General Norm Inequalities of Trapezoid Type for Fréchet Differentiable Functions in Banach Spaces

S S symmetry

Article General Norm Inequalities of Trapezoid Type for Fréchet Differentiable Functions in Banach Spaces

Silvestru Sever Dragomir 1,2

1 Mathematics, College of Engineering & Science, Victoria University, P.O. Box 14428, Melbourne 8001, Australia; [email protected] 2 DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science & Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa

Abstract: In this paper we establish some error bounds in approximating the integral by general trapezoid type rules for Fréchet differentiable functions with values in Banach spaces.

Keywords: Banach spaces; norm inequalities; midpoint inequalities; Banach algebras

1. Introduction We recall some facts about differentiation of functions between normed vector spaces [1]. Let O be an open subset of a normed vector space E, f a real-valued function defined on O, a ∈ O and u a nonzero element of E. The function fu given by t 7→ f (a + tu) is d fu defined on an open interval containing 0. If the derivative dt (0) is defined, i.e., if the limit Citation: Dragomir, S.S. General Norm Inequalities of Trapezoid Type d f f (a + tu) − f (a) u (0) := lim for Fréchet Differentiable Functions dt t→0 t in Banach Spaces. Symmetry 2021, 13, 1288. https://doi.org/10.3390/ exists, then we denote this derivative by ∇a f (u). It is called the Gâteaux derivative (di- sym13071288 rectional derivative) of f at a in the direction u. If ∇a f (u) is defined and λ ∈ R\{0}, then ∇a f (λu) is defined and ∇a f (λu) = λ∇a f (u). The function f is Gâteaux differentiable at a Academic Editors: Carmen Violeta if ∇a f (u) exists for all directions u. Muraru, Ana-Maria Acu, Marius Let E and F be normed vector spaces, and O be an open subset of E. A function Birou and Radu Voichita Adriana f : O → F is called Fréchet differentiable at x ∈ O if there exists a bounded linear operator A : E → F such that k f (x + h) − f (x) − Ahk Received: 22 June 2021 lim = 0. Accepted: 12 July 2021 khk→0 khk Published: 17 July 2021 If there exists such an operator A, it is unique, so we write D f (x) = A and call it the Fréchet derivative of f at x. Publisher’s Note: MDPI stays neutral A function f that is Fréchet differentiable for any point of O is said to be C1 if the with regard to jurisdictional claims in function O 3 x 7→ D f (x) ∈ B(E, F) is continuous, where B(E, F) is the space of all published maps and institutional affil- bounded linear operators defined on E with values in F. A function Fréchet differentiable iations. at a point is continuous at that point. Fréchet differentiation is a linear operation. If f is Fréchet differentiable at x, it is also Gâteaux differentiable there, and ∇x f (u) = D f (x)(u) for all u ∈ E. We say that the function f : O ⊂ E → F is L-Lipschitzian on O with the constant L > 0 Copyright: © 2021 by the author. if Licensee MDPI, Basel, Switzerland. k f (x) − f (y)k ≤ Lkx − yk for all x, y ∈ O. This article is an open access article distributed under the terms and In [2] we established among others the following midpoint and trapezoid type in- conditions of the Creative Commons equalities for L-Lipschitzian functions f on an open and convex subset C in E Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ Z 1 x + y 1 f ((1 − t)t + ty)dt − f ≤ Lkx − yk (1) 4.0/). 0 2 4

Symmetry 2021, 13, 1288. https://doi.org/10.3390/sym13071288 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 1288 2 of 19

and f (x) + f (y) Z 1 1 − f ((1 − t)t + ty)dt ≤ Lkx − yk (2) 2 0 4 1 for all x, y ∈ C. The constant 4 is best possible in both inequalities (1) and (2). For Hermite–Hadamard’s type inequalities for functions with scalar values, namely

a + b 1 Z b f (a) + f (b) f ≤ f (u)du ≤ , (3) 2 b − a a 2

where f : [a, b] → R is convex on [a, b], see for instance [1,3–21] and the references therein. Some of the integrals used in these papers are taken in the sense of Riemann–Stieltjes while in the current paper we consider only vector valued functions with values in Banach spaces, in the ﬁrst instance, and secondly, with values in Banach algebras where some applications for some fundamental functions like the exponential are also given. In the recent paper [22] we obtained among others the following weighted version of the trapezoid inequality (2).

Theorem 1. Let f : C ⊂ E → F be a function of class C1 on the open and convex subset C of the Banach space E with values into another Banach space F and p : [0, 1] → [0, ∞) a Lebesgue integrable and symmetric function on [0, 1], namely p(1 − t) = p(t) for all t ∈ [0, 1]. Then for all x, y ∈ C Z 1 f (x) + f (y) Z 1 p(t)dt − p(t) f ((1 − t)x + ty)dt (4) 0 2 0 Z 1/2Z 1/2 ≤ p(s)ds kD f ((1 − t)x + ty)(y − x)kdt 0 t Z 1 Z t + p(s)ds kD f ((1 − t)x + ty)(y − x)kdt 1/2 1/2

Z 1 Z 1/2 = p(s)ds kD f ((1 − t)x + ty)(y − x)kdt 0 t =: B( f , p, x, y).

Moreover, we have the upper bounds

1 Z 1 Z 1 B( f , p, x, y) ≤ p(s)ds kD f ((1 − t)x + ty)(y − x)kdt (5) 2 0 0 1 Z 1 ≤ p(s)ds sup kD f ((1 − t)x + ty)(y − x)k, 2 0 t∈[0,1]

1 Z 1 Z 1 1 B( f , p, x, y) ≤ p(s)ds − sgn t − tp(t)dt (6) 2 0 0 2 × sup kD f ((1 − t)x + ty)(y − x)k t∈[0,1] 1 Z 1 ≤ p(s)ds sup kD f ((1 − t)x + ty)(y − x)k 2 0 t∈[0,1] Symmetry 2021, 13, 1288 3 of 19

and

" #1/r Z 1/2Z 1/2 r Z 1 Z t r B( f , p, x, y) ≤ p(s)ds dt + p(s)ds dt (7) 0 t 1/2 1/2

Z 1 1/q × kD f ((1 − t)x + ty)(y − x)kqdt 0 " #1/r Z 1/2Z 1/2 r Z 1 Z t r ≤ p(s)ds dt + p(s)ds dt 0 t 1/2 1/2 × sup kD f ((1 − t)x + ty)(y − x)k t∈[0,1]

1 1 for r, q > 1 with r + q = 1.

