View of High order approximation theory for Banach space valued functions

J. Numer. Anal. Approx. Theory, vol. 46 (2017) no. 2, pp. 113–130 ictp.acad.ro/jnaat

HIGH ORDER APPROXIMATION THEORY FOR BANACH SPACE VALUED FUNCTIONS

GEORGE A. ANASTASSIOU^∗

Abstract. Here we study quantitatively the high degree of approximation of sequences of linear operators acting on Banach space valued differentiable func- tions to the unit operator. These operators are bounded by real positive linear companion operators. The Banach spaces considered here are general and no pos- itivity assumption is made on the initial linear operators whose we study their approximation properties. We derive pointwise and uniform estimates which im- ply the approximation of these operators to the unit assuming differentiability of functions. At the end we study the special case where the high order deriva- tive of the on hand function fulfills a convexity condition resulting into sharper estimates.

MSC 2010. 41A17, 41A25, 41A36.

Keywords. Banach space valued differentiable functions, positive linear oper- ator, convexity, modulus of continuity, rate of convergence.

1. MOTIVATION

Let (X,k·k) be a Banach space, N ∈ N. Consider g ∈ C([0,1]) and the classic Bernstein polynomials

(1.1) BeNg(t) =

N

X

k=0

g _N^{k N}_kt^k(1−t)^N−k, ∀t∈[0,1].

Let also f ∈ C([0,1], X) and define the vector valued in X Bernsein linear operators

(1.2) B_Nf(t) =

N

X

k=0

f _N^{k N}_kt^k(1−t)^N−k, ∀t∈[0,1]. That is (BNf) (t)∈X.

Clearly herekfk ∈C([0,1]).

∗Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA, e-mail: [email protected].

We notice that (1.3)

k(B_Nf) (t)k ≤

N

X

k=0

f _N^{k N}_kt^k(1−t)^N−k= BeN(kfk)(t),∀t∈[0,1]

The property

(1.4) k(B_Nf) (t)k ≤ BeN(kfk)(t), ∀t∈[0,1],

is shared by almost all summation/integration similar operators and motivates our work here.

Iff(x) =c∈X the constant function, then

(1.5) (BNc) =c.

Ifg∈C([0,1]) andc∈X, thencg∈C([0,1], X) and (1.6) (B_N(cg)) =cB^e_N(g).

Again (1.5), (1.6) are fulfilled by many summation/integration operators.

In fact here (1.6) implies (1.5), wheng≡1.

The above can be generalized from [0,1] to any interval [a, b]⊂R. All this discussion motivates us to consider the following situation.

Let LN : C([a, b], X) ,→ C([a, b], X), (X,k·k) a Banach space, LN is a linear operator, ∀ N ∈ N, x0 ∈ [a, b]. Let also L^eN : C([a, b]) ,→ C([a, b]), a sequence of positive linear operators, ∀N ∈N.

We assume that

(1.7) k(L_N(f)) (x0)k ≤ LeN kfk(x0),

∀ N ∈N,∀ x0 ∈X,∀ f ∈C([a, b], X). When g∈C([a, b]), c∈X, we assume that (1.8) (LN(cg)) =cL^eN(g). The special case of

(1.9) Le_N(1) = 1,

implies

(1.10) L_N(c) =c, ∀c∈X.

We callLeN the companion operator ofLN.

Based on the above fundamental properties we study the high order ap- proximation properties of the sequence of linear operators{L_N}_N∈

N, i.e. their high speed convergence to the unit operator. No kind of positivity property of {L_N}_N∈

Nis assumed. Other important motivation comes from [1], [2], [3]-[6].

Our vector valued differentiation here resembles completely the numerical one, see [7, pp. 83-84].

2. MAIN RESULTS

We need vector Taylor’s formula

Theorem 2.1. [7, pp. 93]. Let n ∈ N and f ∈ Cⁿ([a, b], X), where [a, b]⊂Rand X is a Banach space. Then

(2.1) f(b) =

n−1

X

i=0 (b−a)ⁱ

i! f⁽ⁱ⁾(a) +_(n−1)!¹ Z b

a

(b−t)ⁿ⁻¹f⁽ⁿ⁾(t)dt.

