On functions continuous on certain classes of `thin' sets
by
Wolfgang Ring, Peter Sch
opf and Jens Schwaiger (Graz)
Abstract. Continuity properties of functions with continuous restrictions to certain
classes of `thin' sets are considered. By construction of an explicit counterexample it
is shown that continuity of the restriction to neither of the following classes of sets is
sucient to guarantee continuity of a real-valued function on the plane: (real) analytic
images of compact intervals, zero sets of real analytic functions in two variables and
regular C 1;-images of compact intervals. However, continuity of a real-valued function
in two variables along all regular C 1 -curves implies its continuity. In addition it is shown
that a function which is continuous on the boundary of some ball of positive radius in
a Hilbert space, or even | in the nite dimensional case | on the boundary of a
bounded open set must be continuous if it satises a Cauchy-type functional equation.
In dimension two the same result can be obtained if the set on which continuity is
required is connected and contains three noncollinear points.
1. Introduction
In [Z] F. Zorzitto showed that every function f : C ! C which satises the Cauchy
equation
f (x + y) = f (x)f (y)
(1)
and is continuous on circles has necessarily to be continuous globally. He wrote:
\The above proof leads one to speculate, out of mere curiosity, whether any function
f : C ! C, having continuous restrictions to all circles in C, can still possess a point of
discontinuity."
In Section 2 we deal with the problem of constructing several discontinuous functions
which have continuous restrictions to certain classes of `thin' sets, all of them including
the class of circles Zorzitto was curious about. More specically we give an example
of a discontinuous function f : IR2 ! IR with continuous restrictions f j(I ) (f jN resp.)
for every analytic image (I ) of a compact interval I and for every zero-set Na of
an arbitrary (non vanishing) real analytic function a(x; y) in two variables. Under
the additional regularity assumption that _ (t) 6= (0; 0) for all t 2 I , we can drop the
requirement of analyticity for assuming only that is taken from some Holder-class
C 1; with > 0.
In contrast to these counterexamples we show in Section 3 that any function f : IR2 !
IR, for which all restrictions to arbitrary regular C 1 -images of compact intervals are
continuous, has to be continuous itself.
The last Section 4 is devoted to the study of continuity properties of functions satisfying
a generalized form of the functional equation (1). Problems of similar structure, dealing
with Jensen-convex functions instead of solutions of the Cauchy equation (1), were
considered by Ger [G1],[G2] and Kuczma [K].
a
{1{
2. The counterexamples
We consider the well-known function ' : IR ! IR, dened by
,x,1 ) x > 0
'(x) := exp(
0
x 0.
(2)
This function is of class C 1 with '(n) (0) = 0 for all n. We also note separately the
important property that for all n 2 IN0
'(x)
lim
! xn = 0:
x 0
x=0
6
(This is used to establish the relations '(n) (0) = 0 mentioned above.)
Using this we may dene : IR ! IR by
'(x2 )
(x) := 1 , '(x2) +
'(1 , x2 ) :
(3)
Again, is (well-dened and) of class C 1 . Moreover
0 1;
(0) = 1; and
outside [,1; 1]:
(x) = 0
This is used for the denition of the function f : IR2 ! IR:
(
f (x; y) :=
0
2
( )
'x
,
y , '(x)
3
2
x0
x > 0.
Remark 1. By the properties of it follows easily that f vanishes outside the closed
set A dened by
A := f(x; y) 2 IR2 j x 0; '(x) y 2'(x)g:
(4)
Moreover, 0 f 1, f (0; 0) = 0 and f (x; 32 '(x)) = 1 for all x > 0.
Lemma 1. The function f is of class C 1 on IR2 nf(0; 0)g and discontinuous at (0; 0).
Proof. Obviously f is C 1 on the open set IR2 n A since it is zero there. Moreover it
is clear from the deniton that f is C 1 in the right halfplane. Therefore f is C 1 on
the union of these two open sets, which is IR2 n f(0; 0)g.
