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A q-Analog of the Class of Completely Convex Functions and Lidstone Series

A q-Analog of the Class of Completely Convex Functions and Lidstone Series axioms Article A q-Analog of the Class of Completely Convex Functions and Lidstone Series 1, 2 Maryam Al-Towailb * and Zeinab S. I. Mansour Department of Computer Science and Engineering, College of Applied Studies and Community Service, King Saud University, Riyadh 11451, Saudi Arabia Department of Mathematics, Faculty of Science, Cairo University, Giza 12613, Egypt; zsmansour@cu.edu.eg * Correspondence: mtowaileb@ksu.edu.sa Abstract: This paper introduces a q-analog of the class of completely convex functions. We prove specific properties, including that q-completely convex functions have convergent q-Lidstone series expansions. We also provide a sufficient and necessary condition for a real function to have an absolutely convergent q-Lidstone series expansion. Keywords: quantum calculus; q-series; q-Lidstone polynomials; completely convex functions MSC: 05A30; 41A58; 39A70; 40A05 1. Introduction In 1929, Lidstone [1] introduced a generalization of Taylor ’s theorem that approximates an entire function f in a neighborhood of two points instead of one. That is ¥ h i (2n) (2n) f (x) = f (1)L (x) + f (0)L (1 x) , (1) å n n n=0 Citation: Al-Towailb, M.; Mansour, where L (x) is a unique polynomial of degree 2n + 1, and called a Lidstone polynomial. Z.S.I. A q-Analog of the Class of In [2], Whittaker proved that an entire function of an exponential type of less than p has a Completely Convex Functions and convergent Lidstone series expansion in any compact set of the complex plane. Buckholtz Lidstone Series. Axioms 2023, 12, 412. and Shaw [3] provided some conditions for (1) to hold. Other authors worked on this https://doi.org/10.3390/ problem (see, e.g., [4–10]). They presented different sufficient and necessary conditions axioms12050412 for the representation of functions by this series. We mention, in particular, the result of Academic Editors: Ivanka Stamova, Widder [10]. He proved that if f is a real-valued function satisfying Gheorghe Oros and Georgia Irina Oros k (2k) (1) f (x)  0 (k 2 N ) (2) Received: 14 February 2023 in an interval of length greater than p, then it has a Lidstone series expansion (1) (such a Revised: 8 March 2023 function is known as completely convex). Furthermore, he defined the class of minimal Accepted: 19 March 2023 completely convex functions, and then he proved that a real-valued function f (x) could be Published: 24 April 2023 expanded in an absolutely convergent Lidstone series if and only if it is the difference of two minimal completely convex functions. Recently, the Lidstone expansion theorem was generalized in quantum calculus (as Copyright: © 2023 by the authors. can be seen in [11–17]). The quantum calculus (Jackson calculus or q-calculus [18]) is Licensee MDPI, Basel, Switzerland. an extension of the traditional calculus, and it has been used by many researchers in This article is an open access article different branches of science and engineering (as can be seen in, e.g., [19–24]). It has a lot distributed under the terms and of applications in different mathematical areas such as orthogonal polynomials, number conditions of the Creative Commons theory, hypergeometric functions, theory of finite differences, gamma function theory, Attribution (CC BY) license (https:// Sobolev spaces, Bernoulli and Euler polynomials, operator theory, and quantum mechanics. creativecommons.org/licenses/by/ For the basic definitions and notations applicable in the q-calculus, see Section 2. 4.0/). Axioms 2023, 12, 412. https://doi.org/10.3390/axioms12050412 https://www.mdpi.com/journal/axioms Axioms 2023, 12, 412 2 of 13 In [11], Ismail and Mansour proved the following q-analog of the Lidstone expansion theorem. Theorem 1. Assume that the function f (z) is an entire function of q -exponential growth of order 1 and a finite type a less than x , or it is an entire function of q -exponential growth of an order of less than 1. Then, f (z) has a convergent q-Lidstone representation ¥ h i 2n 2n f (z) = D f (1) A (z) D f (0)B (z) , (3) å 1 n 1 n q q n=0 where ( A ) and (B ) are the q-Lidstone polynomials defined, respectively, by the generating n n n n functions E (zw) E (zw) q q 2n = A (z)w , (4) E (w) E (w) q q n=0 E (zw)E (w) E (zw)E (w) q q q q = B (z) . (5) E (w) E (w) [n] ! q q q n=0 Moreover, A (z) = z, B (z) = 1 z, and for n 2 N, A (z) and B (z) satisfy the q-difference 0 0 n n equation D y (z) = y (z) with y (0) = y (1) = 0. (6) n n n 1 n1 In [16], AL-Towailb and Mansour proved that the condition n n D f (0) = o(x ) as n ! ¥ (7) is both sufficient and necessary for expanding an entire function f (z) in the q-Lidstone series 2 2 f (1) A (z) f (0)B (z) + D f (1) A (z) D f (0)B (z) + . . . , 0 0 1 1 1 1 q q and we noted that Condition (7) is insufficient for the convergence of the following arrange- ment of the q-Lidstone series: ¥ ¥ 2n 2n D f (1) A (z) D f (0)B (z), n n å 1 å 1 q q n=0 n=0 and not necessary for the convergence of (3). This paper aimed to obtain a sufficient and necessary condition for a real-valued function to have an absolutely convergent q-Lidstone series expansion (3). To achieve this aim, we introduced generalizations for the class of completely convex functions (2) on a closed interval of form [0, a] (a > 0), and the class of minimal completely convex functions on the interval [0, 1]. This paper is organized as follows. The following section gives the essential notions and basic definitions of q-calculus. Section 3 contains some properties and basic results on q-Lidstone polynomials, which we need in our investigation. In Section 4, we define a q-analog of the class of completely convex functions for the difference operator D 1 . Then, we study the relation of this class to a problem of the representation of functions by the q-Lidstone series. In Section 5, we provide a necessary and sufficient condition for a real function to have an absolutely convergent q-Lidstone series expansion. 2. Preliminaries In this section, we recall some definitions, notations, and results in the q-calculus, which we need in our investigations (see [25]). Throughout this paper, q is a positive number less than one, and we use the following standard notations: N := f1, 2, 3, . . .g, N := f0, 1, 2, . . .g = N[f0g. 0 Axioms 2023, 12, 412 3 of 13 The sets A and A are defined by A := fq : n 2 N g and A := A [f0g. For q q 0 q q q a 2 C, n 2 N , (a; q) (a; q) = (1 aq ), (a; q) := , ¥ Õ n (aq ; q) j=0 and the q-numbers [n] and q-factorial [n] ! are defined by q q n n 1 q [n] = , [n] ! = [k] . q q q 1 q k=1 Let m 2 C. A set A  C is called m-geometric set if mz 2 A for any z 2 A. If f is a function defined on a q-geometric set A, then Jackson’s q-difference operator is defined by f (z) f (qz) , z 2 Af0g; D f (z) = (8) (1 q)z f (0), z = 0, provided that f is differentiable at zero. Furthermore, Jackson [26] introduced the following q-integrals for a function f defined on a q-geometric set A: Z Z Z b b a f (t) d t := f (t)d t f (t) d t (a, b 2 R), q q q a 0 0 where n n f (t) d t := (1 q) zq f (zq ), n=0 provided that the series converges at z = a and z = b. Jackson’s q-trigonometric functions Sin z and Cos z are defined by q q ¥ n(2n+1) n 2n+1 Sin z := (1) (z(1 q)) , (q; q) 2n+1 n=0 (9) ¥ n(2n1) n 2n Cos z := (1) (z(1 q)) , (q; q) 2n n=0 where E () is one of Jackson’s q-exponential function defined by n(n1) (z(1 q)) E (z) = q = (z(1 q); q) (z 2 C). (10) q å ¥ (q; q) n=0 We use fx g to denote the positive zeros of Sin z arranged in increasing order of k k2N q 3/2 magnitude. One can verify that Sin z has no zeroes on jzj < q , i.e., the first positive 3/2 zeros x > q . Lemma 1. For any x 2 [0, 1], we have Sin x x  x x. (11) q 1 1 Proof. Let f (x) = x x Sin x x, x 2 [0, 1]. Then, D 1 f (x) = x (1 Cos x x)  0. 