Above the integral is the usual vector valued Riemann integral, see [7, p.

86].

We also need

Theorem 2.2. Let n∈Nand f ∈Cⁿ([a, b], X), where [a, b]⊂Rand X is a Banach space. Then

(2.2) f(a) =

n−1

X

i=0 (a−b)ⁱ

i! f⁽ⁱ⁾(b) +_(n−1)!¹ Z a

b

(a−t)ⁿ⁻¹f⁽ⁿ⁾(t)dt.

Proof. Let

F(x) :=

n−1

X

i=0 (a−x)ⁱ

i! f⁽ⁱ⁾(x), x∈[a, b]. Here F ∈C([a, b], X).Notice thatF(a) =f(a), and

(2.3) F(b) =

n−1

X

i=0 (a−b)ⁱ

i! f⁽ⁱ⁾(b). We have

F⁰(x) = ^(a−x)_(n−1)!ⁿ⁻¹f⁽ⁿ⁾(x), ∀ x∈[a, b]. Clearly F⁰∈C([a, b], X).

By [7, pp. 92] we get

(2.4) F(b)−F(a) =

Z b a

F⁰(t)dt.

That is we have (2.5)

n−1

X

i=0 (a−b)ⁱ

i! f⁽ⁱ⁾(b)−f(a) = Z b

a

(a−t)ⁿ⁻¹

(n−1)! f⁽ⁿ⁾(t)dt=− Z a

b

(a−t)ⁿ⁻¹

(n−1)! f⁽ⁿ⁾(t)dt,

proving (2.2).

Based on the above Theorems 2.1, 2.2, we have

Corollary 2.3. Let (X,k·k) be a Banach space and f ∈ Cⁿ([a, b], X), then we have the vector valued Taylor’s formula

(2.6) f(y)−

n−1

X

i=0

f⁽ⁱ⁾(x)^(y−x)

i

i! = _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹f⁽ⁿ⁾(t)dt,

∀ x, y∈[a, b]or (2.7) f(y)−

n

X

i=0

f⁽ⁱ⁾(x)^(y−x)_i! ⁱ = _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt,

∀ x, y∈[a, b]. We need

Definition 2.4. Let f ∈ C([a, b], X), where (X,k·k) is a Banach space.

We define

(2.8) ω1(f, δ) := sup

x,y:

|x−y|≤δ

kf(x)−f(y)k, 0< δ≤b−a,

the first modulus of continuity of f.

Remark 2.5. We study the remainder of (2.7):

(2.9) Rn(x, y) := _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt,

∀ x, y∈[a, b].

We estimate R_n(x, y).

Case of y≥x. We have kR_n(x, y)k= _(n−1)!¹

Z y x

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt

[7, p. 88]

≤

≤ _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt (2.10)

≤ _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹ω₁ f⁽ⁿ⁾,|t−x|dt^let=^h>0

= _(n−1)!¹ Z y

x

(y−t)ⁿ⁻¹ω₁ f⁽ⁿ⁾,^|t−x|_h hdt

≤ ^ω1(f⁽ⁿ⁾,h)

(n−1)!

Z y x

(y−t)ⁿ⁻¹ 1 +^|t−x|_h dt (2.11)

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!

Z y x

(y−t)ⁿ⁻¹ 1 +^(t−x)_h dt

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!

hZ y x

(y−t)ⁿ⁻¹dt+ ¹_h Z y

x

(y−t)ⁿ⁻¹(t−x)²⁻¹dtⁱ

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!

h_(y−x)n

n + ¹_h^Γ(n)Γ(2)_Γ(n+2) (y−x)ⁿ⁺¹ⁱ (2.12)

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!

h_(y−x)n

n +^(y−x)_n(n+1)hⁿ⁺¹ⁱ= ^ω1(f⁽ⁿ⁾,h)

n! (y−x)^nh1 +_(n+1)h^{(y−x) i}. We have found that

(2.13) kR_n(x, y)k ≤ ^ω1(f⁽ⁿ⁾,h)

n! (y−x)^nh1 +_(n+1)h^{(y−x) i}, fory≥x, and h >0.