Motivated by the above results, in this paper we establish some upper bounds for the quantity Z 1 Z 1 p(s)ds − γ f (y) + γ f (x) − p(t) f ((1 − t)x + ty)dt 0 0

in the case that f : C ⊂ E → F is Fréchet differentiable on the open and convex subset C of the Banach space E with values into another Banach space F, x, y ∈ C, p : [0, 1] → C is a Lebesgue integrable function and γ ∈ C. Some particular cases of interest for different choices of γ are given. Applications for Banach algebras are also provided.

2. Some Identities of Interest Consider a function f : C ⊂ E → F that is deﬁned on the open and convex set C. We have the following properties for the auxiliary function

ϕ(x,y)(t) := f ((1 − t)x + ty), t ∈ [0, 1],

where x, y ∈ C.

Lemma 1. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Then for all x, y ∈ C the auxiliary function ϕ(x,y) is differentiable on (0, 1) and

0 ϕ(x,y)(t) = D f ((1 − t)x + ty)(y − x). (8)

Also 0 ϕ(x,y)(0+) = D f (x)(y − x) (9) and 0 ϕ(x,y)(1−) = D f (y)(y − x). (10)

Proof. Let t ∈ (0, 1) and h 6= 0 small enough such that t + h ∈ (0, 1). Then

ϕ(x,y)(t + h) − ϕ(x,y)(t) (11) h f ((1 − t − h)x + (t + h)y) − f ((1 − t)x + ty) = h f ((1 − t)x + ty + h(y − x)) − f ((1 − t)x + ty) = . h Symmetry 2021, 13, 1288 4 of 19

Since f is Fréchet differentiable, hence by taking the limit over h → 0 in (11) we get

0 ϕ(x,y)(t + h) − ϕ(x,y)(t) ϕ(x,y)(t) = lim h→0 h f ((1 − t)x + ty + h(y − x)) − f ((1 − t)x + ty) = lim h→0 h = D f ((1 − t)x + ty)(y − x),

which proves (8). Additionally, we have

0 ϕ(x,y)(h) − ϕ(x,y)(0) ϕ(x,y)(0+) = lim h→0+ h f ((1 − h)x + hy) − f (x) = lim h→0+ h f (x + h(y − x)) − f (x) = lim = D f (x)(y − x) h→0+ h

since f is assumed to be Fréchet differentiable in x. This proves (9). The equality (10) follows in a similar way.

We have the following identity for the Riemann–Stieltjes integral:

Lemma 2. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Let u : [0, 1] → C be of bounded variation on [0, 1] and s ∈ [0, 1]. Then for all x, y ∈ C and any γ, µ ∈ C,

[u(1) − µ]ϕ(x,y)(1) + [γ − u(0)]ϕ(x,y)(0) + (µ − γ)ϕ(x,y)(s) (12) Z 1 − ϕ(x,y)(t)du(t) 0 Z s Z 1 0 0 = [u(t) − γ]ϕ(x,y)(t)dt + [u(t) − µ]ϕ(x,y)(t)dt. 0 s In particular, for µ = γ we have

Z 1 [u(1) − γ]ϕ(x,y)(1) + [γ − u(0)]ϕ(x,y)(0) − ϕ(x,y)(t)du(t) (13) 0 Z 1 0 = [u(t) − γ]ϕ(x,y)(t)dt. 0

Proof. Using integration by parts rule for the Riemann–Stieltjes integral, we have

Z s 0 [u(t) − γ]ϕ(x,y)(t)dt = [u(s) − γ]ϕ(x,y)(s) − [u(0) − γ]ϕ(x,y)(0) 0 Z s − ϕ(x,y)(t)du(t) 0 and

Z 1 0 [u(t) − µ]ϕ(x,y)(t)dt = [u(1) − µ]ϕ(x,y)(1) − [u(s) − µ]ϕ(x,y)(s) s Z 1 − ϕ(x,y)(t)du(t) s

for any s ∈ [0, 1]. Symmetry 2021, 13, 1288 5 of 19

If we add these two equalities, then we get

Z s Z 1 0 0 [u(t) − γ]ϕ(x,y)(t)dt + [u(t) − µ]ϕ(x,y)(t)dt 0 s

= [u(1) − µ]ϕ(x,y)(1) + [γ − u(0)]ϕ(x,y)(0) + [µ − u(s)]ϕ(x,y)(s) Z s Z 1 + [u(s) − γ]ϕ(x,y)(s) − ϕ(x,y)(t)du(t) − ϕ(x,y)(t)du(t) 0 s

= [u(1) − µ]ϕ(x,y)(1) + [γ − u(0)]ϕ(x,y)(0) + (µ − γ)ϕ(x,y)(s) Z 1 − ϕ(x,y)(t)du(t) 0

for any s ∈ [0, 1], which proves the desired equality (12).

