VOL. 14 NO. 3 (1991) 431-434
ON STABILITY OF ADDITIVE MAPPINGS
ZBIGNIEW GAJDA InstituteofMathematics
SilesianUniversity 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 ofhisresultprovedin[3].
KEY WORDS AND PHRASES.
Additivemappings,linearmappings,Banachspaces, stability.1980 AMS
SUBJECT CLASSIFICATION CODES.
Primary39B70, Secondary 39C05.1.
INTRODUCTION.
In
connection with a problem posed by Ulam(cf. [5];
see also[2])
Th. M. Rassias[3]
proved the following theoremon stability oflinearmappingsin Banachspaces.THEOREM 1.
(see [3])
LetE
and E2 be two(real)
Banach spaces and letf:
E E2be amapping such that for eachfixed xeE thetransformation
R f(tx)
is continuous.Moreover,
assumethat thereexist eE
[0, oo)
andpc[O,
such thatf(
/v)- f()- f()II -< (
p/P) (1.1)
for all z,
yeE
1. ThenthereexistsauniquelinearmappingT:E
--,E
2such thatf()- r()II -<
p(-)
for all
zeE1,
where:
2e2_2p"
As
was mentioned by Th. M. Rassias[4],
the proofpresented in[3]
reveals that, in fact, it works for everypfrom theinterval(-
oo,1) and,
therefore,the theorem holdstruefor all suchp’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 thatf
is approximated by an additive mappingT
which means thatT
satisfiesthefollowingequation
T(x + y) T(x) + T(y) (1.3)
for all
x,yE
1. Finally, it should be noticed that the completeness of the spaceE
may be removed from the assumptions of Theorem 1.However,
there is stillonenon-trivial(as
itseems)
question concerning a possibleextension of the range of validity of Theorem 1. Namely, onecan ask whether thesameresult holds true under the hypothesis that pis taken from theinterval
[1,oo)
432 Z. GAJDA
problem was raised by Th. M. Rassias during the 27th International Symposium on Functional Equations whichwas held in Bielsko-Biala, Katowice and Krokowin
August
1989. Thegoalof the presentnote is togiveacompletesolutiontothisproblem.2.
MAIN RESULTS.
First, let us realizewhy theproofofTheorem in itsoriginal form
(see [3])
doesnot work forp
_>
1. The fundamental rolein thisproofisplayedbythe sequence{n f(2nx)" ne} (2.1)
which, under the assumptions of Theorem
(in
fact aslong aspc(-
cxz,1))
is convergent for eachfixedxeE1. ThenT:
E E
2defined by the formulaT(z): =nlim n f(2nx), :reEl (2.2)
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 proofovertothis case,one has tochangethe argument itselforthe definition of themappingT. It turnsout that,for p>
1, thelattermodification oftheproofis possible. As aresult weobtainthe followingextensionofTheorem 1:THEOREM 2. Let E and
E
2 be two(real)
normed linear spaces and assume thatE
2 is complete. Letf:E E
2 be a mapping for which there exist two constants eE[0,o)
andpE
R\{1 }
such thatf(x + 9)- f(x)- f(9)11 < e(
x p+
9P) (2.3)
for all z,
9eE
1. Then thereexistsauniqueadditivemappingT:E
E2such thatf(z)- T(x)II <
6(2.4)
for all
xcE1,
where2 for
<
1,6=
2-2P
P2 for p
>
1.2P-2
Moreover,
is for eachx qE the transformationR9f(tx)
is continuous, thenthe mappingTis linear.PROOF. In
view of what has been said so far, it remains to consider the case p>
1. Themain innovation in comparison with the case p
<
1 consists in defining the mappingT
by the formulaT(x)" =nlim 2nf(#),
xcE(2.5)
instead of
(2.2).
Obviously, one has to verify the convergence of the sequence occurring on the right-handsideof(2.5).
Putting inplace ofz andy ininequality
(2.3),
weobtainf(x)-
2f()II <
221 Pc
x pfor allx
E
1. Henceforeachn
andeveryxEl,
wehavef(x)-
2nf(#)II < f(x)- 2f()II +
2f()- 2f(2- ) + +
2n11 f(2n
X2f(#)I]
<21 P[[xlIP+2.21 Pll[[P+...+2
n1.21 Pll2n_l[[
p(21
p+ 22(1 p) + + 2n(1 p))
z pwhere$is thesumofthefollowing convergentseries:
c
2-(1-P)
2Z
2P_2Now,
fixanzcE andchse arbitrarym,n
suchthatm>
n. Then2mf()- 2"f()[[
2a2m -nf(2-n "#)- f()II
2"
2"(1 p)g ,
which becomes arbitrarily smM1 as n
. On
account of the completeness of the spaceE
2, this implies that the sequence{2"f()" n}
isconvergentfor eachzE
1. ThusT
is correctlydefinedby
(2.5). Morver,
itsatisfies condition(2.4)
whichresultsonletting, in(2.6).