Case of y≤x.

Then

kR_n(x, y)k= _(n−1)!¹

Z y x

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt

= _(n−1)!¹

Z x y

(t−y)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt (2.14)

≤ _(n−1)!¹ Z x

y

(t−y)ⁿ⁻¹f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt

≤ _(n−1)!¹ Z x

y

(t−y)ⁿ⁻¹ω₁ f⁽ⁿ⁾,^|t−x|_h hdt

≤ ^ω1(f⁽ⁿ⁾,h)

(n−1)!

Z x y

(t−y)ⁿ⁻¹ 1 +^(x−t)_h dt

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!

Z x y

(t−y)ⁿ⁻¹dt+_h¹ Z x

y

(x−t)²⁻¹(t−y)ⁿ⁻¹dt (2.15)

= ^ω1(^f(n),h)

(n−1)!

h_(x−y)n

n + ^(x−y)

n+1

n(n+1)h

i

= ^ω1(^f(n),h)

n! (x−y)^nh1 +_(n+1)h^{(x−y) i}. Hence

(2.16) kR_n(x, y)k ≤ ^ω1(f⁽ⁿ⁾,h)

n! (x−y)^nh1 +_(n+1)h^{(x−y) i}, when y≤x,h >0.

We have proved that kR_n(x, y)k=

1 (n−1)!

Z y x

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x)dt

≤

≤ ^ω1(f⁽ⁿ⁾,h)

n! |x−y|^nh1 +_(n+1)h^{|x−y| i}, (2.17)

∀ x, y∈[a, b], h >0.

We have established

Theorem 2.6. Let (X,k·k) be a Banach space and f ∈Cⁿ([a, b], X), n∈ N. Then

(2.18)

f(y)−

n

X

i=0

f⁽ⁱ⁾(x)^(y−x)_i! ⁱ

≤ ^ω1(f⁽ⁿ⁾,h)

n! |x−y|^nh1 +_(n+1)h^{|x−y| i},

∀ x, y∈[a, b], h >0.

It follows our first main result

Theorem 2.7. Let N ∈ N and LN : C([a, b], X) → C([a, b], X), where (X,k·k)is a Banach space andL_N is a linear operator. Let the positive linear operators L^e_N :C([a, b]),→C([a, b]), such that

(2.19) k(L_N(f)) (x₀)k ≤ Le_N(kfk)(x₀),

∀ N ∈N,∀ f ∈C([a, b], X), ∀ x0 ∈[a, b]. Furthermore assume that

(2.20) L_N(cg) =cL^e_N(g), ∀ g∈C([a, b]), ∀ c∈X.

Let n∈N, here we deal withf ∈Cⁿ([a, b], X). Then

1)

(L_N(f)) (x₀)−

n

X

i=0

fⁿ⁽ⁱ⁾(x0)

i! Le_N (· −x₀)ⁱ(x₀)

≤

≤ ^{ω1 f}

(n), e^LN(^|·−x0|ⁿ⁺¹)(x0)_n+1¹

n!

Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )

×

×^h LeN(1)(x0)

1

n+1 +_n+1^{1 i}, (2.21)

2)

k(L_N(f)) (x₀)−f(x₀)k ≤

≤ kf(x0)k LeN(1)(x0)−1+ +

n

X

k=0

kf^(k)(x0)k

k! Le_N(| · −x₀|^k)(x₀)+ +ω₁ f⁽ⁿ⁾, L^e_N |· −x₀|ⁿ⁺¹(x₀)

1 n+1

n! L^e_N · −x₀

n+1

(x₀)(_n+1ⁿ )×

×^h Le_N(1)(x₀)

1

n+1 +_n+1^{1 i}, (2.22)

from(2.22), and as Le_N(1)(x₀)→1, Le_N · −x₀

n+1

(x₀)→0, we obtain (LN(f)) (x0)→f(x0), as N → ∞,

3) if f^(k)(x₀) = 0, k= 0,1, ..., n, we get that k(L_N(f)) (x₀)−f(x₀)k ≤

≤ ^{ω1 f}

(n), ^LeN(^|·−x0|ⁿ⁺¹)(x0)_n+1¹

n! ×

× LeN(| · −x₀|ⁿ⁺¹)(x0)⁽

n n+1)h

LeN(1)(x0)

1

n+1 +t_n+1^{1 i}, (2.23)

an extreme high speed of convergence,

4) one also derives

kk(L_N(f))−fkk_∞,[a,b]≤

≤ kkfkk_∞,[a,b] L^e_N(1)−1

∞,[a,b]+ +

n

X

k=0

kk^f(k)kk_∞,[a,b]

k!

LeN | · −x₀|^k(x0)

∞,x0∈[a,b]+ +

ω1 f⁽ⁿ⁾,

^LeN(|·−x0|ⁿ⁺¹)(x0)

1 n+1

∞,x0∈[a,b]

n! ×

× Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )

∞,x0∈[a,b]

eL_N(1)

1

n+1 +_n+1¹ , (2.24)

if Le_N(1)→^u 1, uniformly, and Le_N |· −x₀|ⁿ⁺¹(x₀)→^u 0, uniformly in x₀ ∈ [a, b], by (2.24), we obtain L_N(f)→^u f, uniformly, asN → ∞.

Proof. 1) One can rewrite (2.18) as follows (2.25) f(·)−

n

X

i=0

f⁽ⁱ⁾(x₀)^(·−x_i!⁰⁾ⁱ≤ ^ω1(f⁽ⁿ⁾,h)

n!

h|· −x₀|ⁿ+ ^|·−x_(n+1)h^0|n+1i, for a fixedx₀∈[a, b], h >0.

We observe that (N ∈N)

(L_N(f)) (x₀)−

n

X

i=0 f⁽ⁱ⁾(x0)

i! Le_N (· −x₀)ⁱ(x₀)

=

=

L_N^hf(·)−

n

X

i=0

f⁽ⁱ⁾(x₀)^(·−x_i!⁰⁾ⁱⁱ(x₀) (2.26)

≤LeN

f(·)−

n

X

i=0

f⁽ⁱ⁾(x0)^(·−x_i!⁰⁾ⁱ(x0)

(by (2.25))

≤

≤ ^ω1(f⁽ⁿ⁾,h)

n!

h

Le_N(|· −x₀|ⁿ)(x₀) + ^LeN(^|·−x0|ⁿ⁺¹)^(x0) (n+1)h

i=: (ξ₁). (2.27)

Above notice that f(·)−^Pn_i=0^f(i)_i!^(x0)(· −x0)ⁱ∈C([a, b], X).

By H¨older’s inequality and Riesz representation theorem we obtain (2.28)

LeN(|· −x0|ⁿ)(x0)≤ LeN |· −x0|ⁿ⁺¹(x0)(_n+1ⁿ )

LeN(1)(x0)

1 (n+1). Therefore

(ξ1)≤ ^{ω1 f}

(n),h

n!

LeN · −x₀

n+1

(x0)(_n+1ⁿ )

LeN(1)(x0)

1 (n+1)

+_h^{1 eLN}(^|·−x0|ⁿ⁺¹)(x0) (n+1)

=: (ξ2). (2.29)

We choose

(2.30) h:= Le_N |· −x₀|ⁿ⁺¹(x₀)

1 (n+1), in case of LeN |· −x0|ⁿ⁺¹(x0)>0.