Remark 1. From the equality (13) we have for s ∈ [0, 1] and γ = u(s) that

Z 1 [u(1) − u(s)]ϕ(x,y)(1) + [u(s) − u(0)]ϕ(x,y)(0) − ϕ(x,y)(t)du(t) (14) 0 Z 1 0 = [u(t) − u(s)]ϕ(x,y)(t)dt 0 and, in particular

1 1 u(1) − u ϕ (1) + u − u(0) ϕ (0) (15) 2 (x,y) 2 (x,y) Z 1 − ϕ(x,y)(t)du(t) 0 Z 1 1 0 = u(t) − u ϕ(x,y)(t)dt. 0 2

u(0)+u(1) Additionally, if m ∈ [0, 1] is such that u(m) = 2 , then from (14) we get

ϕ(x,y)(1) + ϕ(x,y)(0) Z 1 [u(1) − u(0)] − ϕ(x,y)(t)du(t) (16) 2 0 Z 1 0 = [u(t) − u(m)]ϕ(x,y)(t)dt. 0

Now, if we take γ = (1 − α)u(0) + αu(1), α ∈ [0, 1] in (13), then we get

h i Z 1 [u(1) − u(0)] (1 − α)ϕ(x,y)(1) + αϕ(x,y)(0) − ϕ(x,y)(t)du(t) (17) 0 Z 1 0 = [u(t) − (1 − α)u(0) − αu(1)]ϕ(x,y)(t)dt 0 and, in particular

ϕ(x,y)(1) + ϕ(x,y)(0) Z 1 [u(1) − u(0)] − ϕ(x,y)(t)du(t) (18) 2 0 Z 1 u(0) + u(1) 0 = u(t) − ϕ(x,y)(t)dt. 0 2

The case of weighted integrals is as follows: Symmetry 2021, 13, 1288 6 of 19

Corollary 1. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Let p : [0, 1] → C be Lebesgue integrable on [0, 1] and s ∈ [0, 1]. Then for all x, y ∈ C and any γ, µ ∈ C,

Z 1 p(s)ds − µ ϕ(x,y)(1) + γϕ(x,y)(0) + (µ − γ)ϕ(x,y)(s) (19) 0 Z 1 − p(t)ϕ(x,y)(t)dt 0 Z sZ t Z 1Z t 0 0 = p(τ)dτ − γ ϕ(x,y)(t)dt + p(τ)dτ − µ ϕ(x,y)(t)dt. 0 0 s 0

In particular, for µ = γ we have

Z 1 Z 1 p(s)ds − γ ϕ(x,y)(1) + γϕ(x,y)(0) − p(t)ϕ(x,y)(t)dt (20) 0 0 Z 1Z t 0 = p(τ)dτ − γ ϕ(x,y)(t)dt. 0 0

R t The proof follows by Lemma2 applied for the function u : [0, 1] → C, u(t) = 0 p(s)ds that is absolutely continuous on [0, 1] and therefore of bounded variation and

Z 1 Z 1 ϕ(x,y)(t)du(t) = p(t)ϕ(x,y)(t)dt. 0 0

Remark 2. With the assumptions of Corollary1 and by utilizing Remark1 we get

Z 1 Z s Z 1 p(τ)dτ ϕ(x,y)(1) + p(τ)dτ ϕ(x,y)(0) − p(t)ϕ(x,y)(t)dt (21) s 0 0 Z 1Z t 0 = p(τ)dτ ϕ(x,y)(t)dt 0 s

and, in particular

Z 1 Z 1/2 p(τ)dτ ϕ(x,y)(1) + p(τ)dτ ϕ(x,y)(0) (22) 1/2 0 Z 1 − p(t)ϕ(x,y)(t)dt 0 Z 1Z t 0 = p(τ)dτ ϕ(x,y)(t)dt. 0 1/2

R m 1 R 1 Additionally, if m ∈ [0, 1] is such that 0 p(τ)dτ = 2 0 p(τ)dτ, then

ϕ(x,y)(1) + ϕ(x,y)(0) Z 1 Z 1 p(τ)dτ − p(t)ϕ(x,y)(t)dt (23) 2 0 0 Z 1Z t 0 = p(τ)dτ ϕ(x,y)(t)dt. 0 m

Now, for α ∈ [0, 1] we get

h i Z 1 Z 1 (1 − α)ϕ(x,y)(1) + αϕ(x,y)(0) p(τ)dτ − p(t)ϕ(x,y)(t)dt (24) 0 0 Z 1Z t Z 1 0 = p(τ)dτ − α p(τ)dτ ϕ(x,y)(t)dt 0 0 0 Symmetry 2021, 13, 1288 7 of 19

and, in particular

ϕ(x,y)(1) + ϕ(x,y)(0) Z 1 Z 1 p(τ)dτ − p(t)ϕ(x,y)(t)dt (25) 2 0 0 Z 1Z t Z 1 1 0 = p(τ)dτ − p(τ)dτ ϕ(x,y)(t)dt. 0 0 2 0

3. Main Results We have:

Theorem 2. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Let p : [0, 1] → C be Lebesgue integrable on [0, 1] and s ∈ [0, 1]. Then for all x, y ∈ C and any γ, µ ∈ C, Z 1 p(s)ds − µ f (y) + γ f (x) + (µ − γ) f ((1 − s)x + sy) (26) 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0

Z s Z t ≤ p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt 0 0

Z 1 Z t + p(τ)dτ − µ kD f ((1 − t)x + ty)(y − x)kdt s 0 =: B( f , p, x, y, γ, µ).

In particular, for µ = γ we have Z 1 Z 1 p(s)ds − γ f (y) + γ f (x) − p(t) f ((1 − t)x + ty)dt (27) 0 0

Z 1 Z t ≤ p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt 0 0 =: B( f , p, x, y, γ).

Moreover, we have the upper bounds

B( f , p, x, y, γ, µ) (28)  n o R t ( ) − R t ( ) −  max supt∈[0,s] 0 p τ dτ γ , supt∈[s,1] 0 p τ dτ µ  h i  × R 1k (( − ) + )( − )k  0 D f 1 t x ty y x dt     h r r i1/r  R s R t p(τ)dτ − γ dt + R 1 R t p(τ)dτ − µ dt  0 0 s 0 ≤ 1/q × R 1 − + − q  0 kD f ((1 t)x ty)(y x)k dt   r, q > 1, 1 + 1 = 1  r q    h i  R s R t p(τ)dτ − γ dt + R 1 R t p(τ)dτ − µ dt  0 0 s 0   × supt∈[0,1]kD f ((1 − t)x + ty)(y − x)k Symmetry 2021, 13, 1288 8 of 19

and

B( f , p, x, y, γ) (29)  R t hR 1 i  sup p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt  t∈[0,1] 0 0    1/r  hR 1 R t r i  p(τ)dτ − γ dt ≤ 0 0 1/q  × R 1k (( − ) + )( − )kq > 1 + 1 =  0 D f 1 t x ty y x dt r, q 1, r q 1    h i  R 1 R t ( ) − k (( − ) + )( − )k  0 0 p τ dτ γ dt supt∈[0,1] D f 1 t x ty y x .

