We give a simple proof of it for the sake of completeness.Lemma 2.
Letf:I→R be a function such that f(c)=f(d)=0, for some c,d∈I with c≤d. Then, (11)M≔sup⁡c≤y≤d⁡fy≤ω2f;d-c2.Proof.
Assume thaty∈(c,(c+d)/2], the case y∈[(c+d)/2,d) being similar. Set y~=y+(y-c)=2y-c∈[c,d]. Then, (12)fy=12fc-2fy+fy~-fy~≤12ω2f;d-c2+M2.The proof is complete.For any0<ɛ≤(b-a)/3, denote by Lɛ(I) the set of functions in L(I) whose set of nodes Nɛ={xi:i=-(m+1),…,k+1} satisfies(13)ɛ≤min⁡⁡xi-xi-1:i=-m,…,k+1.Lemma 3.
Letg∈Lɛ(I), for some 0<ɛ≤(b-a)/3. Then, (14)ω2g;τ=τmax⁡-m≤i≤k⁡δci,0≤τ≤ɛ.Proof.
Let0≤h≤τ≤ɛ and i=-m,…,k. Denote by si(y)=(y-xi)+. We claim that(15)Δh2siy=h-y-xi+≕qiy,y∈a+h,b-h.Formula (15) is obvious if y-xi≥h; suppose that y-xi<h. If y∈(xi-h,xi]∩[a+h,b-h], then (16)Δh2siy=y+h-xi+=h-y-xi+,whereas if y∈[xi,xi+h)∩[a+h,b-h], then (17)Δh2siy=-2y-xi+y+h-xi=h-y-xi+,thus showing claim (15). By virtue of (10), formula (15) is also true if we replace si by any one of the functions s¯i(y)=(y-xi)- or s¯¯iy=y-xi/2, y∈I. We therefore have from (8) and (15)(18)Δh2gy=∑i=-mkδciqiy,y∈a+h,b-h.Since qi(y)=0 for y∈[a,xi-1]∪[xi+1,b], as follows from (13), we have from (18) (19)sup⁡a≤y≤x-m⁡Δh2gy=sup⁡a≤y≤x-m⁡δc-mq-my=δc-mq-mx-m=δc-mh.Similarly, (20)sup⁡xk≤y≤b⁡Δh2gy=δckh,and, for i=-m,…,k-1(21)sup⁡xi≤y≤xi+1⁡Δh2gy=sup⁡xi≤y≤xi+1⁡δciqiy+δci+1qi+1y≤max⁡⁡δci,δci+1qiy+qi+1y=hmax⁡⁡δci,δci+1,thus showing that(22)ω2g;τ≤τmax⁡-m≤i≤k⁡δci.By assumption (13), xi∈[a+h,b-h]. We thus have from (18) (23)Δh2gxi=δciqixi=hδci.This shows the converse inequality to (22) and completes the proof.Remark 4.
If assumption (13) is dropped, Lemma 3 is no longer true. To see this, consider the function s(y)=(y-x)+, y∈I, where a<x<(b+a)/2. Then,(24)ω2s,τ=min⁡⁡τ,x-a,0≤τ≤b-a2.Actually, let 0≤h≤τ. If x-a≤h, we have from (15) (25)sup⁡a+h≤y≤b-h⁡Δh2sy=sup⁡a+h≤y≤b-h⁡h-y-x+=x-a,whereas if h≤x-a, we have (26)sup⁡a+h≤y≤b-h⁡Δh2sy=sup⁡a+h≤y≤b-h⁡h-y-x+=h=Δh2sx,thus showing (24).We close this section with the following auxiliary result concerning the symmetric functions(27)φy=∑i=1∞y-i+,ψy=12y+φy,y∈R.For any y∈R, let ⌊y⌋ and ⌈y⌉ be the floor and the ceiling of y, respectively; that is, (28)y=sup⁡⁡k∈Z:k≤y,y=inf⁡⁡k∈Z:k≥y.Lemma 5.
Letφ and ψ be as in (27). Then, (29)φy≤12y211,∞y,ψy≤12y2+18,y∈R.Proof.
Lety≥0. Then,(30)φy=∑i=1yy-i=y22y-1+y≤y2y11,∞y≤y2211,∞y.Thanks to (30), the second inequality in Lemma 5 is equivalent to(31)ηy≔y2y-1+y≤y-122≕νy,y≥0.It is easily checked that (32)ηm+12=νm+12,η′m+12=ν′m+12,m=0,1,….These equalities imply (31), since ν is convex and η is linear in each interval [m,m+1], m=0,1,…. The proof is complete.