Then it holds (ξ₂) =

ω1

f⁽ⁿ⁾, ^LeN(^|·−x0|ⁿ⁺¹)^(x0)_(n+1)¹ (2.31) n!

×

LeN |· −x0|ⁿ⁺¹(x0)(_n+1ⁿ )

LeN(1)(x0)

1 (n+1)

+ ^LeN(^|·−x0|ⁿ⁺¹)(x0)(_n+1ⁿ )

(n+1)

=

=

ω1

f⁽ⁿ⁾, LeN(^|·−x0|ⁿ⁺¹)(x0)_(n+1)¹

n!

× Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )^h

Le_N(1)(x₀)

1

(n+1) +_n+1^{1 i}. (2.32)

We have proved that

(LN(f)) (x0)−

n

X

i=0 f⁽ⁱ⁾(x0)

i! LeN (· −x0)ⁱ(x0)

≤

≤ ^ω1

f⁽ⁿ⁾, ^LeN(^|·−x0|ⁿ⁺¹)(x0)_(n+1)¹

n!

× LeN |· −x0|ⁿ⁺¹(x0)(_n+1ⁿ )^h

LeN(1)(x0)

1

(n+1) + _n+1^{1 i}. (2.33)

By Riesz representation theorem we have (2.34) Le_N(g)(x₀) =

Z

[a,b]

g(t)dµ_x0(t), ∀g∈C([a, b]), whereµ_x0 is a positive finite measure on [a, b].

That is

(2.35) LeN(1)(x0) =µx0([a, b]) =:M.

We have that µ_x0([a, b]) > 0, because otherwise, if µ_x0([a, b]) = 0, then Le_n(g)(x₀) = 0, ∀g∈C([a, b]), and the whole theory here becomes trivial.

Therefore it holds Le_N(1)(x₀)>0.

In case of

(2.36) LeN |· −x0|ⁿ⁺¹(x0) = 0, we have

(2.37)

Z

[a,b]

|t−x₀|ⁿ⁺¹dµ_x0(t) = 0.

The last implies |t−x₀|ⁿ⁺¹ = 0, a.e, hence |t−x₀|= 0, a.e, then t−x₀ = 0 a.e., and t = x₀, a.e. on [a, b]. Consequently µ_x0({t∈[a, b] :t6=x₀}) = 0.

That is µx0 = δx0M, where δx0 is the Dirac measure at {x₀}. In that case holds

(2.38) Le_N(g)(x₀) =g(x₀)M, ∀g∈C([a, b]).

Under (2.36), the right hand side of (2.33) equals zero. Furthermore it holds (2.39)

LeN

f(·)−

n

X

i=0

f⁽ⁱ⁾(x0)^(·−x_i!⁰⁾ⁱ(x0)^(2.38)= kf(x0)−f(x0)kM = 0.

So that by (2.26) to have

(LN(f)) (x0)−

n

X

i=0 f⁽ⁱ⁾(x0)

i! LeN (· −x0)ⁱ(x0)

=

=k(L_N(f)) (x0)−f(x0)Mk= 0, (2.40)

also implying

(2.41) (LN(f)) (x0) =M f(x0).

So we have proved that inequality (2.33) will be always true.

2) Next we see that

(LN(f)) (x0)−f(x0)=

=

(L_N(f)) (x₀)−

n

X

k=0

f^(k)(x0)

k! Le_N · −x₀^k(x₀) +

+

n

X

k=0

f^(k)(x0)

k! LeN (· −x0)^k(x0)−f(x0)

≤ (2.42)

≤

(LN(f)) (x0)−

n

X

k=0

f^(k)(x0)

k! LeN (· −x0)^k(x0)

+ +

n

X

k=1

f^(k)(x0)

k! Le_N (· −x₀)^k(x₀) +f(x₀) Le_N 1(x₀)−f x₀

(2.33)

≤

≤ kf(x₀)k Le_N(1)(x₀)−1+

n

X

k=1

kf^(k)(x0)k

k! Le_N |· −x₀|^k(x₀) + +

ω1

f⁽ⁿ⁾, e^LN(^|·−x0|ⁿ⁺¹)(x0)_(n+1)¹

n!

