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 2 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 ra f (u). It is called the Gâteaux derivative (di- sym13071288 rectional derivative) of f at a in the direction u. If ra f (u) is defined and l 2 Rnf0g, then ra f (lu) is defined and ra f (lu) = lra f (u). The function f is Gâteaux differentiable at a Academic Editors: Carmen Violeta if ra 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 2 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) 2 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 rx f (u) = D f (x)(u) for all u 2 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 2 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 2 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 2 [0, 1]. Then for all x, y 2 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 t2[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 t2[0,1] 1 Z 1 ≤ p(s)ds sup kD f ((1 − t)x + ty)(y − x)k 2 0 t2[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 t2[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 − g f (y) + g 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 2 C, p : [0, 1] ! C is a Lebesgue integrable function and g 2 C. Some particular cases of interest for different choices of g 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 j(x,y)(t) := f ((1 − t)x + ty), t 2 [0, 1], where x, y 2 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 2 C the auxiliary function j(x,y) is differentiable on (0, 1) and 0 j(x,y)(t) = D f ((1 − t)x + ty)(y − x). (8) Also 0 j(x,y)(0+) = D f (x)(y − x) (9) and 0 j(x,y)(1−) = D f (y)(y − x). (10) Proof. Let t 2 (0, 1) and h 6= 0 small enough such that t + h 2 (0, 1). Then j(x,y)(t + h) − j(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 j(x,y)(t + h) − j(x,y)(t) j(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 j(x,y)(h) − j(x,y)(0) j(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.