## 3. Main Results

Denote byC(I) the set of convex functions in M(I). Given 0<ɛ≤(b-a)/3 and x∈I°, we consider the set(33)Aɛ,x=xi=x+iɛ,i=-m,…,k,m=x-aɛ-1,k=b-xɛ-1.If I=[a,∞), the preceding set should be defined as (34)Aɛ,x=xi=x+iɛ,i=-m,…,0,1,2,…,m=x-aɛ-1,and analogously if I=(-∞,b] or I=R. Observe that x-m∈(a,a+ɛ] and xk∈[b-ɛ,b). We define the function(35)gɛ,xy=r02y-xɛ+∑i=1kriy-xi+ɛ+∑i=-m-1riy-xi-ɛ,y∈I,where(36)ri=1a+ɛ,b-ɛxi,i=-m,…,k.Note that gɛ,x∈Lɛ(I)∩M(I) and its set of nodes is(37)Nɛ,x=Aɛ,x∩a+ɛ,b-ɛ∪a,b.If x∈(a,a+ɛ), then x=x-m and therefore x∉Nɛ,x. The same is true if x∈(b-ɛ,b). Since ɛ≤(b-a)/3, we see that Aɛ,x∩[a+ɛ,b-ɛ]≠∅ and therefore Nɛ,x has at least three nodes. From (35), (36), and Lemma 3, we have(38)ω2gɛ,x;τ=τɛ,0≤τ≤ɛ.Finally, letY be a random variable taking values in I such that(39)EY=x,EY-x2<∞.Since EY=x, we have from (10)(40)EY-x+=EY-x-=12EY-x.With these notations, we enunciate our first main result.Theorem 6.
Let0<ɛ≤(b-a)/3 and x∈I°. Then one has the following. (a)
Iff∈M(I), then (41)EfY-fx≤ω2f;ɛEgɛ,xY+ω2f;ɛ2+ω2f;x-m-aPY<x-m+ω2f;b-xkPY>xk.(b)
Iff∈C(I), then (42)EfY-fx≤ω2f;ɛEgɛ,xY+ω2f;ɛ2+ω2f;x-m-aPY<x-m+ω2f;ɛ2+ω2f;b-xkPY>xk.Proof.
Fix0<ɛ≤(b-a)/3 and x∈I°. Let f~∈Lɛ(I) be the function having representation (8), whose set of nodes is Nɛ,x, as defined in (37), and satisfying the following properties: (a)
f~(xi)=f(xi),  xi∈Aɛ,x,   i=-m,…,k;(b)
ifx-m=a+ɛ, then f~ is linear in [a,a+ɛ] and f~(a)=f(a); if x-m∈(a,a+ɛ), then f~ is linear in [a,x-m+1] (in such a case, it could happen that f~(a)≠f(a));(c)
ifxk=b-ɛ, then f~ is linear in [b-ɛ,b] and f~(b)=f(b); if xk∈(b-ɛ,b), then f~ is linear in [xk-1,b] (in such a case, it could happen that f~(b)≠f(b)).
These properties, together with (8) and (39), allow us to write(43)Ef~Y-fx=r0δc02EY-x+∑i=1kriδciEY-xi++∑i=-m-1riδciEY-xi-,where ri is defined in (36) and (44)δci=fxi+ɛ-fxiɛ-fxi-fxi-ɛɛ,xi∈Aɛ,x∩a+ɛ,b-ɛ.We therefore have from (35) and (43)(45)Ef~Y-fx≤ω2f;ɛEgɛ,xY.On the other hand, applying Lemma 2 to the function f⋆≔f-f~, we have(46)sup⁡xi≤y≤xi+1⁡f⋆y≤ω2f;ɛ2,i=-m,…,k-1.If y∈[a,x-m), we set y⋆=x-m+(x-m-y)∈[x-m,x-m+1] and obtain thanks to (46)(47)f⋆y=f⋆y-2f⋆x-m+f⋆y⋆-f⋆y⋆≤ω2f;x-m-a+ω2f;ɛ2.In the same way,(48)sup⁡xk<y≤b⁡f⋆y≤ω2f;b-xk+ω2f;ɛ2.Thus, we have from (46)–(48) (49)EfY-Ef~Y≤ω2f;ɛ2+ω2f;x-m-aPY<x-m+ω2f;b-xkPY>xk.This, together with (45), shows part (a).