× LeN |· −x0|ⁿ⁺¹(x0)(_n+1ⁿ )^h

LeN(1)(x0)

1

(n+1) +_n+1^{1 i}. (2.43)

We have proved that

k(L_N(f)) (x₀)−f(x₀)k

≤ kf(x0)kLeN(1)(x0)−1+ +

n

X

k=1

kf^(k)(x0)k

k! Le_N · −x₀

k (x₀) + +^{ω1 f}

(n), e^LN(^|·−x0|ⁿ⁺¹)(x0)_(n+1)¹ (2.44) n!

× Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )^h

Le_N(1)(x₀)

1

(n+1) + _n+1^{1 i}. By H¨older’s inequality for k= 1, ..., n,we obtain

(2.45)

Le_N · −x₀

k

(x₀)≤ Le_N · −x₀

n+1

x₀(_n+1^k )

Le_N(1)(x₀)(^n+1−k_n+1 ) . Clealry by (2.44) and (2.45), when Le_N(1)(x₀)→1 and Le_N ·−x₀

n+1

(x₀)→ 0, we obtain (LN(f)) (x0) → f(x0), as N → ∞. Notice that LeN(1)(x0) will be bounded.

3) Iff^(k)(x0) = 0, k= 0,1, ..., n,we get that

k(L_N(f)) (x0)−f(x0)k⁽⁵⁴⁾≤

≤ ^ω1

f⁽ⁿ⁾, LeN(^|·−x0|ⁿ⁺¹)(x0)_(n+1)¹

n!

× Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )h

Le_N 1(x₀)

1

(n+1) +_n+1^{1 i}, (2.46)

an extreme high speed of convergence.

4) One also derives from (2.44) that kk(L_N(f))−fkk_∞,[a,b]≤

≤ kkfkk_∞,[a,b]Le_N(1)−1

∞,[a,b]+ +

n

X

k=1

kkf^(k)kk_∞,[a,b]

k!

Le_N |· −x₀|^k(x₀)

∞,x0∈[a,b]+ +

ω1

f⁽ⁿ⁾,

e^LN(^|·−x0|ⁿ⁺¹)(x0)

1 (n+1)

∞,x0∈[a,b]

n!

× Le_N |· −x₀|ⁿ⁺¹(x₀)(_n+1ⁿ )

∞,x0∈[a,b]

kLe_N(1)k⁽ⁿ⁺¹⁾¹ +_n+1¹ . (2.47)

Inequality (2.45), fork= 1, ..., n, implies

Le_N |· −x₀|^k(x₀)

∞,x0∈[a,b]

≤ LeN |· −x0|ⁿ⁺¹(x0)(_n+1^k )

∞,x₀∈[a,b]

LeN(1)(^n+1−k_n+1 )

∞,x₀∈[a,b]. (2.48)

Consequently, if Le_N(1) →^u 1, uniformly, and Le_N |· −x₀|ⁿ⁺¹(x₀) →^u 0, uniformly inx₀ ∈[a, b], by (2.47) and (2.48), we obtainL_N(f)→^u f, uniformly, asN → ∞.

Here the assumption LeN(1) →^u 1, uniformly, as N → ∞, implies that

eL_N(1) is bounded.

The proof of the theorem now is complete.

We make

Remark2.8. Let (X,k·k) be a Banach space andf ∈Cⁿ([a, b], X),n∈N, and x0 ∈(a, b) be fixed. Then

f(y)−

n

X

i=0

f⁽ⁱ⁾(x₀)^(y−x_i!⁰⁾ⁱ = _(n−1)!¹ Z y

x0

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x₀)dt

=:Rn(x0, y) , ∀y∈[a, b]

(2.49)

We assume thatg(t) :=f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0) is convex int∈[a, b].