1 q 1 1 q 1 Therefore, by using (8), we obtain f (x)  f ( ) (x 2 [0, 1]), which implies f (x)  lim f (q x) = 0. Then, Inequality (11) holds. n!¥ Axioms 2023, 12, 412 4 of 13 3. Some Results on q-Lidstone Polynomials We start this section by recalling some properties of the q-Lidstone polynomials A (x) and B (x) from [14,16,17], for which we need to prove the main results. Proposition 1 ([16]). Let fx g be the sequence of the positive zeros of Sin (x) and m 2 N . k k2N q 0 Then, 2Sin (x x) q 1 (2n+1) n1 (1) A (x) = +O(x ); (12) 2n+1 0 x Sin (x ) q 1 Sin (x x)Cos (x ) q 1 q 1 n1 2n m (1) B (x) = +O(x (2n) ), (13) 0 1 2n+1 (1 q)(x ) Sin (x ) 1 1 for a sufficiently large n. 2n Proposition 2 ([17]). If f 2 C ([0, 1]), then n1 h i 2m 2m 2n 2 f (x) = D f (1) A (x) D f (0)B (x) + G (x, qt)D f (q t) d t, (14) m m n q å 1 1 1 q q q m=0 where qt(1 x), 0  t < x  1; G(x, t) = G (x, t) = (15) qx (1 t), 0  x < t  1, G (x, qt) = G(x, qy) G (qy, qt) d y (n 2 N). (16) n n1 q Moreover, G (x, qt) d t = A (x) B (x) (n 2 N). (17) n q n n Remark 1 ([14]). For x 2 [0, 1] and n 2 N , we have n n1 (1) A (x)  0 and (1) B (x)  0. (18) n n Proposition 3. Let x be the smallest positive zero of Sin (x). Then, there exist some constants M and M and a positive integer n such that the following inequalities hold 1 2 0 0  (1) A (x)  ; (19) 2n n1 0  (1) B (x)  , (20) 2n for all x 2 [0, 1] and n  n . Proof. From (12), there is a positive real number C and n 2 N such that 1 0 Sin (x x) C q 1 n1 (1) A (x) 2  , (21) 2n+1 0 2n x Sin (x ) x q 2 for all x 2 [0, 1] and n  n . Consequently, Sin (x x) q 1 0  (1) A (x)  2 . (22) 2n 2n+1 0 x Sin (x ) q 1 2 1 Axioms 2023, 12, 412 5 of 13 Note that x < x and Sin (x x) is bounded on [0, 1]. Then, from (22), we obtain 1 2 q 1 Sin (x x) C 2 q 1 0  (1) A (x)  + 2n 2n+1 x Sin (x ) q 1 1 1 (23) C C M 1 2 1 + = . 2n 2n 2n x x x 1 1 1 Similarly, we obtain (20) from (13). Proposition 4. There exists a constant M such that 0  (1) G (x, qt) d t  . n q 2n 0 x Proof. The proof follows immediately from Equation (17) and Proposition 3. Proposition 5. For any fixed point x 2 (0, 1) and sufficiently large n, there exist some constants M and M such that (1) A (x )  ; (24) n 0 2n n1 (1) B (x )  . (25) n 0 2n Proof. From (12), we obtain n 2n+1 2n+1 (1) A (x)x = L(x) +O(( ) ) (n ! ¥), 2Sin (x x) q 1 where L(x) = . Notice, for any fixed x 2 (0, 1), L(x ) > 0 and 0 0 0 Sin (x ) n 2n+1 lim (1) A (x )x = L(x ). n 0 0 n!¥ n 2n+1 This implies that the sequence (1) A (x )x is bounded below by a positive n 0 number. I.e., (24) holds. Similarly, we obtain the Inequality (25) from (13). Now, using the previous results, we prove the following theorem. Theorem 2. If the series S = a A (x) + b B (x) + a A (x) + b B (x) + . . . (26) 0 0 0 0 1 1 1 1 h i a +b ¥ n n n converges for a single value x 2 (0, 1), then the series å (1) is absolutely convergent. n=0 2n Proof. Since the series (26) converges for x 2 (0, 1), we have lim a A (x ) = 0, lim b B (x ) = 0. n n 0 n n 0 n!¥ n!¥ Then, from the inequalities (24) and (25), we obtain 2n 2n a = O(x ) and b = O(x ). (27) n n 1 1 Axioms 2023, 12, 412 6 of 13 From (12), (13), and (27), we conclude that the series ( ) h n i h n i 2(1) Sin (x x ) (1) Cos x Sin (x x ) q q q 1 0 1 1 0 S = a A (x ) + + b B (x ) + 1 å n n 0 n n 0 2n+1 0 2n+1 0 x Sin (x ) (1 q)x Sin (x ) n=0 q 1 q 1 1 1 converges absolutely. This implies that S S is also convergent. Notice that h i ¥ n n 2Sin (x x ) Cos x Sin (x x ) (1) (1) q 1 0 q 1 q 1 0 S S = a + b 1 n n 0 0 2n 2n x Sin (x ) x (1 q)x Sin (x ) x 1 q 1 1 q 1 n=0 1 1 ¥ h n n i 2Sin (x x ) (1) (1) q 1 0 > a + b . n n 2n 2n x Sin (x ) x x 1 q 1 n=0 1 1 Therefore, we obtain the result. 4. A q-Analog of Completely Convex Function In this section, by C [0, a], we mean the space of all functions defined on [0, a] such that D f (x) is defined and continuous at zero. Definition 1. A real-valued function f , defined on the interval [0, a] (a > 0), is said to be a q-completely convex function if f 2 C [0, a] and n 2n k (1) D f (aq )  0 (for all fn, kg  N ). (28) 1 0 Example 1. The functions f (x) = Sin x x, defined in (9), are q-completely convex on the interval [0, 1]. Indeed, one can verify that n 2n n 2n 2n (1) D f (x) = (1) D Sin x x = x Sin (x x) > 0, (29) 1 1 q 1 q 1 q q for all x 2 [0, 1] and n 2 N . In the following, we prove certain properties of q-completely convex functions. Proposition 6. If a function f 2 C [0, a] is q-completely convex, then n 2n (1) D f (0)  0 (n 2 N ). (30) 1 0 2n Proof. The proof follows directly by taking the limit as k ! ¥ in (28) and using that D f is continuous at zero for all n 2 N . Proposition 7. Let f 2 C (0, 1) be a q-completely convex function on [0, 1]. Then, for a suffi- ciently large n, we have 2n 2n D f (0) = O(x ); (31) 2n 2n D f (1) = O(x ). (32) Proof. From Proposition 1 and Inequality (28), every term of (14) is non-negative. Therefore, 2n 0  A (x) D f (0)  f (x); (33) 2n 0  (B (x))D f (1)  f (x) (x 2 [0, 1]; n 2 N ). (34) n 1 0 q Axioms 2023, 12, 412 7 of 13 Thus, by using (24) and (33), we obtain f (x ) n 2n 2n 0  (1) D f (0)   Kx (n ! ¥), q n (1) A (x ) n 0 for some constant K > 0 and x 2 (0, 1). Then, we have (31). Similarly, we obtain the asymptotic behavior in (32). Proposition 8. Let f be a q-completely convex function on [0, 1]. Then, there exists a positive constant C such that for all x 2 A 2n n 2n 0  (1) D f (x)  C , (35) where x is the smallest positive zero of Sin (x). Proof. If f is q-completely convex on [0, 1], then it is q-completely convex on [0, x] for all x 2 A . Consequently, the function f (t) := f (xt) is q-completely convex on [0, 1]. Therefore, from Proposition (7), we have n 2n n 2n 2n 2n 0  (1) D f (1) = (1) x D f (x) = O(x ), 1 1 q q which is nothing else but (35). Lemma 2. Let f (x) and D f (x) be non-negative on A , and continuous at 0. Assume that 1 q there exists a number x 2 A such that f (x )  a (a 2 R). Then, 0 q 0 (1 + q)a f (x)  , for all x 2 A . (1 q)x Proof. First, let x 2 A and x  x . Then, by using the assumption D f (x)  0, we have q 0 1 D f ( ) d t  0. Therefore, D f (x)  D f (x ), and q q 0 D f (t) d t  (x x )D f (x ) (x 2 A , x  x). (36) q q 0 q 0 0 Since f (x)  0 on A , from (8) and Inequality (36), we obtain (x x ) x x q a 0 0 f (x)  f (x ) + f (x ) = f (x ) < , (37) 0 0 0 (1 q)x (1 q)x (1 q)x 0 0 0 for all x 2 A and x  x. Similarly, if x 2 A and x < x , then q 0 q 0 x x f (qx ) 0 0 f (x)  f (qx ) < . (38) (1 q)x (1 q)x 0 0 On the other hand, since D f (x)  0, we have (1 + q) f (qx)  q f (x) + f (q x) (x 2 A ). q Axioms 2023, 12, 412 8 of 13 Therefore, from the condition f (x)  0, we obtain (1 + q) f (qx)  q f ( ) + f (qx) > f (qx) (x 2 A ). (39) So, from the inequalities (38) and (39), we obtain (1 + q)a f (x) < (x 2 A , x < x ). (40) (1 q)x Hence, the relations (37) and (40) yield the required result. Corollary 1. If f 2 C [0, 1] is a q-completely convex function, then there exists a positive constant M such that n 2n 2n 0  (1) D f (x)  Mx (n 2 N , x 2 A ). (41) 1 0 1 q Proof. The proof follows from Proposition 8 and Lemma 2 by taking x = 1 and M = 1+q C. 1q Lemma 3. If f 2 C [0, 1] is a q-completely convex function on [0, 1], then there exists a constant K > 0 such that n n jD f (x)j  Kx (x 2 A ), (42) 1 q where x is the smallest positive zero of Sin (z). 1 q Proof. From Corollary 1, it suffices to prove (42) when n is an odd integer. We set g(x) = n 2n (1) D f (x). Since f (x) is a q-completely convex on 0  x  1, again from Corollary 1, there exists the constant M > 0 (independent of n) such that for all x 2 A 2n 0  g(x)  Mx , (43) 2 2n+2 0  D g(x)  Mx . q 1 Therefore, for every x 2 A f1g, we have Z 2 2 2n+2 0  D g(t) d t  Mq(q x)x . 1 q q 1 qx So, by using the fundamental theorem of the q-calculus, we obtain n 2n+1 n 2n+1 2n+2 0  (1) D f (x) (1) D f (1)  Mx , 1 1 q q and hence, n 2n+1 n 2n+1 n 2n+1 2n+2 (1) D f (1)  (1) D f (x)  (1) D f (1) + Mx , 1 1 1 q q q for all x 2 A f1g. Consequently, 2n+1 2n+1 2n+2 jD f (x)j  jD f (1)j + Mx . (44) 1 1 q q On the other hand, since D g(x) < 0, one can verify that for all x 2 A 1 q x x (1 + q)g( )  g(x) + qg( ), q q Axioms 2023, 12, 412 9 of 13 and then qg( )  (1 + q)g(x) g(qx) (x 2 A ). Thus, if x = 1, we obtain 1 1 (1 + q) 2n n 2n 2n (D f )( ) = (1) (D f )( )  D f (1) . (45) 1 1 1 q q q q q q Hence, from (8), (43) and (45), we have jg(1)j +jg(1/q)j 2q + 1 2n+1 2n D f (1) = jD g(1)j   Mx . (46) q 1 1/q 1 q 3/2 However, x > q , this implies 2n+1 2n+1 D f (1)  q(2q + 1) Mx . (47) By substituting (47) in (44), we obtain 2n+1 2n+1 2n+2 2n+1 jD f (x)j  q(2q + 1) Mx + Mx  M Mx , 1 1 1 1 1 3/2 for all n 2 N and x 2 A , where M = q(2q + 1) + q . 2n+1 2n+1 2n+1 Since D f (x) is continuous at zero, then we obtain D f (x) = O(x ) for a 1 1 q q sufficiently large n. This completes the proof. Theorem 3. Let f 2 C [0, 1] be a q-completely convex on [0, 1]. If f is analytic at zero, then the following q-Lidstone series expansion holds for all x 2 [0, 1]. ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . (48) 1 n 1 n q q n=0 Moreover, f (x) is the restriction of an entire function of q -exponential growth of order 1 and a finite type less than x and the expansion (48) holds for all x on the entire complex plane. Proof. Since f is analytic at 0, there exists 0 < c < 1 and the open interval W = (c, c) such that f (x) has the Maclaurin series expansion ¥ (n) ¥ D f (0) n(n1) 1 f (0) n n f (x) = x = q x (x 2 W ). (49) å å n! [n] ! n=0 n=0 From Lemma 3, there exists a constant K such that ¥ D f (0) ¥ n n(n1) n(n1) (x x) n 1 2 2 f (x)  q x  K q = K E (x x), (50) q 1 å å [n] ! [n] ! q q n=0 n=0 where E (.) is Jackson’s q-exponential function defined in (10). Notice that, by the known properties of E (.) (see [11]), E (x) is an entire function that has a q -exponential growth of q q order 1, and it converges everywhere in the complex plane. Therefore, f (x) is the restriction of an entire function of q -exponential growth of order 1 and a finite type less than x . So, according to Theorem 1, we obtain the result. 5. A q-Analog of Minimal Completely Convex Function Definition 2. A real-valued function f 2 C [0, 1] is a minimal q-completely convex on [0, 1] if it is q-completely convex in the interval [0, 1], and if the function g(x) = f (x) e Sin x x is not q 1 q-completely convex for any e > 0. Axioms 2023, 12, 412 10 of 13 For example, the function f (x) = Sin x is a minimal q-completely convex in 0  x  1 while the function f (x) = Sin x x is not because for any 0 < e < 1 and x 2 (0, 1), n 2n 2n (1) D Sin x x e Sin x x = (1 e)x Sin (x x) > 0. 1 q 1 q 1 q 1 Theorem 4. Let n 2 N , (a ) and (b ) be two sequences of non-negative integers. Assume that 0 n n n n the series ¥ h i n n (1) a A (x) (1) b B (x) å n n n n n=0 converges to a function f (x), 0  x  1. Then, f (x) is a minimal q-completely convex on the interval [0, 1]. Proof. From the assumption, we have h i n n f (x) = (1) a A (x) (1) b B (x) , 0  x  1. (51) n n n n n=0 Taking the q -derivative for (51) 2k times and using (6), we obtain k 2k nk nk (1) D f (x) = (1) a A (x) (1) b B (x) 1 å n nk n nk n=k (52) m m = (1) a A (x) (1) b B (x). m m å m+k m+k m=0 From Proposition 5, since (a ) and (b ) are positive sequences, the right-hand side n n n n of Equation (52) is non-negative, and f (x) is q-completely convex in [0, 1]. On the other hand, from Proposition 3 and Equation (52), there exists a constant M > 0 such that h i ¥ ¥ a + b n n k 2k 2m 2k (1) D f (x)  M a + b x = Mx . (53) 1 å m+k m+k å q 1 2n m=0 n=k 1 ¥ a +b n n According to Theorem 2, the power series T = å converges to zero as k 2n n=k k ! ¥. Hence, for given e > 0 and x 2 A , there exists an integer k 2 N such that 0 q 0 MT e Sin (x x ) < 0 (k  k ). q 1 0 0 This implies from (53) that the function k 2k k 2k 2k (1) D f (x) e Sin (x x) = (1) D f (x) e x Sin (x x) 1 q 1 1 q 1 q q is negative at x . Therefore, the function f is a minimal q-completely convex in [0, 1]. Theorem 5. If f (x) is a minimal q-completely convex function on [0, 1], then it can be expanded into a convergent q-Lidstone series: 2 2 f (x) = f (1) A (x) f (0)B (x) + D f (1) A (x) D f (0)B (x) + . . . . (54) 1 1 0 0 1 1 q q Proof. We denote by S (x) the nth partial sum of the series (54). Then, from the hypothesis on f (x) and Equation (14), we obtain S (x)  f (x) (0  x  1, n 2 N ). n 0 Axioms 2023, 12, 412 11 of 13 Moreover, for each x, S (x) is a non-decreasing function of n. Thus, lim S (x) exists n n n!¥ and tends towards some function. To prove the result, we prove that lim S (x) = f (x) (x 2 [0, 1]). n!¥ Suppose the contrary, and assume that for some x 2 [0, 1] f (x ) lim S (x ) = 4 > 0. 0 n 0 n!¥ Then, by using Equation (14), we have 2n 2 f (x ) S (x ) = G (x , qt)D f (q t) d t  4 (n 2 N). (55) n q 0 2n 0 0 1 Since f (x) is a minimal q-completely convex function on [0, 1], then f (x) e Sin x x q 1 is not q-completely convex in 0  x  1 for any e > 0. That is, there exists n 2 N and t 2 A , 0 q 2n 2n 0 0 0 (1) D f (t ) e x Sin (x t ) < 0. 1 0 q 1 0 From Inequality (11), we have 2n 2n +1 n 0 0 (1) D f (t ) < e x t . 0 0 2n n 0 By applying Lemma 2 on the function g(x) = (1) D f (x), we obtain 1 + q 2n 2n +1 n 0 0 (1) D f (t)  e x (t 2 A ). 