Suppose thatf∈C(I). By substracting an affine function, if necessary, we can assume without loss of generality that f(y)≥f(x)=0, y∈I. The convexity of f implies that (50)fy≤f~y+fy-f~y1a,x-my+1xk,by,y∈I.We therefore have from (45), (47), and (48) (51)EfY≤Ef~Y+Ef⋆Y1a,x-mY+1xk,bY≤ω2f;ɛEgɛ,xY+ω2f;ɛ2+ω2f;x-m-aPY<x-m+ω2f;ɛ2+ω2f;b-xkPY>xk.The proof is complete.Let0<ɛ≤(b-a)/3 and x∈[a+ɛ,b-ɛ]. Denote by Lɛ,x(I) the set of functions g∈Lɛ(I) having a node at x and not being linear in I (in other words, ω2(g;τ)>0, τ>0). It turns out that the function gɛ,x defined in (35) is a maximal function in Lɛ,x(I), as shown in the following result.Theorem 7.
Let0<ɛ≤b-a/3 and x∈[a+ɛ,b-ɛ]. Then, (52)sup⁡g∈Lɛ,xI⁡EgY-gxω2g;ɛ=Egɛ,xY.Proof.
Letg∈Lɛ,x(I) with representation (8) and set of nodes a=x¯-(m¯+1)<x¯-m¯<⋯<x¯-1<x¯0=x<x¯1<⋯<x¯k¯<x¯k¯+1=b, for some m¯,k¯=0,1,…. From (39) and Lemma 3, we have(53)EgY-gx=δc02EY-x+∑i=1k¯δciEY-x¯i++∑i=-m¯-1δciEY-x¯i-≤ω2g;ɛɛEY-x+∑i=1k¯EY-x¯i++∑i=-m¯-1EY-x¯i-.Let gɛ,x be as in (35) with set of nodes Nɛ,x as defined in (37). By assumption, x¯i-x¯i-1≥ɛ,   i=-m¯,…,k¯+1. Therefore, (54)x¯i≥xi,i=1,…,k¯,x¯i≤xi,i=-m¯,…,-1.This implies, by virtue of (35) and (53), that (55)EgY-gx≤ω2g;ɛEgɛ,xY.This, in conjunction with (38), completes the proof.In order to apply Theorems6 and 7 to concrete examples, we need to estimate the expectation Egɛ,x(Y) and the tail probabilities of the random variable Y under consideration. With regard to the first question, we give the following.Theorem 8.
Let0<ɛ≤(b-a)/3. Then one has the following. (a)
Ifx∈[a+ɛ,b-ɛ], then (56)Egɛ,xY≤12EY-xɛ2+18.(b)
Ifx∈I°∖[a+ɛ,b-ɛ], then (57)Egɛ,xY≤12EY-xɛ211,∞Y-xɛ.Proof.
Suppose thatx∈[a+ɛ,b-ɛ]. Using definitions (33)–(36) and Lemma 5, we have (58)gɛ,xy=12y-xɛ+∑i=1kriy-xɛ-i++∑i=-m-1riy-xɛ-i-≤ψy-xɛ≤12y-xɛ2+18,y∈I.Part (a) follows by replacing y by Y in the preceding inequality and then taking expectations. Part (b) follows in a similar manner, by noting that if x∈I°∖[a+ɛ,b-ɛ], then (59)gɛ,xy≤φy-xɛ≤12y-xɛ211,∞y-xɛ,y∈I,as follows from Lemma 5. This completes the proof.Theorem8 gives an upper bound for Egɛ,x(Y) in terms of the variance of the random variable Y, which is easy to compute in many usual examples. Such an upper bound also suggests the choice (60)ɛ=ɛx=E1/2Y-x2,x∈I°.


## 4. Example  1: The Szàsz Operator

Let(Nλ,λ≥0) be the standard Poisson process, that is, a stochastic process starting at the origin, having independent stationary increments such that(61)PNλ=k=λke-λk!,k=0,1,2,…,λ≥0.Let n=1,2,… and x≥0. Thanks to (61), the classical Szàsz-Mirakyan operator Ln can be written as(62)Lnfx=∑k=0∞fknnxke-nxk!=EfNnxn,where f∈M([0,∞)). It is well known that(63)ENnxn=x,ENnxn-x2=xn.Concerning the tail probabilities of the standard Poisson process, we give the following lemma.Lemma 9.
