On Some Inequalities and Stability

(C) 1998 OPA (Overseas Publishers Association) N.V. Published by license under the Gordon and Breach Science Publishers imprint. J. oflnequal. & Appl...
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(C) 1998 OPA (Overseas Publishers Association) N.V. Published by license under the Gordon and Breach Science Publishers imprint.

J. oflnequal. & Appl., 1998, Vol. 2, pp. 373-380 Reprints available directly from the publisher Photocopying permitted by license only

Printed in India.

On Some Inequalities and Stability Results Related to the Exponential Function CLAUDI ALSINA a,, and ROMAN GER b a

Sec. Matemtiques Informtica, Univ. Polit+cnica de Catalunya, Diagonal 649, 08028 Barcelona, Spain; blnstitute of Mathematics, Silesian University, ul. Bankowa 14, 40-007 Katowice, Poland (Received 12 November 1997, Revised 20 February 1998)

Some inequalities related to the exponential function are solved and the stability of the functional equationsf’(x)-f(x) and (f(y)-f(x))/(y-x)=f((x + y)/2) is studied. Keywords." Inequalities; Exponential function; Hyers-Ulam stability; Functional equation AMS 1991 Subject Classification." 39C05

One of the most classical characterizations of the real exponential function f(x)- e is the fact that the exponential function is the only (modulo a multiplicative constant) nontrivial solution of the differential equation f’=f Our aim in this note is to study the Hyers-Ulam stability of this equation, i.e. to solve for a given c > 0 the inequality

}f’(x)-f(x)l _< e,


and to study also the related inequality (for all x =/= y)

(2) Corresponding author. E-mail: [email protected] 373



In dealing with (1) and (2) we will solve several inequalities which have their own interest. In what follows I will stand for any real interval and R + for the set of all nonnegative real numbers. A function f will be termed Jensen concave if f satisfies the inequality f((x+y)/2)>_ (f(x)+f(y))/2 and f will be said to be k-lipschitz whenever [f(x)-f(y)] 0 such that x + h E I we have

hf(x +) >_f(x) + hf(x)

f(x + h) >_f(x) +

because, clearly, fhas to be nondecreasing. By an obvious induction we get

f(x + ih) >_ (1 + h)if(x)

(1 + h)f(x),



whenever x + ih E I and E N. Thus, for an arbitrarily fixed n N, for every x < y from I, one eventually obtains

f (y)

f(x + n Y-x)_> (1+



and if we let n tend to infinityf(y) _> ey-Xf(x), i.e., the function i: I-+ R + defined by i(x)=f(x)e is nondecreasing. Conversely, if we have the representation f(x) i(x)e x, x I, with i: I-+ 1R + nondecreasing then, since for x < y we have i(x) i(x)

that is,

i(y)ey-x- i(x) >

i(x +2 Y)(e



and multiplying both terms by eX/(y-x) with the aid of Lemma 2 we have ey i(y)ey -i(x)e x > {x + y’

f (y) f (x)









i(x +2 y)e(X+y)/2 =f(x +2 y)

i.e. (11) follows. Moreover, f is nonnegative because so is i.

THEOREM 2 Given an c > 0 let f: I---+ IR + be a function such that f(x) >_ for all x in L Then f satisfies the inequality

f(x +2 y)

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