1 q 1q Therefore, by choosing e < 4, where M is the constant of Proposition 4, we (1+q)x M obtain 2n 2 0  G (x , qt)D f (q t) d t < 4, n 0 q which contradicts Inequality (55), and then the result is proved. The following theorem is the main result of this section. Theorem 6. A real function f (x) can be represented by an absolutely convergent q-Lidstone series if and only if it is the difference of two minimal q-completely convex functions on [0, 1]. Proof. First, assume that f (x) = g(x) h(x), where g(x) and h(x) are both minimal q-completely convex functions on [0, 1]. According to Theorem 5, we have h i 2n 2n g(x) = D g(1) A (x) D g(0)B (x) , (56) å 1 n 1 n q q n=0 h i 2n 2n h(x) = D h(1) A (x) D h(0)B (x) . (57) å 1 n 1 n q q n=0 Notice that each series only has positive terms. Thus, by subtracting (57) from (56), we obtain an absolutely convergent q-Lidstone series whose sum is f (x). Conversely, assume that f (x) can be represented by an absolutely convergent q- Lidstone series ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . (58) n n å 1 1 q q n=0 Axioms 2023, 12, 412 12 of 13 2n 2n Set a = D f (1), b = D f (0), and n n 1 1 q q ¥ h i n n n+1 n g(x) = (1) fja j (1) a g A (x) + (1) fjb j (1) b g B (x) , (59) n n n n n n n=0 ¥ h i n n+1 h(x) = (1) ja j A (x) + (1) jb j B (x) . (60) n n n n n=0 Since series in (58) is absolutely convergent, then the two series in (59) and (60) both converge. Furthermore, note that every term of these series is positive. Hence, by using Theorem 4, g(x) and h(x) are minimal q-completely convex functions on [0, 1]. Since f (x) = h(x) g(x), the proof is complete. 6. Conclusions We introduced the class of q-completely convex functions in the interval [0, a], with the functions satisfying the inequality n 2n k (1) D f (aq )  0 (fn, kg  N )). 1 0 This class of functions is a generalization of the class of completely convex functions introduced by Widder [10]. First, we presented some properties of a q-completely convex function, and then we proved that such a function could be expanded in a convergent q-Lidstone series: ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . å 1 n 1 n q q n=0 Furthermore, we obtained a necessary and sufficient condition for a function f (x) to have an absolutely convergent q-Lidstone series expansion by introducing the class of minimal q-completely convex functions. Author Contributions: M.A.-T. and Z.S.I.M. equally and significantly contributed to writing this article. All authors have read and agreed to the published version of the manuscript. Funding: Research Center of the Female Scientific and Medical Colleges, Deanship of Scientific Research, King Saud University. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: This research project was supported by a grant from the ”Research Center of the Female Scientific and Medical Colleges”, Deanship of Scientific Research, King Saud University. The authors would like to thank the editor and the referees for their helpful comments and suggestions that improved this article. Conflicts of Interest: The authors declare no conflicts of interest. References 1. Lidstone, G. Notes on the extension of Aitken’s theorem (for polynomial interpolation) to the Everett types. Proc. Edinb. Math. Soc. 1929, 2, 16–19. [CrossRef] 2. Whittaker, J.M. On Lidstone’ series and two-point expansions of analytic functions. Proc. Lond. Math. Soc. 1934, 2, 451–469. [CrossRef] 3. Buckholtz, J.D.; Shaw, J.K. On functions expandable in Lidstone series. J. Math. Anal. Appl. 1974, 47, 626–632. [CrossRef] 4. Boas, R.P. Representation of functions by Lidstone series. Duke Math. J. 1943, 10, 239–245. 5. Boas, R.P.; Buck, R.C. Polynomial Expansions of Analytic Functions, 2nd ed.; Springer: Berlin, Germany, 1964. 6. Golightly, G.O. Coefficients in sine series expansions of special entire functions. Huston J. Math. 1988, 14, 365–410. 7. Leeming, D.; Sharma, A. A generalization of the class of completely convex functions. Symp. Inequalities 1972, 3, 177–199. Axioms 2023, 12, 412 13 of 13 8. Portisky, H. On certain polynomial and other approximations to analytic functions. Proc. Natl. Acad. Sci. USA 1930, 16, 83–85. [CrossRef] 9. Schoenberg, I. On certain two-point expansions of integral functions of exponential type. Bull. Am. Math. Soc. 1936, 42, 284–288. [CrossRef] 10. Widder, D. Completely convex functions and Lidstone series. Trans. Am. Math. Soc. 1942, 51, 387–398. [CrossRef] 11. Ismail, M.; Mansour, Z.S. q-analogs of Lidstone expansion theorem, two point Taylor expansion theorem, and Bernoulli polyno- mials. Anal. Appl. 2018, 17, 1–47. [CrossRef] 12. AL-Towailb, M. A generalization of the q-Lidstone series. AIMS Math. J. 2022, 7, 9339–9352. [CrossRef] 13. AL-Towailb, M.; Mansour, Z.S. The q-Lidstone series involving q-Bernoulli and q-Euler polynomials generated by the third Jackson q-Bessel function. Khayyam J. Math. 2022, accepted. 14. Mansour, Z.S.; AL-Towailb, M. The Complementary q-Lidstone Interpolating Polynomials and Applications. Math. Comput. Appl. 2020, 25, 34. [CrossRef] 15. Al-Towailb, M. A q-Difference Equation and Fourier Series Expansions of q-Lidstone Polynomials. Symmetry 2022, 14, 782. [CrossRef] 16. AL-Towailb, M.; Mansour, Z.S. Conditional expanding of functions by q-Lidstone series. Axiom 2023, 12, 22. [CrossRef] 17. Mansour, Z.S.; AL-Towailb, M. q-Lidstone polynomials and existence results for q-boundary value problems. Bound Value Probl. 2017, 2017, 178. [CrossRef] 18. Jackson, F.H. On q-functions and a certain difference operator. Trans. Roy. Soc. Edinb. 1908, 46, 64–72. [CrossRef] 19. Ayman Mursaleen, M.; Serra-Capizzano, S. Statistical Convergence via q-Calculus and a Korovkin’s Type Approximation Theorem. Axioms 2022, 11, 70. [CrossRef] 20. Kac, V.; Cheung, P. Quantum Calculus; Universitext; Springer: New York, NY, USA, 2002. 21. Hadid, S.B.; Ibrahim, R.W.; Shaher, M. Multivalent functions and differential operator extended by the quantum calculus. Fractal Fract. 2022, 6, 354. [CrossRef] 22. Ali, I.; Malghani, Y.A.K.; Hussain, S.M.; Khan, N.; Ro, J.-S. Generalization of k-Uniformly Starlike and Convex Functions Using q-Difference Operator. Fractal Fract. 2022, 6, 216. [CrossRef] 23. Aldawish, I.; Ibrahim, R.W. Solvability of a new q-differential equation related to q-differential inequality of a special type of analytic functions. Fractal Fract. 2021, 5, 228. [CrossRef] 24. Vivas-Cortez, M.; Aamir Ali, M.; Kashuri, A.; Bashir Sial, I.; Zhang, Z. Some New Newton’s Type Integral Inequalities for Co-Ordinated Convex Functions in Quantum Calculus. Symmetry 2020, 12, 1476. [CrossRef] 25. Gasper, G.; Rahman, M. Basic Hypergeometric Series, 2nd ed.; Cambridge University Press: Cambridge, UK, 2004. 26. Jackson, F.H. On q-definite integrals. Quart. J. Pure Appl. Math. 1910, 41, 193–203. Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Axioms Multidisciplinary Digital Publishing Institute

A q-Analog of the Class of Completely Convex Functions and Lidstone Series

Al-Towailb, Maryam; Mansour, Zeinab S. I.