Finally, f is discontinuous at (0; 0), since f (0; 0) = 0 and limx!0 f (x; 23 '(x)) = 1.
+
{2{
The function f is constructed in such a way that a curve which is either an analytic
image of an interval or part of the zero set of an analytic function in two variables and
which passes through the origin cannot do so from within A. This property of f is the
essential point in the proof of the following two theorems.
Theorem 1. Let I D IR with D open and I compact. Let furthermore : D ! IR2
be (real) analytic and denote by (I ) the image of I under . Then the restriction f j(I )
is continuous.
Proof. We write (t) = (1(t); 2(t)), where 1; 2 : D ! IR are real analytic. Since f is
continuous on IR2 nf(0; 0)g the assertion of the theorem is obviously true if (0; 0) 62 (I ).
2
Thus we
may suppose that (0; 0) 2 (I ). If, in this case, (I ) IR n A we conclude
that f (I ) = 0 and thus continuous. Here A denotes the interior of A. So we are left
with the case (I ) \ A 6= ;.
This means that both 1 and 2 are dierent from zero and that there is some t~ 2 I
such that 1(t~) > 0 and '(t~) < 2(t~) < 2'(t~). In particular 1; 2 are not constant.
Since I is compact and 1 is analytic, there are only nitely many zeroes t1; t2; : : : ; tr
of 1 in I . Now, for 1 i r we have representations
1(t) = (t , ti ) 1;i(t , ti ); 2(t) = (t , ti ) 2;i(t , ti ) (i 2 IN; i 2 IN0); (5)
where 1;i(0); 2;i(0) 6= 0 for 1 i r.
Now, choose m 2 IN such that mi > i for all i. Due to the continuity of the functions
j;i there are positive reals M1 ; M2; 0 such that for all 1 i r and all t 2 I we have
that jt , ti j < 0 implies
j2;i(t , ti )j > M2 and j1;i(t , ti )j < M1:
Then, for 0 < < min(1; M2=M1m ; 0) and 0 < jt , ti j < j2(t)j = jt , ti j j2;i(t , ti )j jt , ti jm ,1 M2 >
> jt , ti jm ,1 (M1m) > jt , ti jm M1m >
> jt , ti jm j1;i(t , ti )jm = j1 (t)jm :
j1 (t)j < "g. Then T" is contained in the union K :=
For
"
>
0
put
T
:=
f
t
2
I
"
Sr
i=1 ]ti , ; ti + [ provided that " is suciently small. (For otherwise we could construct
a sequence (sn) of reals contained in the compact set I n K , converging to some t0 2
I n K , such that limn!1 1(sn ) = 0 = 1 (t0) contradicting the fact that t1; t2; : : : ; tr
is the set of all zeroes of 1 in I .)
Taking into account the properties of the function ' we may choose 0 < " < 1 such
that
T" K and 2'(jxj) jxjm (jxj "):
Thus t 2 I and j1 (t)j < " implies t 2 T" and
j2(t)j > j1(t)jm 2' (j1 (t)j) :
But this means that (I ) \ (] , 1; "[IR) \ A = ;. Accordingly f is continuous on
(I ) \ (] , 1; "[IR) as it vanishes on this set. But obviously f is also continuous on
(I ) \ (]0; 1[IR). So f (I ) is continuous.
i
i
i
i
i
i
i
{3{
Now we consider the restrictions of f to locally analytic sets.
Theorem 2. Let D IR2 be an open set and assume that a : D ! IR is a real analytic
function, a 6 0. Moreover let
Na := f(x; y) 2 D a(x; y) = 0g:
Then f jN is continuous.
Proof. The assertion is obviously true whenever (0; 0) 62 Na . Thus we suppose that
(0; 0) 2 Na .
If, in this case, Na is contained in the complement of A, we have that f vanishes
identically on Na , implying continuity.
