ON STABILITY OF ADDITIVE MAPPINGS

In this paper we answer a question of Th. M. Rassias concerning an extension of 
validity of his result proved in [3].


Institute of Mathematics
Silesian University Bankowa 14 40-007 Katowice Poland (Received January 26, 1990) ABSTRACT.In this paper we answer a question of Th.M. Rassias concerning an extension of validity of his result proved in [3].
In connection with a problem posed by Ulam (cf.[5]; see also [2]) Th.M. Rassias [3] proved the following theorem on stability of linear mappings in Banach spaces.
THEOREM 1. (see [3]) Let E and E 2 be two (real) Banach spaces and let f: E E 2 be a mapping such that for each fixed xeE the transformation R f(tx) is continuous.Moreover, assume that there exist e E [0, oo) and pc[O, such that f( / v)-f()-f()II -< ( p / P) (1.1) for all z, yeE 1.Then there exists a unique linear mapping T: E --, E 2 such that f()-r()II -< p (-) for all zeE1, where : 2e As was mentioned by Th.M. Rassias [4], the proof presented in [3] reveals that, in fact, it works for every p from the interval (oo, 1) and, therefore, the theorem holds true for all such p's.
It is also readily seen that the only purpose of assuming that all the transformations of the form t--, f(tx) are continuous is to guarantee the real homogeneity of the mapping T. Without this assumption one can show that f is approximated by an additive mapping T which means that T satisfies the following equation T(x + y) T(x) + T(y) (1.3) for all x,yE 1.Finally, it should be noticed that the completeness of the space E may be removed from the assumptions of Theorem 1.However, there is still one non-trivial (as it seems) question concerning a possible extension of the range of validity of Theorem 1. Namely, one can ask whether the same result holds true under the hypothesis that p is taken from the interval [1,oo) (obviously in this case the constant should have been defined in a different manner).Such a problem was raised by Th.M. Rassias during the 27th International Symposium on Functional Equations which was held in Bielsko-Biala, Katowice and Krokow in August 1989.The goal of the present note is to give a complete solution to this problem.

MAIN RESULTS.
First, let us realize why the proof of Theorem in its original form (see [3]) does not work for p _> 1.The fundamental role in this proof is played by the sequence {n f(2nx)" ne} (2.1) which, under the assumptions of Theorem (in fact as long as pc(-cxz, 1)) is convergent for each fixed xeE 1.Then T: E E 2 defined by the formula T(z): =nlim n f(2nx), :reEl is the desired linear mapping approximating f.The argument ensuring the convergence of sequence (2.1) is no longer valid when p becomes greater or equal to 1, so in order to carry the proof over to this case, one has to change the argument itself or the definition of the mapping T. It turns out that, for p > 1, the latter modification of the proof is possible.As a result we obtain the following extension of Theorem 1: THEOREM 2. Let E and E 2 be two (real) normed linear spaces and assume that E 2 is complete.Let f:E E 2 be a mapping for which there exist two constants e E [0,o) and p E R\{1 } such that f(x + 9)-f(x)-f(9)11 < e( x p + 9 P) (2.3) for all z, 9eE 1.Then there exists a unique additive mapping T: E E 2 such that f(z)-T(x)II < 6 (2.4) for all xcE1, where 2 for < 1, 6= 2-2P P 2 for p > 1. 2P-2 Moreover, is for each x q E the transformation R 9 f(tx) is continuous, then the mapping T is linear.
PROOF.In view of what has been said so far, it remains to consider the case p > 1.The main innovation in comparison with the case p < 1 consists in defining the mapping T by the formula T(x)" =nlim 2nf(#), xcE   (2.5) instead of (2.2).Obviously, one has to verify the convergence of the sequence occurring on the right-hand side of (2.5).
Putting in place of z and y in inequality (2.3), we obtain f(x)-2 f()II < 2 21 Pc x p for all x E 1. Hence for each n and every x El, we have 2"(1 p)g , which becomes arbitrarily smM1 as n .On account of the completeness of the space E 2, this implies that the sequence {2"f()" n} is convergent for each z E 1. Thus T is correctly defined by (2.5).Morver, it satisfies condition (2.4) which results on letting, in (2.6).
FinMly, replacing z by =d by in (2.3) =d then multiplying both sides of the resulting inequMity by 2n, we get 2nf()-2nf()-2nf()II 2n(X P)( + P), for z,V E E 1. Since the right-hand side of this inequality tends to zero as n cx, it becomes apparent that the mapping T defined by (2.5) is additive.The proof of the homogeneity of T (under the supplementary assumption that f(tz) is continuous for each z E El) needs no essential alterations in comparison with the case p < 1.It is also clear what has to be changed in the proof of the uniqueness of T.
Theorem 2 leaves the case p undecided.This is not a mere coincidence.It turns out that 1 is the only critical value of p to which Theorem 2 can not be extended.In fact, we shall show that e > 0 one can find a function f:R R such that f(x + y)-f(x)-f(Y) < e(lx + Yl) (2.7) for all x, yeR, but, at the same time, there is no constant 6 [0,cx) and no additive function T:R R satisfying the condition If(x)-T(z) < 6lz for all zeR.. (2.8)This singularity is illustrated by the following: EXAMPLE.Fix e > 0 and put/: .First we define a function :R R by for [1, (x): x for xe 1,1), Evidently, is continuous d I(z) zR.Therefore, a function f: R is correctly defined by the formula f(z): ff(2n) n:0 2n Since f is defined by means of a uniformly convergent series of continuous functions, f itself is continuous.Moreover, If(z)l < -2, x.We are going to show that f satisfies (2.7).

F2nz
x 2n----n-=Nu>6+ Ixl, n=O n=O which yields a contradiction.Thus the function f provides a good example to the effect that Theorem 2 fails to hold for p 1.