Proof. We have by (19) that

Z 1 p(s)ds − µ f (y) + γ f (x) + (µ − γ) f ((1 − s)x + sy) (30) 0 Z 1 − p(t) f ((1 − t)x + ty)dt 0 Z sZ t = p(τ)dτ − γ D f ((1 − t)x + ty)(y − x)dt 0 0 Z 1Z t + p(τ)dτ − µ D f ((1 − t)x + ty)(y − x)dt s 0

for all x, y ∈ C and any γ, µ ∈ C. By taking the norm in (30), we get Z 1 p(s)ds − µ f (y) + γ f (x) + (µ − γ) f ((1 − s)x + sy) (31) 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0 Z s Z t ≤ p(τ)dτ − γ D f ((1 − t)x + ty)(y − x)dt 0 0 Z 1 Z t + p(τ)dτ − µ D f ((1 − t)x + ty)(y − x)dt s 0 Z s Z t ≤ p(τ)dτ − γ D f ((1 − t)x + ty)(y − x) dt 0 0 Z 1 Z t + p(τ)dτ − µ D f ((1 − t)x + ty)(y − x) dt s 0

Z s Z t = p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt 0 0

Z 1 Z t + p(τ)dτ − µ kD f ((1 − t)x + ty)(y − x)kdt s 0 = B( f , p, x, y, γ, µ). Symmetry 2021, 13, 1288 9 of 19

By using Hölder’s inequality we have

Z s Z t p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt 0 0  R t R s  sup p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt  t∈[0,s] 0 0    1/r  s t r s q 1/q  R R p(τ)dτ − γ dt R kD f ((1 − t)x + ty)(y − x)k dt ≤ 0 0 0 1 1  r, q > 1, r + q = 1     k (( − ) + )( − )k R s R t ( ) −  supt∈[0,s] D f 1 t x ty y x 0 0 p τ dτ γ dt

and

Z 1 Z t p(τ)dτ − µ kD f ((1 − t)x + ty)(y − x)kdt s 0  R t R 1  sup p(τ)dτ − µ kD f ((1 − t)x + ty)(y − x)kdt  t∈[s,1] 0 s    1/r  1 t r 1 q 1/q  R R p(τ)dτ − µ dt R kD f ((1 − t)x + ty)(y − x)k dt ≤ s 0 s 1 1  r, q > 1, r + q = 1     k (( − ) + )( − )k R 1 R t ( ) −  supt∈[s,1] D f 1 t x ty y x s 0 p τ dτ µ dt.

Therefore

B( f , p, x, y, γ, µ)  R t R s  sup p(τ)dτ − γ kD f ((1 − t)x + ty)(y − x)kdt  t∈[0,s] 0 0    1/r  s t r s q 1/q  R R p(τ)dτ − γ dt R kD f ((1 − t)x + ty)(y − x)k dt ≤ 0 0 0 1 1  r, q > 1, r + q = 1     k (( − ) + )( − )k R s R t ( ) −  supt∈[0,s] D f 1 t x ty y x 0 0 p τ dτ γ dt  R t R 1  sup p(τ)dτ − µ kD f ((1 − t)x + ty)(y − x)kdt  t∈[s,1] 0 s    1/r  1 t r 1 q 1/q  R R p(τ)dτ − µ dt R kD f ((1 − t)x + ty)(y − x)k dt + s 0 s 1 1  r, q > 1, r + q = 1     k (( − ) + )( − )k R 1 R t ( ) −  supt∈[s,1] D f 1 t x ty y x s 0 p τ dτ µ dt. Symmetry 2021, 13, 1288 10 of 19

 n o max sup R t p(τ)dτ − γ , sup R t p(τ)dτ − µ  t∈[0,s] 0 t∈[s,1] 0  h i  × R skD f ((1 − t)x + ty)(y − x)kdt + R 1kD f ((1 − t)x + ty)(y − x)kdt  0 s    1/r  hR s R t p R 1 R t r i  p(τ)dτ − γ dt + p(τ)dτ − µ dt  0 0 s 0  1/q  × R sk (( − ) + )( − )kq + R 1k (( − ) + )( − )kq ≤ 0 D f 1 t x ty y x dt s D f 1 t x ty y x dt  > 1 + 1 =  r, q 1, r q 1    n  kD f (( − t)x + ty)(y − x)k  max supt∈[0,s] 1 ,  o  kD f (( − t)x + ty)(y − x)k  supt∈[s,1] 1  h i  × R s R t ( ) − + R 1 R t ( ) − 0 0 p τ dτ γ dt s 0 p τ dτ µ dt  n o R t ( ) − R t ( ) −  max supt∈[0,s] 0 p τ dτ γ , supt∈[s,1] 0 p τ dτ µ  h i  × R 1k (( − ) + )( − )k  0 D f 1 t x ty y x dt     h r r i1/r  R s R t p(τ)dτ − γ dt + R 1 R t p(τ)dτ − µ dt  0 0 s 0 = 1/q × R 1 − + − q  0 kD f ((1 t)x ty)(y x)k dt   r, q > 1, 1 + 1 = 1  r q    h i  R s R t p(τ)dτ − γ dt + R 1 R t p(τ)dτ − µ dt  0 0 s 0   × supt∈[0,1]kD f ((1 − t)x + ty)(y − x)k,

which proves the inequality (27).