Otherwise we have Na \ A 6= ;. It is enough to show continuity of f N at (0; 0). We
can assume that the power series representation of a at (0; 0) given by
a
a
a(x; y) =
X
i;j 0
ai;j xi yj ;
(6)
is absolutely and uniformly convergent on Q = [,; ][,; ] for some > 0. Without
loss of generality we may also assume that a (as given in (6)) is neither divisible by x nor
by y. If, for example, a(x; y) is divisible by x, we could write a(x; y) = x` b(x; y) with
` 1 and b(0; y) 6 0. Putting c(x; y) := x` this would imply Na \ Q = (Nb [ Nc ) \ Q .
But Nc = f0g IR and f is continuous on the y-axes (and, as for the divisibility by y,
on the x-axes as well). Thus f is continuous on Na at (0; 0) i it is continuous on Nb
at (0; 0). This and a(0; 0) = 0 implies that
a0;0 = 0 and a0;`; an;0 6= 0 for some `; n 2 IN:
(7)
We will now show that there is some integer m 1 and some " > 0 such that
jxj < " and a(x; y) = 0 implies jyj jxjm :
This can be seen as follows. The properties (7) imply that
X
mn
am;0xm = ,
X
i;j )2IN0 IN
ai;j xi yj :
(8)
(
for all (x; y) 2 Na \ Q , where n 1 is minimal with an;0 6= 0. We may rewrite the
left-hand side of (8) as
xn (an;0 + an+1;0x + ) = xn (an;0 + g(x));
where an;0 6= 0 and jg(x)j < jan;0=2j for all jxj < , if is chosen suciently small.
Hence for jxj < we have
an;0 n
n
n
jx (an;0 + g(x))j jxj (jan;0j , jg(x)j) jxj 2 :
(9)
{4{
Now we consider the right-hand side of (8). This expression is not zero since a0;` 6= 0
and it does not contain any pure power of x. Thus there is a maximal k 1 such that
the right-hand side of (8) is divisible by yk . This yields
X
i
j
a
x
y
i;j
(i;j )2IN0 IN
=
k
y
X
i
j
,
k
ai;j x y (i;j )2IN0 IN
j
jyjk jh(x; y)j ;
(10)
k
with h continuous on Q . Let K > 0 be the maximum of jhj on the compact set Q .
Combining (9), (10) and (8) gives
an;0 n
jxj 2 jxn (an;0 + g(x))j = yk h(x; y) jyjk K
,
for all (x; y) 2 Na \ Q . If we set M := jan;0 =2j K ,1 1=k we get
jyj M jxjn=k ;
and for m 2 IN, m (n=k) + 1, and < min(1; M ) we nally get
jyj jxjn=k jxj(n=k)+1 jxjm
provided that (x; y) 2 Na \ Q .
Now the proof can be nished as in Theorem 1.
Remark 2.
1) The compactness of the interval I in Theorem 1 is necessary. This can be seen by
considering the curve
: IR ! IR2 ; (t) = (cos(t) + cos(!t); sin(t) + sin(!t))
with irrational !. The image of this curve is dense in the disk of radius 2 centered
at the origin (and the origin is in the image). Thus f restricted to (IR) cannot be
continuous. Note, however, that although the restriction f j (IR) is discontinuous,
the composition f is continuous.
2) Theorems 1 and 2 are not independent. It can be proved that every analytic zero
set is locally the union of nitely many analytically parametrized curves (c.f. [KP]
Theorem 3.2.3, p.84 and Theorem 5.2.1 p.153), thus, in fact, Theorem 2 follows from
Theorem 1. However, the proof for this fact requires deep methods from the theory
of real analytic functions and is restricted to dimension n=2, so we preferred to
give elementary proofs for both cases. It is not known to the authors whether every
analytic curve is locally the zero set of an analytic function, so conversely Theorem 2
would imply Theorem 1. A counterexample to an analogous situation is known for
2-dimensional analytically parametrized surfaces in IR3 . (c.f. [N] Example 1.4.16,
p.28.)