FinMly, replacingz by =d by in
(2.3)
=d then multiplying bothsidesof the resulting inequMityby2n,
weget2nf()- 2nf()- 2nf()II 2n(X P)( + P),
for z,VE
E
1. Since the right-hand side of this inequality tends to zero as n cx, it becomes apparentthat the mappingT
definedby(2.5)
isadditive.The proof of the homogeneity of
T (under
the supplementary assumption thatf(tz)
iscontinuousfor each zE
El)
needsnoessential alterations incomparisonwith thecase p<
1. Itis also clear what has to bechangedin theproofof the uniqueness ofT.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 whichTheorem 2 can not be extended.
In
fact, we shall show thate>
0one canfindafunctionf:R R
such thatf(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:RR
satisfyingthe conditionIf(x)- T(z) < 6lz
for allzeR.. (2.8)
Thissingularityisillustratedbythe following:
EXAMPLE.
Fix e>
0andput/:.
Firstwedefineafunction:R R
byfor
[1,
(x):
x forxe1,1),
Evidently, is continuousd
I(z)
zR. Therefore, afunctionf: R
is correctlydefinedbytheformula
f(z): ff(2n)
n:0
2n
Since
f
is definedby
means of a uniformly convergent series of continuous functions,f
itself iscontinuous.
Moreover,
If(z)l <
n=0-2, x.
434 Z. GAJDA
Ifx y 0, then
(2.7)
istriviallyfulfilled. Next assume that 0< ]x[ + [y[ <
1. Then thereexistsan
Ne
such thatne, 12u-l<, IN-luI<a 12N--(,+U)I--<ZU--(II+IUl)<,
wiimplies that for each n
{0,1,
N-1}
the numbers2nz,
2ny and2n(z + y)
remainin the interval(- 1,1).
Since is linearon thisinterval,weinfer that(2"( + u)) (2") (2"u) o
for n 0,1, N-1. Asa result,weget
(Ixl + lyl)
n=N2"(Ixl + lYl)
(2n(x + y)) (2"x) (2ny)
Finally, assumethat
[x[ + ]y[ >
1. Thenmerely byvirtueofthe boundedness off
wehavef(x + y)- f(x)- f(r)
< 6#
e.Il+lul
Thusweconcludethat
f
satisfies(2.7)
forallreal xand y.Now,
contrarytowhat weclaim,suppose thatthereexista([0, oo)
andanadditivefunction T:R--,R
such that(2.8)
holds true.Hence,
from the continuity off
itfollows thatT
is bounded on someneighbourhood ofzero. Then, bya classical result(see
e.g.[1],
2.1.1., Theorem1)
thereexists arealconstantcsuch that
Hence,
whichimplies that
T(x)
cx, xOn the other hand, we can choose an Ne[ so large that
Np >
g+ [x[.*
Then picking out anz from the interval(0,
12N 1)’
we have 2nx(0,1)
for each n{0,1
N1}.
Consequently, for suchanxwehavef(x)> (2nx) F2nz
x n=O
2n----
n=On-=Nu>6+ Ixl,
which yields a contradiction. Thus the function
f
provides a good example to the effect that Theorem 2failstohold forp 1.REFERENCES.
1.
ACZEL,
J. Lectures o__n Functional Equations and their Applicatio.ns, AcademicPress,
New York- SanFrancisco-London,
1966.2.
HYERS,
D.H. On the stability of the linear functional equation,Proc..Nat.
Acad. Sci.,U.S.A.,
27(1941),
222-224.3.
RASSIAS,
TH. M. On the stability of the linear mapping in Banach spaces, Proc. Amer.Math. Soc.72
(1978),
297-300.4.
RASSIAS, TH. M.
Communication, 27t.___h.h International Symposium onFunctional Equations, Bielsko-Biala, Katowice,Krokow,
Poland, 1989.5.
ULAM,
S.M. Problems in modern mathematics, ChapterVI,
Science Editions, Wiley, New York, 1960.Mathematical Problems in Engineering
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