Corollary 2. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Let p : [0, 1] → C be Lebesgue integrable on [0, 1] and s ∈ [0, 1]. Then for all x, y ∈ C we have Z 1 Z s Z 1 p(τ)dτ f (y) + p(τ)dτ f (x) − p(t) f ((1 − t)x + ty)dt (32) s 0 0  R t hR 1 i  sup p(τ)dτ kD f ((1 − t)x + ty)(y − x)kdt  t∈[0,1] s 0    1/r  hR 1 R t r i  p(τ)dτ dt ≤ 0 s 1/q  × R 1k (( − ) + )( − )kq > 1 + 1 =  0 D f 1 t x ty y x dt r, q 1, r q 1    h i  R 1 R t ( ) k (( − ) + )( − )k  0 s p τ dτ dt supt∈[0,1] D f 1 t x ty y x . Symmetry 2021, 13, 1288 11 of 19

In particular,

1 ! Z 1 Z 2 p(τ)dτ f (y) + p(τ)dτ f (x) (33) 1 2 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0  R t R 1  sup p(τ)dτ kD f ((1 − t)x + ty)(y − x)kdt  t∈[0,1] 1/2 0    1/r  hR 1 R t r i  p(τ)dτ dt ≤ 0 1/2 1/q  × R 1k (( − ) + )( − )kq > 1 + 1 =  0 D f 1 t x ty y x dt , r, q 1, r q 1     R 1 R t ( ) k (( − ) + )( − )k  0 1/2 p τ dτ dt supt∈[0,1] D f 1 t x ty y x .

R s The proof follows by Theorem2 on choosing γ = 0 p(τ)dτ, s ∈ [0, 1].

Corollary 3. Assume that the function f : C ⊂ E → F is Fréchet differentiable on the open and convex set C. Let p : [0, 1] → C be Lebesgue integrable on [0, 1]. Then for all x, y ∈ C Z 1 Z 1 p(τ)dτ [(1 − α) f (y) + α f (x)] − p(t) f ((1 − t)x + ty)dtdt (34) 0 0  R t ( ) − R 1 ( )  supt∈[0,1] 0 p τ dτ α 0 p τ dτ   × R 1kD f (( − t)x + ty)(y − x)kdt  0 1    1/r  h 1 t 1 r i  R R p(τ)dτ − α R p(τ)dτ dt ≤ 0 0 0 R 1 q 1/q  × kD f ((1 − t)x + ty)(y − x)k dt r, q > 1, 1 + 1 = 1  0 r q    t  R 1 R p(τ)dτ − α R 1 p(τ)dτ dt  0 0 0   × supt∈[0,1]kD f ((1 − t)x + ty)(y − x)k

for all α ∈ [0, 1]. In particular, we have the trapezoid type inequalities Z 1 f (y) + f (x) Z 1 p(τ)dτ − p(t) f ((1 − t)x + ty)dtdt (35) 0 2 0  R t ( ) − 1 R 1 ( )  supt∈[0,1] 0 p τ dτ 2 0 p τ dτ   × R 1kD f (( − t)x + ty)(y − x)kdt  0 1    1/r  h 1 t 1 r i  R R p(τ)dτ − 1 R p(τ)dτ dt ≤ 0 0 2 0 R 1 q 1/q  × kD f ((1 − t)x + ty)(y − x)k dt r, q > 1, 1 + 1 = 1  0 r q    t  R 1 R p(τ)dτ − 1 R 1 p(τ)dτ dt  0 0 2 0   × supt∈[0,1]kD f ((1 − t)x + ty)(y − x)k.

Remark 3. Since the Fréchet derivative satisﬁes the condition

kD f (a)(b)k ≤ kD f (a)kkbk Symmetry 2021, 13, 1288 12 of 19

for a ∈ C and b ∈ E, then for all x, y ∈ C we also have the chain of inequalities Z 1 Z s p(τ)dτ f (y) + p(τ)dτ f (x) (36) s 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0  R t hR 1 i  sup p(τ)dτ kD f ((1 − t)x + ty)kdt  t∈[0,1] s 0    1/r  hR 1 R t r i  p(τ)dτ dt ≤ k − k × 0 s y x 1/q  × R 1k (( − ) + )kq > 1 + 1 =  0 D f 1 t x ty dt r, q 1, r q 1    h i  R 1 R t ( ) k (( − ) + )k  0 s p τ dτ dt supt∈[0,1] D f 1 t x ty .

In particular,

1 ! Z 1 Z 2 p(τ)dτ f (y) + p(τ)dτ f (x) (37) 1 2 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0

 R t R 1  sup p(τ)dτ kD f ((1 − t)x + ty)kdt  t∈[0,1] 1/2 0    1/r  hR 1 R t r i  p(τ)dτ dt ≤ k − k × 0 1/2 y x 1/q  × R 1k (( − ) + )kq > 1 + 1 =  0 D f 1 t x ty dt , r, q 1, r q 1     R 1 R t ( ) k (( − ) + )k  0 1/2 p τ dτ dt supt∈[0,1] D f 1 t x ty . If α ∈ [0, 1], then for all x, y ∈ C we also have the chain of inequalities Z 1 Z 1 p(τ)dτ [(1 − α) f (y) + α f (x)] − p(t) f ((1 − t)x + ty)dtdt (38) 0 0  R t ( ) − R 1 ( )  supt∈[0,1] 0 p τ dτ α 0 p τ dτ   × R 1kD f (( − t)x + ty)kdt  0 1    1/r  h 1 t 1 r i  R R p(τ)dτ − α R p(τ)dτ dt ≤ ky − xk 0 0 0 R 1 q 1/q  × kD f ((1 − t)x + ty)k dt , r, q > 1, 1 + 1 = 1  0 r q    t  R 1 R p(τ)dτ − α R 1 p(τ)dτ dt  0 0 0   × supt∈[0,1]kD f ((1 − t)x + ty)k.