Let(Nλ,λ≥0) be as in (61). Then, (a)
one has(64)sup⁡1≤λ≤4⁡PNλ<λ-λ≤e-1,sup⁡4<λ≤9⁡PNλ<λ-2λ≤e-1;(b)
for anyλ>1, one has(65)PNλ<λ≤e-τλ,τλ=λ-λ1+12log⁡⁡λ,τ(λ) being strictly increasing in (1,∞).Proof.
(a) Suppose that1≤λ≤4. Denote by λ0=(3+5)/2 the solution to the equation λ-λ=1. If 1≤λ≤λ0, we obviously have from (61) (66)PNλ<λ-λ=PNλ=0=e-λ≤e-1,whereas if λ0≤λ≤4, we have (67)PNλ<λ-λ=PNλ≤1=e-λ1+λ≤e-λ01+λ0=0.263⋯<e-1.Suppose that 4<λ≤9. We have from (63) and Markov’s inequality (68)PNλ<λ-2λ≤PNλ-λ>2λ≤14λENλ-λ2=14<e-1.
(b) Letu>0. Again by Markov’s inequality, we have (69)PNλ<λ=Pe-uNλ>e-uλ≤euλEe-uNλ=eλe-u-λ+uλ.It suffices to choose u=log⁡λ in the preceding inequality. The proof is complete.Theorem 10.
LetLn be as in (62), n=1,2,…, and let f∈M([0,∞)). Then, (a)
if0<x<1/n, then (70)Lnfx-fx≤121-nxe-nxω2f;xn+ω2f;12xn+e-nxω2f;x;(b)
if1/n≤x≤9/n, then (71)Lnfx-fx≤58+1eω2f;xn+ω2f;12xn;(c)
if9/n<x, then (72)Lnfx-fx≤58+e-τnxω2f;xn+ω2f;12xn;whereτ(·) is defined in (65).Proof.
For anyn=1,2,… and x>0, denote by ɛ=x/n and λ=nx. In view of (62), we will apply Theorems 6 and 8 with Y=Nnx/n.
(a) Ifx<1/n, then x∈(0,ɛ) and x-m=x, as follows from (33). Thus, we have from Theorem 8(b) and (63) (73)Egɛ,xNnxn≤12ENλ-λλ211,∞Nλ-λλ≤12ENλ-λλ210,∞Nλ=121-λe-λ,as well as (74)PNnxn<x-m=PNλ<λ=PNλ=0=e-λ.Therefore, the conclusion follows from Theorem 6(a).
(b) If1/n≤x≤9/n, we see that x∈[ɛ,∞). By Theorem 8(a) and (63), we have(75)Egɛ,xNnxn≤12ENλ-λλ2+18=58.If x=1/n, then x-m=x, as follows from (33). Thus, (76)PNnxn<x-m=PN1=0=e-1.If 1/n<x≤4/n, then x-m=x-ɛ, again by (33). We therefore have from Lemma 9(a) (77)PNnxn<x-m=PNλ<λ-λ<e-1.Similarly, if 4/n<x≤9/n, then x-m=x-2ɛ. Again by Lemma 9(a), we have (78)PNnxn<x-m=PNλ<λ-2λ<e-1.In any of the previous cases, we always have x-m≤ɛ and therefore(79)ω2f;x-m≤ω2f;ɛ.In view of the preceding discussion, part (b) follows from Theorem 6(a).
(c) Ifx>9/n, we have as in (75) (80)Egɛ,xNnxn≤58.As in part (b), x-m≤ɛ and inequality (79) holds. Also, we have from Lemma 9(b) (81)PNnxn<x-m≤PNnxn<ɛ=PNλ<λ≤e-τλ.By Theorem 6(a), this shows part (c) and completes the proof.Theorem10 could also be stated for functions f∈C([0,∞)) using Theorem 6(b) instead of Theorem 6(a). In such a case, we obtain better estimates. For instance, if x>9/n, we get (82)Lnfx-fx≤58+e-τnxω2f;xn+e-τnxω2f;12xn.Observe that, for fixed x>0, the constant e-τ(nx) exponentially decreases to zero, as n→∞, as follows from (65).