{5{
A modied version of the function f can be be shown to have continuous restrictions
to all C 1; -images of compact intervals. We briey recall the denition of the Holder
space C 1; (I; IR2): Let I be a compact interval and let 0 < 1. Then C 1;(I; IR2)
consists of all curves : I ! IR2 ; (t) = (1(t); 2(t)) for which i 2 C 1 (I; IR) and
there exists a constant M = M ( ) such that
j_ i (t) , _ i (s)j M jt , sj
(11)
for all t; s 2 I and i = 1; 2. An important property of C 1; -functions lies in the fact
that the remainder in the rst-order Taylor expansion is of order O(1 + ). Actually
we have
Z th
i
i (t) = i (s) + _ i (s)(t , s) +
_ i ( ) , _ i (s) d;
(12)
and due to (11),
s
Z
t
s
_ i ( ) , _ i (s)
d M jt , sj1+
(13)
for i = 1; 2 and s; t 2 I .
A curve 2 C 1 (I; IR) is said to be regular if j_ (t)j 6= 0 for all t 2 I .
Let be as dened in (3) and let p : [0; 1[! [0; 1[ be given by
p(x) = ,
x
Z
0
ds
ln s
h
i
for x 2 0; 12 ;
and suppose that p is continued outside [0; 21 ] in such a way that p is monotone and
C 1 -smooth on [0; 1[. It follows easily by de l' Hospitals rule that
p(x) = 1
lim
x!0 x1+
(14)
+
for all > 0 and
p(x) = 0:
lim
x!0 x
We now dene the function f~ : IR2 ! IR in analogy to f as
+
(
0
f~(x; y) := 2
( )
px
,
y , p(x)
3
2
x0
x > 0.
(15)
(16)
Note that, by denition, f~(x; y) = 0 if x 0 or if y 62]p(x); 2p(x)[.
Theorem 3. For any 2]0; 1] the function f~ has continuous restrictions to all regular
C 1;-images of compact intervals.
Proof. Let : I ! IR2 be a regular curve of class C 1;. Obviously f~j (I ) is continuous
at every point (x; y) 2 (I ) which is not the origin. We therefore assume that there
exists t0 2 I such that (t0) = (0; 0) and, as is regular, _ (t0) = (v1 ; v2) 6= (0; 0).
{6{
Let us rst suppose that v2 6= 0. If we choose some neighborhood Vt of t0 small enough
that jt , t0 j minf1; j2vMj g is satised for all t 2 Vt , we can conclude from (12) and
(13) that
j2(t)j jv2 j jt , t0 j , M jt , t0 j1+ jv22 j jt , t0 j;
and
j1(t)j (jv1 j + M )jt , t0 j
holds for all t 2 Vt . Consequently we obtain
j1(t)j 2(jv jvj 2+j M ) j1(t)j
(17)
1
on Vt . Now since
p(j1(t)j) ! 0 for t ! t ;
0
j (t)j
0
2
0
0
0
1
due to (11), we have
(18)
p(j1(t)j) 4(jv jvj 2+j M ) j1(t)j
1
on some neighborhood Wt Vt of t0 . Combining (17) and (18) we nd
j2(t)j 2p(j1(t)j)
and therefore f~(1(t); 2(t)) = 0 on Wt . Hence f is continuous in this case.
Now assume that v2 = 0. Then v1 6= 0 since is regular. Let the neighborhood Vt be
choosen in such a way that jt , t0 j j2vMj on Vt Then, using again (12) and (13) we
obtain
j2(t)j M jt , t0 j1+
and
1+
j1(t)j M jv2 j j1(t)j1+:
1
By (14) and since 1(t) ! 0 as t ! t0, we can choose a neighborhood Wt of t0 so that
2 1+
p(j1(t))j) M jv j
j1(t)j1+:
1
for all t 2 Wt . We thus have
j2(t)j p(j1(t)j)
and accordingly f~(1(t); 2(t)) = 0 if t 2 Wt . Hence also in this case we obtain that
f~ is continuous at t0.