In particular, we have the trapezoid type inequalities Z 1 f (y) + f (x) Z 1 p(τ)dτ − p(t) f ((1 − t)x + ty)dtdt (39) 0 2 0 Symmetry 2021, 13, 1288 13 of 19

 R t ( ) − 1 R 1 ( )  supt∈[0,1] 0 p τ dτ 2 0 p τ dτ   × R 1kD f (( − t)x + ty)kdt  0 1    1/r  h 1 t 1 r i  R R p(τ)dτ − 1 R p(τ)dτ dt ≤ ky − xk 0 0 2 0 R 1 q 1/q  × kD f ((1 − t)x + ty)k dt r, q > 1, 1 + 1 = 1  0 r q    t  R 1 R p(τ)dτ − 1 R 1 p(τ)dτ dt  0 0 2 0   × supt∈[0,1]kD f ((1 − t)x + ty)k.

4. Some Examples for Banach Algebras Let B be an algebra. An algebra norm on B is a map k·k : B→[0, ∞) such that (B, k·k) is a normed space, and, further: kabk ≤ kakkbk for any a, b ∈ B. The normed algebra (B, k·k) is a Banach algebra if k·k is a complete norm. We assume that the Banach algebra is unital, this means that B has an identity 1 and that k1k = 1. Let B be a unital algebra. An element a ∈ B is invertible if there exists an element b ∈ B with ab = ba = 1. The element b is unique; it is called the inverse of a and written a−1 1 or a . The set of invertible elements of B is denoted by InvB. If a, b ∈InvB then ab ∈InvB − and (ab) 1 = b−1a−1. ∞ n Now, by the help of power series f (z) = ∑n=0 αnz we can naturally construct another power series which will have as coefficients the absolute values of the coefficients of the ∞ n original series, namely, fa(z) := ∑n=0|αn|z . It is obvious that this new power series will have the same radius of convergence as the original series. We also notice that if all coefficients αn ≥ 0, then fa = f . The following result holds [23].

∞ n Lemma 3. Let f (z) = ∑n=0 αnz be a function deﬁned by power series with complex coefﬁcients and convergent on the open disk D(0, R) ⊂ C, R > 0. For any x, y ∈ B with kxk, kyk < R we have Z 1 0 k f (y) − f (x)k ≤ ky − xk fa(k(1 − s)x + syk)ds. (40) 0

We also have:

∞ n Lemma 4. Let f (z) = ∑n=0 αnz be a function deﬁned by power series with complex coefﬁcients and convergent on the open disk D(0, R) ⊂ C, R > 0. For any x, y ∈ B with kxk, kyk < R we have 0 kD f ((1 − t)x + ty)(y − x)k ≤ ky − xk fa(k(1 − t)x + tyk) (41) for all t ∈ [0, 1].

Proof. Let kuk, kvk < R. Then there exists δ > 0 such that ku + εvk < R for all ε ∈ (−δ, δ) and by (40) we get

Z 1 0 k f (u + εv) − f (u)k ≤ ku + εv − uk fa(k(1 − s)u + s(u + εv)k)ds 0 Symmetry 2021, 13, 1288 14 of 19

namely

Z 1 f (u + εv) − f (u) 0 ≤ kvk fa(k(1 − s)u + s(u + εv)k)ds, ε 6= 0 ε 0 Z 1 0 = kvk fa(ku + sεvk)ds 0 and by taking ε → 0 we get, by the property of integral, that

0 kD f (u)(v)k ≤ kvk fa(kuk). (42)

Now, if we take in (42) u = (1 − t)x + ty and v = y − x, then we get (41).

We have the following result:

∞ n Theorem 3. Let f (z) = ∑n=0 αnz be a function deﬁned by power series with complex coefﬁcients and convergent on the open disk D(0, R) ⊂ C, R > 0. Also, let p : [0, 1] → C be a Lebesgue integrable function on [0, 1]. For any x, y ∈ B with kxk, kyk < R we have Z 1 Z s Z 1 p(τ)dτ f (y) + p(τ)dτ f (x) − p(t) f ((1 − t)x + ty)dt (43) s 0 0  R t R 1 0  sup p(τ)dτ f (k(1 − t)x + tyk)dt  t∈[0,1] s 0 a    1/r  hR 1 R t r i  p(τ)dτ dt ≤ k − k × 0 s y x 1/q  × R 1[ 0(k( − ) + k)]q > 1 + 1 =  0 fa 1 t x ty dt r, q 1, r q 1    h i  R 1 R t ( ) 0(k( − ) + k)  0 s p τ dτ dt supt∈[0,1] fa 1 t x ty .

In particular,

1 ! Z 1 Z 2 p(τ)dτ f (y) + p(τ)dτ f (x) (44) 1 2 0

Z 1 − p(t) f ((1 − t)x + ty)dt 0  R t R 1 0  sup p(τ)dτ f (k(1 − t)x + tyk)dt  t∈[0,1] 1/2 0 a    1/r  hR 1 R t r i  p(τ)dτ dt ≤ k − k × 0 1/2 y x 1/q  × R 1[ 0(k( − ) + k)]q > 1 + 1 =  0 fa 1 t x ty dt r, q 1, r q 1     R 1 R t ( ) 0(k( − ) + k)  0 1/2 p τ dτ dt supt∈[0,1] fa 1 t x ty . Symmetry 2021, 13, 1288 15 of 19

Also, Z 1 Z 1 p(τ)dτ [(1 − α) f (y) + α f (x)] − p(t) f ((1 − t)x + ty)dt (45) 0 0  R t ( ) − R 1 ( )  supt∈[0,1] 0 p τ dτ α 0 p τ dτ   × R 1 f 0(k( − t)x + tyk)dt  0 a 1    h r i1/r  R 1 R t ( ) − R 1 ( )  0 0 p τ dτ α 0 p τ dτ dt ≤ ky − xk × 1/q R 1 0 q  × [ f (k(1 − t)x + tyk)] dt r, q > 1, 1 + 1 = 1  0 a r q    t  R 1 R p(τ)dτ − α R 1 p(τ)dτ dt  0 0 0  0  × supt∈[0,1] fa(k(1 − t)x + tyk)

for all α ∈ [0, 1]. In particular, we have the trapezoid type inequalities Z 1 f (y) + f (x) Z 1 p(τ)dτ − p(t) f ((1 − t)x + ty)dt (46) 0 2 0  R t ( ) − 1 R 1 ( )  supt∈[0,1] 0 p τ dτ 2 0 p τ dτ   × R 1 f 0(k( − t)x + tyk)dt  0 a 1    h r i1/r  R 1 R t ( ) − 1 R 1 ( )  0 0 p τ dτ 2 0 p τ dτ dt ≤ ky − xk × 1/q R 1 0 q  × [ f (k(1 − t)x + tyk)] dt r, q > 1, 1 + 1 = 1  0 a r q    t  R 1 R p(τ)dτ − 1 R 1 p(τ)dτ dt  0 0 2 0  0  × supt∈[0,1] fa(k(1 − t)x + tyk).