## 5. Example  2: Bernstein Polynomials

Letn=1,2,… and let (Uk)k≥1 be a sequence of independent identically distributed random variables having the uniform distribution on [0,1]. We consider the (uniform) empirical process (Sn(x)/n,0≤x≤1) defined as(83)Snx=∑k=1n10,xUk,0≤x≤1.Observe that the random variable Sn(x) has the binomial law with parameters n and x; that is,(84)PSnx=k=nkxk1-xn-k,k=0,1,…,n,0≤x≤1.Also observe that the paths of the empirical process are nondecreasing, since we have from (83)(85)Snx≤Sny,0≤x≤y≤1.It is well known that(86)ESnxn=x,ESnxn-x2=x1-xn,0≤x≤1.For any function f:[0,1]→R, the Bernstein polynomials of f can be written as(87)Bnfx=∑k=0nnkfknxk1-xn-k=EfSnxn,0≤x≤1.In view of (86), we define(88)εnx=x1-xn,n=3,4,…,0<x<1.The following auxiliary result will be needed.Lemma 11.
Letn=3,4,… and 0<x<1. Let τ(·) and εn(x) be as in (65) and (88), respectively. Then,(89)PSnxn<εnx≤e-τnx,(90)PSnxn>1-εnx≤qnx≔min⁡⁡xn1-x-x2,e-τn1-x.Proof.
Letu>0. By Markov’s inequality and (84), we have(91)PSnxn<εnx=Pe-uSnx>e-unεnx≤eunεnxEe-uSnx=eunx1-x1-x1-e-un≤eunx-nx1-e-u,where we have used the inequality(92)1+y≤ey,y∈R.Inequality (89) follows by choosing u=log⁡nx in (91). On the other hand, the random variables Sn(x) and n-Sn(1-x) have the same law, as follows from (84). We therefore have from (88) and (89)(93)PSnxn>1-εnx=PSn1-xn<εn1-x≤e-τn1-x.Again by Markov’s inequality, (86), and (88), we get (94)PSnxn-x>1-x-εnx≤11-x-εnx2ESnxn-x2=xn1-x-x2.This, together with (93), shows (90) and completes the proof.Denote by(95)rn=max⁡⁡e-n/n+1,7n-32n-3e-5n/2n+1,n=3,4,….Numerical computations show that rn=e-n/n+1, for n≥11.Theorem 12.
Letn=3,4,… and x∈(0,1/2]. Let τ(·), qn(x), and rn be as in (65), (90), and (95), respectively. For any f∈M([0,1]), one has the following. (a)
Ifx<1/(n+1), then (96)Bnfx-fx≤121-nx1-xn-1+qnxω2f;εnx+ω2f;εnx2+1-xnω2f;x.(b)
If1/(n+1)≤x≤9/(n+9), then (97)Bnfx-fx≤58+rn+qnxω2f;εnx+ω2f;εnx2.(c)
If9/(n+9)<x, then (98)Bnfx-fx≤58+qnx+e-τnxω2f;εnx+ω2f;εnx2.Proof.
In view of (87), we will apply Theorems 6 and 8 with Y=Sn(x)/n and ɛ=εn(x), as defined in (88). In the first place, we have from (90)(99)ω2f;1-xkPSnxn>xk≤ω2f;εnxPSnxn>1-εnx≤qnxω2f;εnx.(a) If x<1/(n+1), then x∈(0,εn(x)) and x-m=x, as follows from (33). Thus, we have from Theorem 8(b) and (86)(100)Egɛ,xSnxn≤12ESnx-nxnx1-x211,∞Snx-nxnx1-x≤12ESnx-nxnx1-x210,∞Snx=121-nx1-xn-1,and also (101)ω2f;x-mPSnxn<x-m=ω2f;xPSnx=0=1-xnω2f;x.This, together with (99) and (100), shows part (a).
(b) If1/(n+1)≤x≤9/(n+9), see that x∈[εn(x),1/2]. By Theorem 8(a) and (86), we have(102)Egɛ,xSnxn≤12ESnx-nxnx1-x2+18=58.We distinguish the following subcases.
Case  1. One has x=1/(n+1). In this case, x-m=x, as follows from (33). We thus have from (92) (103)PSnxn<x-m=PSn1n+1=0=1-1n+1n≤e-n/n+1≤rn.
