Functional Equations, Difference Inequalities and Ulam by John Michael Rassias

By John Michael Rassias

This e-book is a discussion board for changing principles between eminent mathematicians and physicists, from many elements of the realm, as a tribute to the 1st centennial birthday anniversary of Stanislaw Marcin ULAM. This assortment consists of exceptional contributions in mathematical and actual equations and inequalities and different fields of mathematical and actual sciences. it really is meant to spice up the co-operation between mathematicians and physicists engaged on a vast number of natural and utilized mathematical components. This transatlantic number of mathematical principles and techniques contains a large sector of purposes within which equations, inequalities and computational options pertinent to their suggestions play a middle position. Ulam's impression has been great on our daily life, simply because new instruments were constructed, and progressive examine effects were completed , bringing scientists of actual sciences even nearer, through fostering the emergence of latest ways, options and views.

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Extra resources for Functional Equations, Difference Inequalities and Ulam Stability Notions (F.U.N.) (Mathematics Research Developments)

Example text

Sz´ekelyhidi, Note on Hyers’s theorem, C. R. Math. Rep. Acad. Sci. Canada 8 (1986), No. 2, 127–129. [9] P. Volkmann, Oral communication. In: Functional Equations, Difference Inequalities... ISBN 978-1-60876-461-7 c 2010 Nova Science Publishers, Inc. Editor: John Michael Rassias, pp. 47-53 Chapter 4 O N A P ROBLEM OF J OHN M. R ASSIAS C ONCERNING THE S TABILITY IN U LAM S ENSE OF E ULER –L AGRANGE E QUATION L. G˘avrut¸a and P. G˘avrut¸a∗ Department of Mathematics, University “Politehnica” Timisoara, Piata Victoriei no.

6) On a Problem of John M. Rassias Concerning the Stability... = f (v + y) + f (y − v) − 2f (v) − 2f (y) ≤ kv a y 2−a . Case 5) x < 0, y < 0. If u = −y, v = −x we have: f (x + y) + f (x − y) − 2f (x) − 2f (y) = f (u + v) + f (v − u) − 2f (v) − 2f (u) ≤ kv a u2−a = k|x|a |y|2−a . Step (III). 2). First we prove that if Q : R → R is a quadratic mapping then Q(2n ) = 4n Q(1), n ∈ Z. From Q(x + y) + Q(x − y) = 2Q(x) + 2Q(y) for x = y we obtain Q(2x) + Q(0) = 4Q(x) For x = 0 it follows 2Q(0) = 4Q(0) =⇒ Q(0) = 0.

20) An analogous reasoning can be applied for (14) (without assuming the continuity of f ). Indeed, with the aid of this estimation for y = x + 2h we get f (x) + f (x + 2h) − 1 [f (x + 2h) − f (x)] ≤ 2ε, h which jointly with (19) (which is valid by Theorem 2) gives us f (x) ≤ (h − 1)(e2h − 1) + 2h · ε, x ∈ I, h > 0, x + h ∈ I, (h − 1)e2h + h + 1 stating that (14) is stable. Again, if additionally I is unbounded from the right, then by letting h tend to +∞ we can see that (20) holds. Finally, (17) applied for y = x + 2h implies that 4hf (x + h) + (h + 3)f (x) + (h − 3)f (x + 2h) ≤ 6hε, which jointly with (18) and (19) (which is valid by Theorem 4) leads to the estimation f (x) ≤ he2h + 4heh + h − 1 · ε, x ∈ I, h > 0, x + h ∈ I, (h − 3)e2h + 4heh + h + 3 stating that (14) is stable in the sense of Hyers–Ulam.

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