So far we have shown that f~ : I ! IR is continuous. It is still to prove that f~
restricted to (I ) is a continuous function. To this aim suppose that (xn ; yn ) ! (x; y)
with (xn ; yn) = (tn) and (x; y) = (t) and assume that
jf~(1(tn )) , f~( (t))j for all n 2 IN:
(19)
By the compactness of I , we can nd a convergent subsequence tn ! s of ftn g. The
continuity of f~ implies that
f~( (tn )) ! f~( (s))
(20):
Since is continuous we also have (tn ) ! (s). Thus (t) = (s) in contradiction to
(19) and (20).
0
0
0
0
1
0
0
0
0
k
k
k
{7{
Remark 3. By means of the functions f and f~ we can easily construct functions F
and F~ respectively which have discontinuities of the above discussed types at all points
of an arbitrary countable set fxn gn2IN IR2 . In fact, we set fn (x) := f (x , xn ) and
we dene
1 1
X
(21)
F (x) := 2n fn (x):
n=1
The series (21) is uniformly convergent on IR2 , and every function fn is continuous on
IR2 n fxn g, hence the limit-function F is continuous at every point x 62 fxn g. Let G be
either an analytic image of a compact interval, or the zero-set of a real analytic function.
Then every restriction fn jG is continuous and therefore F is again a continuous function
2
on G. Finally
P F 1cannot be continuous at xn 2 IR since it is the sum of the continuous
function k6=n 2 fk (x) and the discontinuous fn .
The construction of F~ is completely analoguous.
k
3. Functions continuous on regular C -curves are continuous
1
In contrast to the counterexamples given in Theorem 1 and Theorem 2 we shall now
show that the continuity of f along regular C 1 -curves implies its continuity.
Theorem 4. Suppose D IR2 is open and f : D ! IR. Moreover assume that
f : I ! IR is continuous for every regular curve : I ! D, 2 C 1 (I; IR2) and
I IR compact. Then f is continuous.
Proof. Suppose that f is discontinuous at z0 2 D. Then there is a sequence zn 2 D,
zn 6= z0 , zn ! z0 , and " > 0 such that
jf (zn ) , f (z0)j "
(n 2 IN):
(22)
Our aim is to nd an appropriate subsequence of fzn g which can be interpolated by a
C 1 -curve . Since f is continuous along this would contradict (22).
Before we proceed any further, some technical comments have to be made. In the
following we shall frequently choose subsequences of fzn g which we shall always denote
by the same expression fzn g. Furthermore let t 2 C 1 ([,1; 1]; IR) be given, fullling the
requirements t(1) = 1, t(,1) = ,1, t0 (1) = t0 (,1) = 0. For instance we could choose
t(x) = 21 (3x , x3 ). We set
M = ktk1 = supf jt(x)j , 1 x 1g:
(23)
Using polar coordinates centered at z0 = (x0; y0) we can write zn = (xn ; yn) = (x0 +
rn cos 'n ; y0 + rn sin 'n ) with rn > 0 and 'n 2 T = IR=26 Z. From the compactness
of the torus group T we conclude that there exists a convergent subsequence 'n ! '.
Without loss of generality we assume ' = 0, for otherwise we could rotate the coordinate
system and use f(cos '; sin '); (, sin '; cos ')g as a new basis on IR2 . Since 'n ! 0
it follows that xn , x0 > 0 for suciently large n. By taking a subsequence we can
assume that this property holds for all n.
{8{
We shall now list several properties which can be supposed to hold for some subsequence
of the current (sub)sequence zn = (xn ; yn ) and which remain valid if we take further
subsequences.