Axioms , Volume 12 (5) – Apr 24, 2023
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axioms Article A q-Analog of the Class of Completely Convex Functions and Lidstone Series 1, 2 Maryam Al-Towailb * and Zeinab S. I. Mansour Department of Computer Science and Engineering, College of Applied Studies and Community Service, King Saud University, Riyadh 11451, Saudi Arabia Department of Mathematics, Faculty of Science, Cairo University, Giza 12613, Egypt; zsmansour@cu.edu.eg * Correspondence: mtowaileb@ksu.edu.sa Abstract: This paper introduces a q-analog of the class of completely convex functions. We prove specific properties, including that q-completely convex functions have convergent q-Lidstone series expansions. We also provide a sufficient and necessary condition for a real function to have an absolutely convergent q-Lidstone series expansion. Keywords: quantum calculus; q-series; q-Lidstone polynomials; completely convex functions MSC: 05A30; 41A58; 39A70; 40A05 1. Introduction In 1929, Lidstone [1] introduced a generalization of Taylor ’s theorem that approximates an entire function f in a neighborhood of two points instead of one. That is ¥ h i (2n) (2n) f (x) = f (1)L (x) + f (0)L (1 x) , (1) å n n n=0 Citation: Al-Towailb, M.; Mansour, where L (x) is a unique polynomial of degree 2n + 1, and called a Lidstone polynomial. Z.S.I. A q-Analog of the Class of In [2], Whittaker proved that an entire function of an exponential type of less than p has a Completely Convex Functions and convergent Lidstone series expansion in any compact set of the complex plane. Buckholtz Lidstone Series. Axioms 2023, 12, 412. and Shaw [3] provided some conditions for (1) to hold. Other authors worked on this https://doi.org/10.3390/ problem (see, e.g., [4–10]). They presented different sufficient and necessary conditions axioms12050412 for the representation of functions by this series. We mention, in particular, the result of Academic Editors: Ivanka Stamova, Widder [10]. He proved that if f is a real-valued function satisfying Gheorghe Oros and Georgia Irina Oros k (2k) (1) f (x)  0 (k 2 N ) (2) Received: 14 February 2023 in an interval of length greater than p, then it has a Lidstone series expansion (1) (such a Revised: 8 March 2023 function is known as completely convex). Furthermore, he defined the class of minimal Accepted: 19 March 2023 completely convex functions, and then he proved that a real-valued function f (x) could be Published: 24 April 2023 expanded in an absolutely convergent Lidstone series if and only if it is the difference of two minimal completely convex functions. Recently, the Lidstone expansion theorem was generalized in quantum calculus (as Copyright: © 2023 by the authors. can be seen in [11–17]). The quantum calculus (Jackson calculus or q-calculus [18]) is Licensee MDPI, Basel, Switzerland. an extension of the traditional calculus, and it has been used by many researchers in This article is an open access article different branches of science and engineering (as can be seen in, e.g., [19–24]). It has a lot distributed under the terms and of applications in different mathematical areas such as orthogonal polynomials, number conditions of the Creative Commons theory, hypergeometric functions, theory of finite differences, gamma function theory, Attribution (CC BY) license (https:// Sobolev spaces, Bernoulli and Euler polynomials, operator theory, and quantum mechanics. creativecommons.org/licenses/by/ For the basic definitions and notations applicable in the q-calculus, see Section 2. 4.0/). Axioms 2023, 12, 412. https://doi.org/10.3390/axioms12050412 https://www.mdpi.com/journal/axioms Axioms 2023, 12, 412 2 of 13 In [11], Ismail and Mansour proved the following q-analog of the Lidstone expansion theorem. Theorem 1. Assume that the function f (z) is an entire function of q -exponential growth of order 1 and a finite type a less than x , or it is an entire function of q -exponential growth of an order of less than 1. Then, f (z) has a convergent q-Lidstone representation ¥ h i 2n 2n f (z) = D f (1) A (z) D f (0)B (z) , (3) å 1 n 1 n q q n=0 where ( A ) and (B ) are the q-Lidstone polynomials defined, respectively, by the generating n n n n functions E (zw) E (zw) q q 2n = A (z)w , (4) E (w) E (w) q q n=0 E (zw)E (w) E (zw)E (w) q q q q = B (z) . (5) E (w) E (w) [n] ! q q q n=0 Moreover, A (z) = z, B (z) = 1 z, and for n 2 N, A (z) and B (z) satisfy the q-difference 0 0 n n equation D y (z) = y (z) with y (0) = y (1) = 0. (6) n n n 1 n1 In [16], AL-Towailb and Mansour proved that the condition n n D f (0) = o(x ) as n ! ¥ (7) is both sufficient and necessary for expanding an entire function f (z) in the q-Lidstone series 2 2 f (1) A (z) f (0)B (z) + D f (1) A (z) D f (0)B (z) + . . . , 0 0 1 1 1 1 q q and we noted that Condition (7) is insufficient for the convergence of the following arrange- ment of the q-Lidstone series: ¥ ¥ 2n 2n D f (1) A (z) D f (0)B (z), n n å 1 å 1 q q n=0 n=0 and not necessary for the convergence of (3). This paper aimed to obtain a sufficient and necessary condition for a real-valued function to have an absolutely convergent q-Lidstone series expansion (3). To achieve this aim, we introduced generalizations for the class of completely convex functions (2) on a closed interval of form [0, a] (a > 0), and the class of minimal completely convex functions on the interval [0, 1]. This paper is organized as follows. The following section gives the essential notions and basic definitions of q-calculus. Section 3 contains some properties and basic results on q-Lidstone polynomials, which we need in our investigation. In Section 4, we define a q-analog of the class of completely convex functions for the difference operator D 1 . Then, we study the relation of this class to a problem of the representation of functions by the q-Lidstone series. In Section 5, we provide a necessary and sufficient condition for a real function to have an absolutely convergent q-Lidstone series expansion. 