Case  2. One has 1/(n+1)<x≤4/(n+4). Then x-m=x-εn(x), again by (33), and therefore(104)PSnxn<x-m=PSnx<nx-nεnx.Let x¯0 be the solution in (1/(n+1),1/2] to the equation nx-nεn(x)=1; that is, (105)x¯0=3+5-4/n2n+1.If 1/(n+1)<x<x¯0, we have from (104) (106)PSnxn<x-m=PSnx=0≤1-1n+1n≤e-n/n+1≤rn.If n=3, then x¯0>1/2. If n=4,5,…, then x¯0≤1/2 (actually, x¯0=1/2, for n=4). Thus, we can assume without loss of generality that n≥4. In such a case,(107)x¯0≥52n+1≕x0.If x¯0≤x≤4/(n+4), we have from (85), (92), and (107) (108)PSnxn<x-m=PSnx≤1≤PSnx0≤1=1-x0n+nx01-x0n-1=1-52n+1n7n-32n-3≤7n-32n-3e-5n/2n+1≤rn.
Case  3. One has 4/(n+4)<x≤9/(n+9). Again by (33), we see that x-m=x-2εn(x). Therefore, we have from Markov’s inequality and (86) (109)PSnxn<x-m=PSnx-nx<-2nεnx≤PSnx-nx>2nεnx≤14n2εn2xESnx-nx2=14≤e-n/n+1≤rn.The preceding discussion shows that (110)ω2f;x-mPSnxn<x-m≤rnω2f;εnx.This, in conjunction with (99) and (102), shows part (b).
(c) If9/(n+9)<x, we have as in part (b) (111)Egɛ,xSnxn≤58.By (89), we have (112)ω2f;x-mPSnxn<x-m≤ω2f;εnxPSnxn<εnx≤e-τnxω2f;εnx.Therefore, part (c) follows from (99). The proof is complete.Remark 13.
Forx∈[1/2,1), Theorem 12 remains true if we replace x by 1-x in the right-hand sides of the corresponding inequalities. This is due to the fact that (113)Bnfx-fx=Bng1-x-g1-x,gy=f1-y,y∈0,1.Theorem12 can also be stated for convex functions, the estimates being better. For instance, if f∈C([0,1]) and x∈(9/(n+9),1/2], then(114)Bnfx-fx≤58+qnx+e-τnxω2f;εnx+qnx+e-τnxω2f;εnx2.The proof of (114) follows along the lines of that in Theorem 12(c), using Theorem 6(b) instead of Theorem 6(a).Finally, to illustrate the size of the constants in Theorem12, consider the Lipschitz class Lα([0,1]) consisting of those functions in M([0,1]) such that ω2(f;δ)≤δα, δ≥0, α∈(0,2].Corollary 14.
Letx∈(0,1) and α∈(0,2]. If f∈Lα([0,1]), then(115)limn→∞-⁡Bnfx-fxεnxα≤58+12α.If, in addition, f∈C([0,1])∩Lα([0,1]), then(116)limn→∞-⁡Bnfx-fxεnxα≤58.Proof.
Taking into account Remark13, inequalities (115) and (116) readily follow from Theorems 12(c) and (114), respectively.Observe that the asymptotic constant in (115) is less than or equal to 1 if (117)α≥log⁡⁡8-log⁡⁡3log⁡⁡2=1.415⋯.Also, suppose that g∈Lεn(x),x([0,1]), where n=3,4,…, εn(x) is defined in (88), and x∈[εn(x),1-εn(x)] or, equivalently, 1/(n+1)≤x≤n/(n+1). Then, it follows from Theorems 7 and 8(a) that(118)Bngx-gx≤58ω2g;εnx.To close the paper, let us mention some known results concerning the Bernstein polynomials. Gonska and Zhou [12] showed that there exists a constant 0<c<1 such that, for any 1/2≤a<1, there exists N(a) such that(119)sup⁡1-a≤k/n<a⁡Bnfkn-fkn≤cω2f;1n,n≥Na.On the other hand, Kacsó [13] showed that if f∈C([0,1]), then(120)Bnfkn+1-fkn+1≤58ω2f;1n.Finally, Păltănea [5, Corollary  4.2.1, pp. 93-94] has obtained the uniform estimate for x∈(0,1)(121)Bnfx-fx≤118ω2f;εnx,f∈M0,1,n=1,2,….Theorem12(c) and inequality (115) complete in certain sense inequality (119), with ω2(f;1/n) replaced by ω2(f;εn(x)). Similarly, formulas (114), (116), and (118) add new information to inequality (120). Finally, Theorem 12 and Corollary 14 complete the uniform estimate in (121).



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