The proof follows by Theorem2 and Lemma4. As some natural examples that are useful for applications, we can point out that if

∞ (−1)n 1 f (λ) = λn = ln , λ ∈ D(0, 1); (47) ∑ + n=1 n 1 λ ∞ (−1)n g(λ) = λ2n = cos λ, λ ∈ ; ∑ ( ) C n=0 2n ! ∞ (−1)n h(λ) = λ2n+1 = sin λ, λ ∈ ; ∑ ( + ) C n=0 2n 1 ! ∞ 1 l(λ) = (−1)nλn = , λ ∈ D(0, 1); ∑ + n=0 1 λ Symmetry 2021, 13, 1288 16 of 19

then the corresponding functions constructed by the use of the absolute values of the coefﬁcients are

∞ 1 1 f (λ) = λn = ln , λ ∈ D(0, 1); (48) a ∑ − n=1 n 1 λ ∞ 1 g (λ) = λ2n = cosh λ, λ ∈ ; a ∑ ( ) C n=0 2n ! ∞ 1 h (λ) = λ2n+1 = sinh λ, λ ∈ ; a ∑ ( + ) C n=0 2n 1 ! ∞ 1 l (λ) = λn = , λ ∈ D(0, 1). a ∑ − n=0 1 λ

Other important examples of functions as power series representations with non- negative coefﬁcients are:

∞ 1 exp(λ) = ∑ λn λ ∈ C, (49) n=0 n! 1 1 + λ ∞ 1 ln = λ2n−1, λ ∈ D(0, 1); − ∑ − 2 1 λ n=1 2n 1 ∞ 1 Γ n + 2 sin−1(λ) = ∑ √ λ2n+1, λ ∈ D(0, 1); n=0 π(2n + 1)n!

∞ 1 tanh−1(λ) = λ2n−1, λ ∈ D(0, 1) ∑ − n=1 2n 1 ∞ Γ(n + α)Γ(n + β)Γ(γ) F (α, β, γ, λ) = λn, α, β, γ > 0, 2 1 ∑ ( ) ( ) ( + ) n=0 n!Γ α Γ β Γ n γ λ ∈ D(0, 1).

If we consider the exponential function f (x) = exp x, then for x, y ∈ B

Z 1 Z 1 0 fa(k(1 − t)x + tyk)dt = exp(k(1 − t)x + tyk)dt (50) 0 0 Z 1 ≤ exp((1 − t)kxk + tkyk)dt 0 Z 1 = exp(kxk + t(kyk − kxk))dt 0  k k− k k exp y exp x kyk 6= kxk  kyk−kxk if =   expkxk if kyk = kxk.

=: E1(x, y). Symmetry 2021, 13, 1288 17 of 19

Also

Z 1 Z 1 0 q fa(k(1 − t)x + tyk) dt = exp[q(k(1 − t)x + tyk)]dt (51) 0 0 Z 1 ≤ exp[q((1 − t)kxk + tkyk)]dt 0  ( k k)− ( k k) exp q y exp q x kyk 6= kxk  q(kyk−kxk) if =   exp(qkxk) if kyk = kxk

=: Eq(x, y), q > 1.

Moreover, for x, y ∈ B

0 sup fa(k(1 − t)x + tyk) ≤ sup exp((1 − t)kxk + tkyk) (52) t∈[0,1] t∈[0,1] = max{kxk, kyk}.

By making use of Theorem3 and (50)–(52), we have for p : [0, 1] → C, a Lebesgue integrable function on [0, 1] and any x, y ∈ B and s ∈ [0, 1], that Z 1 Z s p(τ)dτ exp y + p(τ)dτ exp x (53) s 0

Z 1 − p(t) exp((1 − t)x + ty)dt 0  R t  sup p(τ)dτ E1(x, y)  t∈[0,1] s    1/r  h 1 t p i 1/q  R R p(τ)dτ dt E (x, y) , ≤ ky − xk × 0 s q 1 1  r, q > 1, r + q = 1     R 1 R t ( ) {k k k k}  0 s p τ dτ dt max x , y .

In particular,

1 ! Z 1 Z 2 p(τ)dτ f (y) + p(τ)dτ f (x) (54) 1 2 0

Z 1 − p(t) exp((1 − t)x + ty)dt 0  R t  sup p(τ)dτ E1(x, y)  t∈[0,1] 1/2    1/r  h 1 t p i 1/q  R R p(τ)dτ dt E (x, y) , ≤ ky − xk × 0 1/2 q 1 1  r, q > 1, r + q = 1     R 1 R t ( ) {k k k k}  0 1/2 p τ dτ dt max x , y . Symmetry 2021, 13, 1288 18 of 19

Also, Z 1 p(τ)dτ [(1 − α) exp(y) + α exp(x)] (55) 0

Z 1 − p(t) exp((1 − t)x + ty)dtdt 0  R t R 1  sup p(τ)dτ − α p(τ)dτ E1(x, y)  t∈[0,1] 0 0    1/r  h 1 t 1 r i 1/q  R R p(τ)dτ − α R p(τ)dτ dt E (x, y) , ≤ ky − xk × 0 0 0 q 1 1  r, q > 1, r + q = 1     R 1 R t ( ) − R 1 ( ) {k k k k}  0 0 p τ dτ α 0 p τ dτ dt max x , y ,

for all α ∈ [0, 1]. In particular, we have the trapezoid type inequalities Z 1 exp(y) + exp(x) Z 1 p(τ)dτ − p(t) exp((1 − t)x + ty)dtdt (56) 0 2 0

 R t 1 R 1  sup p(τ)dτ − p(τ)dτ E1(x, y)  t∈[0,1] 0 2 0    1/r  h 1 t 1 r i 1/q  R R p(τ)dτ − 1 R p(τ)dτ dt E (x, y) , ≤ ky − xk × 0 0 2 0 q 1 1  r, q > 1, r + q = 1     R 1 R t ( ) − 1 R 1 ( ) {k k k k}  0 0 p τ dτ 2 0 p τ dτ dt max x , y . The interested reader may apply the above inequalities for the other functions listed in (47)–(49). The details are omitted.