Since xn ! x0 and xn > 0 we may choose a subsequence such that
xn+1 , x0 < xn ,2 x0
(n 2 IN):
(24)
(n 2 IN):
(25)
As yn ! y0 there is a subsequence such that
jyn+1 , y0j jyn , y0 j
Since 'n ! 0 we also have tan 'n ! 0 as n ! 1. Therefore xy ,,yx ! 0 and we
n
nd a subsequence for which
n
jyn , y0j 1
xn , x0 n
0
0
(n 2 IN):
(26)
For (xn ; yn ) satisfying properties (24), (25) and (26) we now construct a regular C 1 curve = (1(x); 2(x)) : [x0; x1] ! IR2 such that 2(x0) = y0 and 2(xn ) = yn for all
n 2 IN. We dene
xn+1 ) , 1
`n (x) := 2(xx ,
,x
n
n+1
(x 2 [xn+1 ; xn]; n 2 IN)
(27)
and
1
1
; xn],
(x) := ((xx; ;2y(y)n , yn+1 )t(`n (x)) + 2 (yn+1 + yn )) ifif xx 2=]xxn+1
0
0
0.
(28)
Note that `n : [xn+1 ; xn] ! [,1; 1] is linear, `n (xn+1) = ,1 and `n(xn ) = 1. The
only delicate points with respect to continuity of and _ are the points x = xn for
n 2 IN and the point x = x0. It is easy to see that is continuous at x = xn . For the
derivative we nd
_ (x) = (1; (yn , yn+1 )t0(`n (x)) x ,1x ) (x 2]xn+1 ; xn ]):
(29)
n
n+1
Again we easily nd that the left sided and the right sided derivatives of at x = xn
exist and that they both equal (1; 0). Thus _ (xn) exists and _ is continuous at x = xn .
From (25) we obtain
jyn , yn+1 j 2 jyn , y0 j
(n 2 IN)
(30)
and (24) implies
xn , xn+1 = (xn , x0) , (xn+1 , x0 ) xn ,2 x0 :
{9{
(31)
Consequently
jyn , yn+1 j 4 jyn , y0 j 4
xn , xn+1
xn , x0 n
(n 2 IN):
(32)
for xn+1 < x xn ;
(33)
Due to (26) (using also (23) and (32)) we have
yn
x
, yn+1 t0 (` (x)) 4 M
n
n
n , xn+1
hence (recall (29)) limx!x _ (x) = (1; 0). It remains to prove that is dierentiable at
x = x0 and that _ (x0) = (1; 0). Obviously only the second component 2 of has to
be considered. For x 2]xn+1 ; xn] we have
0
j2(x) , 2(x0 )j j2(x) , 2(xn+1 )j + j2(xn+1 ) , 2(x0)j x , x0
x , x0
2(xn+1 )j + jyn+1 , y0j sup j 0 ( )j + 1 j2(xx) ,
2
,x
x ,x
n+1
4nM + n1
n+1
n+1
0
2[x
n+1
;x
n]
by (26), (29) and (32). But this implies the desired properties of _ .
Now, for x suciently
close to x0 we may assume that (x) 2 D. Thus for some n0 2 IN
the function f [x ;x ] is continuous. This implies f (zn) ! f (z0) as n ! 1 since
zn = (xn) in contradiction to (22).
0
n0
Remark 4.
1) Once the existence of an appropriate subsequence fullling (24){(26) is established,
we can conclude the existence of an interpolating C 1 -curve by applying Whitney's
extension theorem (see [N, p.31]). But since the character of this paper is constructive we preferred however to give an explicit construction of .
2) We stress the fact that the velocity of the parametrization of the curves along which
f is assumed to be continuous is bounded away from zero. In fact j_ (x)j 1 would
be a sucient requirement. By this assumption `trivial' interpolating curves, such
as C 1 -parametrizations of broken lines with vanishing velocity at the vertices, are
excluded. By this argument one can easily see that if we drop the assumtion of
regularity, then it is sutient to claim that f is continuous on all C 1 images of
compact intervals in order to obtain continuity of f .
3) The restriction to the two-dimensional case D IR2 is not essential. With some
minor modications the proof works for D IRn , n 2. However it is not clear
to the authors what happens in the innite dimensional case, since we do not have
compactness of the unit sphere in this situation.