2. Preliminaries In this section, we recall some definitions, notations, and results in the q-calculus, which we need in our investigations (see [25]). Throughout this paper, q is a positive number less than one, and we use the following standard notations: N := f1, 2, 3, . . .g, N := f0, 1, 2, . . .g = N[f0g. 0 Axioms 2023, 12, 412 3 of 13 The sets A and A are defined by A := fq : n 2 N g and A := A [f0g. For q q 0 q q q a 2 C, n 2 N , (a; q) (a; q) = (1 aq ), (a; q) := , ¥ Õ n (aq ; q) j=0 and the q-numbers [n] and q-factorial [n] ! are defined by q q n n 1 q [n] = , [n] ! = [k] . q q q 1 q k=1 Let m 2 C. A set A  C is called m-geometric set if mz 2 A for any z 2 A. If f is a function defined on a q-geometric set A, then Jackson’s q-difference operator is defined by f (z) f (qz) , z 2 Af0g; D f (z) = (8) (1 q)z f (0), z = 0, provided that f is differentiable at zero. Furthermore, Jackson [26] introduced the following q-integrals for a function f defined on a q-geometric set A: Z Z Z b b a f (t) d t := f (t)d t f (t) d t (a, b 2 R), q q q a 0 0 where n n f (t) d t := (1 q) zq f (zq ), n=0 provided that the series converges at z = a and z = b. Jackson’s q-trigonometric functions Sin z and Cos z are defined by q q ¥ n(2n+1) n 2n+1 Sin z := (1) (z(1 q)) , (q; q) 2n+1 n=0 (9) ¥ n(2n1) n 2n Cos z := (1) (z(1 q)) , (q; q) 2n n=0 where E () is one of Jackson’s q-exponential function defined by n(n1) (z(1 q)) E (z) = q = (z(1 q); q) (z 2 C). (10) q å ¥ (q; q) n=0 We use fx g to denote the positive zeros of Sin z arranged in increasing order of k k2N q 3/2 magnitude. One can verify that Sin z has no zeroes on jzj < q , i.e., the first positive 3/2 zeros x > q . Lemma 1. For any x 2 [0, 1], we have Sin x x  x x. (11) q 1 1 Proof. Let f (x) = x x Sin x x, x 2 [0, 1]. Then, D 1 f (x) = x (1 Cos x x)  0. 1 q 1 1 q 1 Therefore, by using (8), we obtain f (x)  f ( ) (x 2 [0, 1]), which implies f (x)  lim f (q x) = 0. Then, Inequality (11) holds. n!¥ Axioms 2023, 12, 412 4 of 13 3. Some Results on q-Lidstone Polynomials We start this section by recalling some properties of the q-Lidstone polynomials A (x) and B (x) from [14,16,17], for which we need to prove the main results. Proposition 1 ([16]). Let fx g be the sequence of the positive zeros of Sin (x) and m 2 N . k k2N q 0 Then, 2Sin (x x) q 1 (2n+1) n1 (1) A (x) = +O(x ); (12) 2n+1 0 x Sin (x ) q 1 Sin (x x)Cos (x ) q 1 q 1 n1 2n m (1) B (x) = +O(x (2n) ), (13) 0 1 2n+1 (1 q)(x ) Sin (x ) 1 1 for a sufficiently large n. 2n Proposition 2 ([17]). If f 2 C ([0, 1]), then n1 h i 2m 2m 2n 2 f (x) = D f (1) A (x) D f (0)B (x) + G (x, qt)D f (q t) d t, (14) m m n q å 1 1 1 q q q m=0 where qt(1 x), 0  t < x  1; G(x, t) = G (x, t) = (15) qx (1 t), 0  x < t  1, G (x, qt) = G(x, qy) G (qy, qt) d y (n 2 N). (16) n n1 q Moreover, G (x, qt) d t = A (x) B (x) (n 2 N). (17) n q n n Remark 1 ([14]). For x 2 [0, 1] and n 2 N , we have n n1 (1) A (x)  0 and (1) B (x)  0. (18) n n Proposition 3. Let x be the smallest positive zero of Sin (x). Then, there exist some constants M and M and a positive integer n such that the following inequalities hold 1 2 0 0  (1) A (x)  ; (19) 2n n1 0  (1) B (x)  , (20) 2n for all x 2 [0, 1] and n  n . Proof. From (12), there is a positive real number C and n 2 N such that 1 0 Sin (x x) C q 1 n1 (1) A (x) 2  , (21) 2n+1 0 2n x Sin (x ) x q 2 for all x 2 [0, 1] and n  n . Consequently, Sin (x x) q 1 0  (1) A (x)  2 . (22) 2n 2n+1 0 x Sin (x ) q 1 2 1 Axioms 2023, 12, 412 5 of 13 Note that x < x and Sin (x x) is bounded on [0, 1]. Then, from (22), we obtain 1 2 q 1 Sin (x x) C 2 q 1 0  (1) A (x)  + 2n 2n+1 x Sin (x ) q 1 1 1 (23) C C M 1 2 1 + = . 2n 2n 2n x x x 1 1 1 Similarly, we obtain (20) from (13). Proposition 4. There exists a constant M such that 0  (1) G (x, qt) d t  . n q 2n 0 x Proof. The proof follows immediately from Equation (17) and Proposition 3. Proposition 5. For any fixed point x 2 (0, 1) and sufficiently large n, there exist some constants M and M such that (1) A (x )  ; (24) n 0 2n n1 (1) B (x )  . (25) n 0 2n Proof. From (12), we obtain n 2n+1 2n+1 (1) A (x)x = L(x) +O(( ) ) (n ! ¥), 2Sin (x x) q 1 where L(x) = . Notice, for any fixed x 2 (0, 1), L(x ) > 0 and 0 0 0 Sin (x ) n 2n+1 lim (1) A (x )x = L(x ). n 0 0 n!¥ n 2n+1 This implies that the sequence (1) A (x )x is bounded below by a positive n 0 number. I.e., (24) holds. Similarly, we obtain the Inequality (25) from (13). Now, using the previous results, we prove the following theorem. Theorem 2. If the series S = a A (x) + b B (x) + a A (x) + b B (x) + . . . (26) 0 0 0 0 1 1 1 1 h i a +b ¥ n n n converges for a single value x 2 (0, 1), then the series å (1) is absolutely convergent. n=0 2n Proof. Since the series (26) converges for x 2 (0, 1), we have lim a A (x ) = 0, lim b B (x ) = 0. n n 0 n n 0 n!¥ n!¥ Then, from the inequalities (24) and (25), we obtain 2n 2n a = O(x ) and b = O(x ). (27) n n 1 1 Axioms 2023, 12, 412 6 of 13 From (12), (13), and (27), we conclude that the series ( ) h n i h n i 2(1) Sin (x x ) (1) Cos x Sin (x x ) q q q 1 0 1 1 0 S = a A (x ) + + b B (x ) + 1 å n n 0 n n 0 2n+1 0 2n+1 0 x Sin (x ) (1 q)x Sin (x ) n=0 q 1 q 1 1 1 converges absolutely. This implies that S S is also convergent. Notice that h i ¥ n n 2Sin (x x ) Cos x Sin (x x ) (1) (1) q 1 0 q 1 q 1 0 S S = a + b 1 n n 0 0 2n 2n x Sin (x ) x (1 q)x Sin (x ) x 1 q 1 1 q 1 n=0 1 1 ¥ h n n i 2Sin (x x ) (1) (1) q 1 0 > a + b . n n 2n 2n x Sin (x ) x x 1 q 1 n=0 1 1 Therefore, we obtain the result. 