We consider 0 < h≤ min (x₀−a, b−x₀). Obviously g(x₀) = 0. Then by Lemma 8.1.1, p. 243 of [1], we obtain

(2.50) g(t)≤ ^ω1(g,h)_h |t−x₀|, ∀ t∈[a, b]. For anyt1, t2∈[a, b] :|t₁−t2| ≤h we get

f⁽ⁿ⁾(t₁)−f⁽ⁿ⁾(x₀)−f⁽ⁿ⁾(t₂)−f⁽ⁿ⁾(x₀)

≤f⁽ⁿ⁾(t₁)−f⁽ⁿ⁾(t₂)≤ω₁(fⁿ, h). (2.51)

That is

(2.52) ω₁(g, h)≤ω₁ f⁽ⁿ⁾, h. The last implies

(2.53) f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0)≤ ^ω1(f⁽ⁿ⁾,h)

h |t−x0|, ∀ t∈[a, b]. We estimate R_n(x₀, y).

Case of y≥x0.We have kR_n(x0, y)k= _(n−1)!¹

Z y x0

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0)dt

([7, pp. 88])

≤

≤ _(n−1)!¹ Z y

x0

(y−t)ⁿ⁻¹f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0)dt

(2.53)

≤

≤ ^ω1(f⁽ⁿ⁾,h)

(n−1)!h

Z y x0

(y−t)ⁿ⁻¹(t−x₀)²⁻¹dt=

= ^ω1(f⁽ⁿ⁾,h)

(n−1)!h

Γ(n)Γ(2)

Γ(n+2) (y−x0)ⁿ⁺¹ = ^ω1(f⁽ⁿ⁾,h)

(n+1)!h (y−x0)ⁿ⁺¹. (2.54)

We proved that

(2.55) kR_n(x0, y)k ≤ ^ω1(f⁽ⁿ⁾,h)

(n+1)!h (y−x0)ⁿ⁺¹, ∀y ≥x0. Case of y≤x₀. Then

kR_n(x0, y)k= _(n−1)!¹

Z y x0

(y−t)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0)dt

= _(n−1)!¹

Z x0

y

(t−y)ⁿ⁻¹ f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x0)dt

≤ _(n−1)!¹ Z x0

y

(t−y)ⁿ⁻¹f⁽ⁿ⁾(t)−f⁽ⁿ⁾(x₀)dt

(2.53)

≤

≤ ^ω1(f⁽ⁿ⁾,h)

h(n−1)!

Z x0

y

(x0−t)²⁻¹(t−y)ⁿ⁻¹dt (2.56)

= ^ω1(f⁽ⁿ⁾,h)

h(n+1)! (x0−y)ⁿ⁺¹. That is proving

(2.57) kR_n(x₀, y)k ≤ ^ω1(f⁽ⁿ⁾,h)

h(n+1)! (x₀−y)ⁿ⁺¹, ∀y∈[a, b] :y ≤x₀. We have established that

(2.58) kR_n(x0, y)k ≤ ^ω1(f⁽ⁿ⁾,h)

h(n+1)! |y−x0|ⁿ⁺¹, ∀ y∈[a, b],

where 0< h≤min (x₀−a, b−x₀),x₀ ∈(a, b), andkf(·)−f(x₀)k is convex over [a, b].

We have proved

Theorem 2.9. Let (X,k·k) be a Banach space and f ∈Cⁿ([a, b], X), n∈ N, and x₀ ∈(a, b) be fixed. Let0< h≤min (x₀−a, b−x₀), and assume that

f⁽ⁿ⁾(·)−f⁽ⁿ⁾(x₀)

is convex over[a, b]. Then (2.59)

f(y)−

n

X

i=0

f⁽ⁱ⁾(x0)^(y−x0)

i

i!

≤ ^ω1(^f(n),h)

h(n+1)! |y−x0|ⁿ⁺¹, ∀ y∈[a, b]. We give our second main result under convexity.