4. Functions with a continuous addition law and continuity on `thin' sets
Here we present a generalization of Zorzitto's result. We start with a lemma.
{ 10 {
Lemma 2. Let (X; h; i) be a real inner product space of dimension greater than 1. Let
e 2 X with kek = 1, and put B := B (e; 1) := fx 2 X kx , ek < 1g. Then there exist
continuous functions u; v : B ! @B , where @B denotes the boundary of B , such that
u + v = idB .
Proof. We rst derive some necessary conditions for the function u : B ! @B Suppose,
we have u = u(z) 2 @B , v = v(z) 2 @B and z = u + v. Since u and v have the same
distance from e, we get that
0 = ku , ek2 ,kv , ek2 = hu; ui,hv; vi,hu , v; 2ei = hu , v; u + v , 2ei = h2u , z; z , 2ei:
Upon dividing by 4 in the above inequality it follows that hu , z2 ; z2 , ei = 0. On the
other hand, if we can nd a continuous function w : B ! X n f0g which satises the
orthogonality condition
hw(z); 2z , ei = 0;
(34)
then the construction
r
z
u(z) := 2 + 1 , k 2z , ek2 kww((zz))k and v(z) := z , u(z)
will do the job. Note that k z2 , ek 21 (kz , ek + kek) < 1, hence the square root in the
denition of u makes sense.
Now let e0 2 X be given such that ke0 k = 1, he0 ; ei = 0, and let P (z) = z , hz; e0 ie0
denote the projection onto the orthogonal complement of the subspace generated by e0 .
We have P (z) 6= 0 for all z 2 B , for otherwise we would have z = e0 and consequently
1 > kz , ek2 = 2 + 1 due to the orthogonality of e0 and e.
To determine the function w, we put
w(z) = (z)e0 + (z)P (z):
The orthogonality condition (34) implies
(z)he0 ; 2z i + (z)hP (z); z2 , ei
which allows the determination of the coecients
(z) = ,hP (z); z2 , ei and (z) = he0 ; z2 i:
To prove continuity of w, we still have to ensure that (z)2 + (z)2 6= 0 for z 2 B .
However (z) = (z) = 0 would imply z ? e0 and
0 = hz , hz; e0 ie0; 2z , ei = 12 kzk2 , 2hz; ei;
leading to the contradiction
1 > kz , ek2 = kzk2 , 2hz; ei + kek2 = 1:
Thus we have constructed a function w which has all the properties we want.
{ 11 {
We have the following generalization of Zorzitto's result.
Theorem 5. Let (X; h; i) be a real inner product space of dimension greater than 1,
let Y be a topological space, g : Y Y ! Y continuous. Let furthermore f : X ! Y
fulll the functional equation
f (x + y) = g(f (x); f (y))
(x; y 2 X ):
(35)
Then f is continuous on X provided that its restriction to the boundary of some ball
of positive radius is continuous.
Proof. At rst we suppose that f j@B(e;1) is continuous, where kek = 1. By Lemma
2, we can write z = u(z) + v(z) where z 2 B (e; 1), u(z); v(z) 2 @B (e; 1), and u; v are
continuous. Then from (35) it follows that
f (z) = f (u(z) + v(z)) = g(f (u(z)); f (v(z)))
which implies continuity of f on B (e; 1).
But the continuity of any solution f of (35) in some point implies global continuity. To
see this consider f (x1 + h) = f ((x0 + h) + (x1 , x0 )) = g(f (x0 + h); f (x1 , x0)) and
f (x1) = g(f (x0); f (x1 ,x0 )). The same arguments show that continuity of f j@B(x ;r) for
xed r implies the continuity of f j@B(x ;r) for all x1 . So if we suppose that f j@B(x ;r)
is continuous, we can conclude that f j@B(0;r) must be continuous. Hence also the
function fr dened by fr (x) := f (rx) has continuous restriction onto @B (0; 1). But fr
also satises (35). Thus fr j@B(e;1) is continuous, and, by the rst part of the proof, we
have continuity on X . Accordingly also f is continuous on X since f (x) = fr (r,1x).