4. A q-Analog of Completely Convex Function In this section, by C [0, a], we mean the space of all functions defined on [0, a] such that D f (x) is defined and continuous at zero. Definition 1. A real-valued function f , defined on the interval [0, a] (a > 0), is said to be a q-completely convex function if f 2 C [0, a] and n 2n k (1) D f (aq )  0 (for all fn, kg  N ). (28) 1 0 Example 1. The functions f (x) = Sin x x, defined in (9), are q-completely convex on the interval [0, 1]. Indeed, one can verify that n 2n n 2n 2n (1) D f (x) = (1) D Sin x x = x Sin (x x) > 0, (29) 1 1 q 1 q 1 q q for all x 2 [0, 1] and n 2 N . In the following, we prove certain properties of q-completely convex functions. Proposition 6. If a function f 2 C [0, a] is q-completely convex, then n 2n (1) D f (0)  0 (n 2 N ). (30) 1 0 2n Proof. The proof follows directly by taking the limit as k ! ¥ in (28) and using that D f is continuous at zero for all n 2 N . Proposition 7. Let f 2 C (0, 1) be a q-completely convex function on [0, 1]. Then, for a suffi- ciently large n, we have 2n 2n D f (0) = O(x ); (31) 2n 2n D f (1) = O(x ). (32) Proof. From Proposition 1 and Inequality (28), every term of (14) is non-negative. Therefore, 2n 0  A (x) D f (0)  f (x); (33) 2n 0  (B (x))D f (1)  f (x) (x 2 [0, 1]; n 2 N ). (34) n 1 0 q Axioms 2023, 12, 412 7 of 13 Thus, by using (24) and (33), we obtain f (x ) n 2n 2n 0  (1) D f (0)   Kx (n ! ¥), q n (1) A (x ) n 0 for some constant K > 0 and x 2 (0, 1). Then, we have (31). Similarly, we obtain the asymptotic behavior in (32). Proposition 8. Let f be a q-completely convex function on [0, 1]. Then, there exists a positive constant C such that for all x 2 A 2n n 2n 0  (1) D f (x)  C , (35) where x is the smallest positive zero of Sin (x). Proof. If f is q-completely convex on [0, 1], then it is q-completely convex on [0, x] for all x 2 A . Consequently, the function f (t) := f (xt) is q-completely convex on [0, 1]. Therefore, from Proposition (7), we have n 2n n 2n 2n 2n 0  (1) D f (1) = (1) x D f (x) = O(x ), 1 1 q q which is nothing else but (35). Lemma 2. Let f (x) and D f (x) be non-negative on A , and continuous at 0. Assume that 1 q there exists a number x 2 A such that f (x )  a (a 2 R). Then, 0 q 0 (1 + q)a f (x)  , for all x 2 A . (1 q)x Proof. First, let x 2 A and x  x . Then, by using the assumption D f (x)  0, we have q 0 1 D f ( ) d t  0. Therefore, D f (x)  D f (x ), and q q 0 D f (t) d t  (x x )D f (x ) (x 2 A , x  x). (36) q q 0 q 0 0 Since f (x)  0 on A , from (8) and Inequality (36), we obtain (x x ) x x q a 0 0 f (x)  f (x ) + f (x ) = f (x ) < , (37) 0 0 0 (1 q)x (1 q)x (1 q)x 0 0 0 for all x 2 A and x  x. Similarly, if x 2 A and x < x , then q 0 q 0 x x f (qx ) 0 0 f (x)  f (qx ) < . (38) (1 q)x (1 q)x 0 0 On the other hand, since D f (x)  0, we have (1 + q) f (qx)  q f (x) + f (q x) (x 2 A ). q Axioms 2023, 12, 412 8 of 13 Therefore, from the condition f (x)  0, we obtain (1 + q) f (qx)  q f ( ) + f (qx) > f (qx) (x 2 A ). (39) So, from the inequalities (38) and (39), we obtain (1 + q)a f (x) < (x 2 A , x < x ). (40) (1 q)x Hence, the relations (37) and (40) yield the required result. Corollary 1. If f 2 C [0, 1] is a q-completely convex function, then there exists a positive constant M such that n 2n 2n 0  (1) D f (x)  Mx (n 2 N , x 2 A ). (41) 1 0 1 q Proof. The proof follows from Proposition 8 and Lemma 2 by taking x = 1 and M = 1+q C. 1q Lemma 3. If f 2 C [0, 1] is a q-completely convex function on [0, 1], then there exists a constant K > 0 such that n n jD f (x)j  Kx (x 2 A ), (42) 1 q where x is the smallest positive zero of Sin (z). 1 q Proof. From Corollary 1, it suffices to prove (42) when n is an odd integer. We set g(x) = n 2n (1) D f (x). Since f (x) is a q-completely convex on 0  x  1, again from Corollary 1, there exists the constant M > 0 (independent of n) such that for all x 2 A 2n 0  g(x)  Mx , (43) 2 2n+2 0  D g(x)  Mx . q 1 Therefore, for every x 2 A f1g, we have Z 2 2 2n+2 0  D g(t) d t  Mq(q x)x . 1 q q 1 qx So, by using the fundamental theorem of the q-calculus, we obtain n 2n+1 n 2n+1 2n+2 0  (1) D f (x) (1) D f (1)  Mx , 1 1 q q and hence, n 2n+1 n 2n+1 n 2n+1 2n+2 (1) D f (1)  (1) D f (x)  (1) D f (1) + Mx , 1 1 1 q q q for all x 2 A f1g. Consequently, 2n+1 2n+1 2n+2 jD f (x)j  jD f (1)j + Mx . (44) 1 1 q q On the other hand, since D g(x) < 0, one can verify that for all x 2 A 1 q x x (1 + q)g( )  g(x) + qg( ), q q Axioms 2023, 12, 412 9 of 13 and then qg( )  (1 + q)g(x) g(qx) (x 2 A ). Thus, if x = 1, we obtain 1 1 (1 + q) 2n n 2n 2n (D f )( ) = (1) (D f )( )  D f (1) . (45) 1 1 1 q q q q q q Hence, from (8), (43) and (45), we have jg(1)j +jg(1/q)j 2q + 1 2n+1 2n D f (1) = jD g(1)j   Mx . (46) q 1 1/q 1 q 3/2 However, x > q , this implies 2n+1 2n+1 D f (1)  q(2q + 1) Mx . (47) By substituting (47) in (44), we obtain 2n+1 2n+1 2n+2 2n+1 jD f (x)j  q(2q + 1) Mx + Mx  M Mx , 1 1 1 1 1 3/2 for all n 2 N and x 2 A , where M = q(2q + 1) + q . 2n+1 2n+1 2n+1 Since D f (x) is continuous at zero, then we obtain D f (x) = O(x ) for a 1 1 q q sufficiently large n. This completes the proof. Theorem 3. Let f 2 C [0, 1] be a q-completely convex on [0, 1]. If f is analytic at zero, then the following q-Lidstone series expansion holds for all x 2 [0, 1]. ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . (48) 1 n 1 n q q n=0 Moreover, f (x) is the restriction of an entire function of q -exponential growth of order 1 and a finite type less than x and the expansion (48) holds for all x on the entire complex plane. Proof. Since f is analytic at 0, there exists 0 < c < 1 and the open interval W = (c, c) such that f (x) has the Maclaurin series expansion ¥ (n) ¥ D f (0) n(n1) 1 f (0) n n f (x) = x = q x (x 2 W ). (49) å å n! [n] ! n=0 n=0 From Lemma 3, there exists a constant K such that ¥ D f (0) ¥ n n(n1) n(n1) (x x) n 1 2 2 f (x)  q x  K q = K E (x x), (50) q 1 å å [n] ! [n] ! q q n=0 n=0 where E (.) is Jackson’s q-exponential function defined in (10). Notice that, by the known properties of E (.) (see [11]), E (x) is an entire function that has a q -exponential growth of q q order 1, and it converges everywhere in the complex plane. Therefore, f (x) is the restriction of an entire function of q -exponential growth of order 1 and a finite type less than x . So, according to Theorem 1, we obtain the result. 5. A q-Analog of Minimal Completely Convex Function Definition 2. A real-valued function f 2 C [0, 1] is a minimal q-completely convex on [0, 1] if it is q-completely convex in the interval [0, 1], and if the function g(x) = f (x) e Sin x x is not q 1 q-completely convex for any e > 0. Axioms 2023, 12, 412 10 of 13 For example, the function f (x) = Sin x is a minimal q-completely convex in 0  x  1 while the function f (x) = Sin x x is not because for any 0 < e < 1 and x 2 (0, 1), n 2n 2n (1) D Sin x x e Sin x x = (1 e)x Sin (x x) > 0. 