5. Conclusions In this paper we established some upper bounds for the quantity Z 1 Z 1 p(s)ds − γ f (y) + γ f (x) − p(t) f ((1 − t)x + ty)dt 0 0

in the case that f : C ⊂ E → F is Fréchet differentiable on the open and convex subset C of the Banach space E with values into another Banach space F, x, y ∈ C, p : [0, 1] → C is a Lebesgue integrable function and γ ∈ C. Some particular cases of interest for different choices of γ are given. The symmetric case for the coefﬁcients of f in the above inequality is analysed. Applications for Banach algebras are also provided.

Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conﬂicts of Interest: The author declares no conﬂict of interest.

References 1. Coleman, R. Calculus on Normed Vector Spaces; Springer: New York, NY, USA, 2012. 2. Dragomir, S.S. Integral inequalities for Lipschitzian mappings between two Banach spaces and applications. Kodai Math. J. 2016, 39, 227–251. [CrossRef] Symmetry 2021, 13, 1288 19 of 19

3. Dragomir, S.S.; Cho, Y.J.; Kim, S.S. Inequalities of Hadamard’s type for Lipschitzian mappings and their applications. J. Math. Anal. Appl. 2000, 245, 489–501. [CrossRef] 4. Dragomir, S.S.; Pearce, C.E.M. Selected Topics on Hermite-Hadamard Inequalities and Applications; RGMIA Monographs. 2000. Available online: http://rgmia.org/monographs/hermite_hadamard.html (accessed on 10 July 2021). 5. Feng, Y.; Zhao, W. Refinement of Hermite-Hadamard inequality. Far East J. Math. Sci. 2012, 68, 245–250. 6. Gao, X. A note on the Hermite-Hadamard inequality. J. Math. Inequal. 2010, 4, 587–591. [CrossRef] 7. Hwang, S.-R.; Tseng, K.-L.; Hsu, K.-C. Hermite-Hadamard type and Fejér type inequalities for general weights (I). J. Inequal. Appl. 2013, 2013, 170. [CrossRef] 8. Kırmacı U.S.; Dikici, R. On some Hermite-Hadamard type inequalities for twice differentiable mappings and applications. Tamkang J. Math. 2013, 44, 41–51. [CrossRef] 9. Muddassar, M.; Bhatti, M.I.; Iqbal, M. Some new s-Hermite-Hadamard type inequalities for differentiable functions and their applications. Proc. Pakistan Acad. Sci. 2012, 49, 9–17. 10. Matić,M.; Peˇcarić,J. Note on inequalities of Hadamard’s type for Lipschitzian mappings. Tamkang J. Math. 2001, 32, 127–130. [CrossRef] 11. Sarikaya, M.Z. On new Hermite Hadamard Fejér type integral inequalities. Stud. Univ. Babe¸s-BolyaiMath. 2012, 57, 377–386. 12. W ˛asowicz,S.; Witkowski, A. On some inequality of Hermite-Hadamard type. Opuscula Math. 2012, 32, 591–600. [CrossRef] 13. Xi, B.-Y.; Qi, F. Some integral inequalities of Hermite-Hadamard type for convex functions with applications to means. J. Funct. Spaces Appl. 2012, 2012.[CrossRef] 14. Zabandan, G.; Bodaghi, A.; Kılıçman, A. The Hermite-Hadamard inequality for r-convex functions. J. Inequal. Appl. 2012, 2012. [CrossRef] 15. Zhao, C.-J.; Cheung, W.-S.; Li, X.-Y. On the Hermite-Hadamard type inequalities. J. Inequal. Appl. 2013, 2013, 228. [CrossRef] 16. Fink, A.M. Bounds on the deviation of a function from its averages. Czechoslov. Math. J. 1992, 42, 298–310. [CrossRef] 17. Anastassiou, G.A. Univariate Ostrowski inequalities, revisited. Monatsh. Math. 2002, 135, 175–189. [CrossRef] 18. Cerone, P.; Dragomir, S.S. Midpoint-type rules from an inequalities point of view. In Handbook of Analytic-Computational Methods in Applied Mathematics; Anastassiou, G.A., Ed.; CRC Press: New York, NY, USA, 2019; pp. 135–200. 19. Cerone, P.; Dragomir, S.S. New bounds for the three-point rule involving the Riemann-Stieltjes integrals. In Advances in Statistics Combinatorics and Related Areas; Gulati, C., Ed.; World Science Publishing: Singapore, 2002; pp. 53–62. 20. Cerone, P.; Dragomir, S.S.; Roumeliotis, J. Some Ostrowski type inequalities for n-time differentiable mappings and applications. Demonstr. Math. 1999, 32, 697–712. 21. Ciurdariu, L. A note concerning several Hermite-Hadamard inequalities for different types of convex functions. Int. J. Math. Anal. 2012, 6, 1623–1639. 22. Dragomir, S.S. Weighted norm inequalities of trapezoid type for Frechet differentiable functions in Banach spaces. RGMIA Res. Rep. Coll. 2020, 23, 19. Available online: https://rgmia.org/papers/v23/v23a22.pdf (accessed on 10 July 2021). 23. Dragomir, S.S. Inequalities of Lipschitz type for power series in Banach algebras. Ann. Math. Sil 2015, 29, 61–83. [CrossRef]