1 q 1 q 1 q 1 Theorem 4. Let n 2 N , (a ) and (b ) be two sequences of non-negative integers. Assume that 0 n n n n the series ¥ h i n n (1) a A (x) (1) b B (x) å n n n n n=0 converges to a function f (x), 0  x  1. Then, f (x) is a minimal q-completely convex on the interval [0, 1]. Proof. From the assumption, we have h i n n f (x) = (1) a A (x) (1) b B (x) , 0  x  1. (51) n n n n n=0 Taking the q -derivative for (51) 2k times and using (6), we obtain k 2k nk nk (1) D f (x) = (1) a A (x) (1) b B (x) 1 å n nk n nk n=k (52) m m = (1) a A (x) (1) b B (x). m m å m+k m+k m=0 From Proposition 5, since (a ) and (b ) are positive sequences, the right-hand side n n n n of Equation (52) is non-negative, and f (x) is q-completely convex in [0, 1]. On the other hand, from Proposition 3 and Equation (52), there exists a constant M > 0 such that h i ¥ ¥ a + b n n k 2k 2m 2k (1) D f (x)  M a + b x = Mx . (53) 1 å m+k m+k å q 1 2n m=0 n=k 1 ¥ a +b n n According to Theorem 2, the power series T = å converges to zero as k 2n n=k k ! ¥. Hence, for given e > 0 and x 2 A , there exists an integer k 2 N such that 0 q 0 MT e Sin (x x ) < 0 (k  k ). q 1 0 0 This implies from (53) that the function k 2k k 2k 2k (1) D f (x) e Sin (x x) = (1) D f (x) e x Sin (x x) 1 q 1 1 q 1 q q is negative at x . Therefore, the function f is a minimal q-completely convex in [0, 1]. Theorem 5. If f (x) is a minimal q-completely convex function on [0, 1], then it can be expanded into a convergent q-Lidstone series: 2 2 f (x) = f (1) A (x) f (0)B (x) + D f (1) A (x) D f (0)B (x) + . . . . (54) 1 1 0 0 1 1 q q Proof. We denote by S (x) the nth partial sum of the series (54). Then, from the hypothesis on f (x) and Equation (14), we obtain S (x)  f (x) (0  x  1, n 2 N ). n 0 Axioms 2023, 12, 412 11 of 13 Moreover, for each x, S (x) is a non-decreasing function of n. Thus, lim S (x) exists n n n!¥ and tends towards some function. To prove the result, we prove that lim S (x) = f (x) (x 2 [0, 1]). n!¥ Suppose the contrary, and assume that for some x 2 [0, 1] f (x ) lim S (x ) = 4 > 0. 0 n 0 n!¥ Then, by using Equation (14), we have 2n 2 f (x ) S (x ) = G (x , qt)D f (q t) d t  4 (n 2 N). (55) n q 0 2n 0 0 1 Since f (x) is a minimal q-completely convex function on [0, 1], then f (x) e Sin x x q 1 is not q-completely convex in 0  x  1 for any e > 0. That is, there exists n 2 N and t 2 A , 0 q 2n 2n 0 0 0 (1) D f (t ) e x Sin (x t ) < 0. 1 0 q 1 0 From Inequality (11), we have 2n 2n +1 n 0 0 (1) D f (t ) < e x t . 0 0 2n n 0 By applying Lemma 2 on the function g(x) = (1) D f (x), we obtain 1 + q 2n 2n +1 n 0 0 (1) D f (t)  e x (t 2 A ). 1 q 1q Therefore, by choosing e < 4, where M is the constant of Proposition 4, we (1+q)x M obtain 2n 2 0  G (x , qt)D f (q t) d t < 4, n 0 q which contradicts Inequality (55), and then the result is proved. The following theorem is the main result of this section. Theorem 6. A real function f (x) can be represented by an absolutely convergent q-Lidstone series if and only if it is the difference of two minimal q-completely convex functions on [0, 1]. Proof. First, assume that f (x) = g(x) h(x), where g(x) and h(x) are both minimal q-completely convex functions on [0, 1]. According to Theorem 5, we have h i 2n 2n g(x) = D g(1) A (x) D g(0)B (x) , (56) å 1 n 1 n q q n=0 h i 2n 2n h(x) = D h(1) A (x) D h(0)B (x) . (57) å 1 n 1 n q q n=0 Notice that each series only has positive terms. Thus, by subtracting (57) from (56), we obtain an absolutely convergent q-Lidstone series whose sum is f (x). Conversely, assume that f (x) can be represented by an absolutely convergent q- Lidstone series ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . (58) n n å 1 1 q q n=0 Axioms 2023, 12, 412 12 of 13 2n 2n Set a = D f (1), b = D f (0), and n n 1 1 q q ¥ h i n n n+1 n g(x) = (1) fja j (1) a g A (x) + (1) fjb j (1) b g B (x) , (59) n n n n n n n=0 ¥ h i n n+1 h(x) = (1) ja j A (x) + (1) jb j B (x) . (60) n n n n n=0 Since series in (58) is absolutely convergent, then the two series in (59) and (60) both converge. Furthermore, note that every term of these series is positive. Hence, by using Theorem 4, g(x) and h(x) are minimal q-completely convex functions on [0, 1]. Since f (x) = h(x) g(x), the proof is complete. 6. Conclusions We introduced the class of q-completely convex functions in the interval [0, a], with the functions satisfying the inequality n 2n k (1) D f (aq )  0 (fn, kg  N )). 1 0 This class of functions is a generalization of the class of completely convex functions introduced by Widder [10]. First, we presented some properties of a q-completely convex function, and then we proved that such a function could be expanded in a convergent q-Lidstone series: ¥ h i 2n 2n f (x) = D f (1) A (x) D f (0)B (x) . å 1 n 1 n q q n=0 Furthermore, we obtained a necessary and sufficient condition for a function f (x) to have an absolutely convergent q-Lidstone series expansion by introducing the class of minimal q-completely convex functions. Author Contributions: M.A.-T. and Z.S.I.M. equally and significantly contributed to writing this article. All authors have read and agreed to the published version of the manuscript. Funding: Research Center of the Female Scientific and Medical Colleges, Deanship of Scientific Research, King Saud University. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: This research project was supported by a grant from the ”Research Center of the Female Scientific and Medical Colleges”, Deanship of Scientific Research, King Saud University. The authors would like to thank the editor and the referees for their helpful comments and suggestions that improved this article. Conflicts of Interest: The authors declare no conflicts of interest. References 1. Lidstone, G. Notes on the extension of Aitken’s theorem (for polynomial interpolation) to the Everett types. Proc. Edinb. Math. Soc. 1929, 2, 16–19. [CrossRef] 2. Whittaker, J.M. On Lidstone’ series and two-point expansions of analytic functions. Proc. Lond. Math. Soc. 1934, 2, 451–469. [CrossRef] 3. 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Axioms – Multidisciplinary Digital Publishing Institute

Published: Apr 24, 2023

Keywords: quantum calculus; q-series; q-Lidstone polynomials; completely convex functions

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