0
1
0
It is clear that this result generalizes Zorzitto's result ([Z]) when we consider C as a
(2-dimensional real) Hilbert space. But in the nite dimensional case an even stronger
result may be derived. In the sequel the inner product space X is replaced by IRn
equipped with the usual euclidean norm.
Theorem 6. Let X = IRn with n 2, Y , f , g be given as in Theorem 5. Suppose
moreover that there exist a bounded, open and connected set O IRn and a compact,
connected set C IRn containing the boundary @O of O such that the restriction f jC
is continuous. Then f is continuous on IRn .
Before we start proving Theorem 6, we formulate a topological lemma which we shall
need henceforth.
Lemma 3. Let O IRn be a bounded, open and connected set and assume that
C IRn is compact and connected satisfying @O C . Then
2O C + C and therefore (C + C ) 6= ;:
Proof. It has to be proved that, for every x 2 O, we can nd u; v 2 C with u + v = 2x.
Let x 2 O. We translate IRn in such a way that x is shifted into the origin and consider
the sets C x = ,x + C and Ox = ,x + O
{ 12 {
We shall prove that
C x \ (,C x ) 6= ;. We choose w from the compact set C x such that
kwk = supfkyk y 2 C x g. Since @Ox C x , we have
kwk supfkyk y 2 @Oxg = supfkyk y 2 Ox g;
implying that ,w 62 Ox , for otherwise we could nd some v 2 Ox with kvk > kwk
contradicting the above inequality. We can also suppose that w 62 @Ox, for otherwise
we would have C x \ (,C x ) =
6 ; and the assertion is proved. Now we choose 2 IR in
such a way that n := w 2 C x and
jj = inf fjj w 2 C x ; 2 IRg:
Again we assume that ,n 62 C x for otherwise we would conclude that C x \ (,C x ) =
6 ;.
We consider [ ,n; n[ , the straight segment joining ,n and n with n excluded. Obviously
0 2 [ ,n; n[ and also 0 2 Ox . Moreover [ ,n; n[ does not intersect C x by the minimality
property of , hence it also does not intersect @Ox. Since [ ,n; n[ is connected, has one
point in Ox and does not intersect the boundary of Ox , it must lie completely within
Ox . (c.f. [R, p.141, Prop. 14.5].) Therefore ,n 2 Ox . Thus we have ,n; ,w 2 ,C x ,
,n 2 Ox and ,w 62 Ox . Employing again [R, p.141, Prop. 14.5] and the fact that
,C x is connected, we conclude that ,C x \ @Ox 6= ; and, since @Ox 2 C x , we obtain
C x \ (,C x ) 6= ;.
Now we choose u0 2 C x \ (,C x ), i.e. u0 2 C x and ,u0 2 C x . By denition of C x there
exist u; v 2 C such that u0 = ,x + u and ,u0 = ,x + v and therefore 2x = u + v, which
completes the proof.
Proof of Theorem 6. Let x 2 O. We want to show that f is continuous at 2x.
Assume that it is not. Then there exists a sequence xn 2 O, xn ! x, such that f (2xn)
does not converge to f (2x). This implies the existence of some neighbourhood V of
f (2x) and of a subsequence x0n of the original one such that
f (2x0n) 62 V
(n 2 IN):
(36)
But every point 2x0n may be written as
2x0n = u0n + vn0
(u0n ; vn0 2 C )
and we get
f (2x0n ) = g(f (u0n); f (vn0 )):
Using the compactness of C we may (by choosing suitable subsequences, if necessary)
assume that u0n ! u 2 C and vn0 ! v 2 C . We also have 2x = u + v. Passing to the
limit and using the continuity of f on C we derive
f (2x0n ) = g(f (u0n); f (vn0 )) ! g(f (u); f (v)) = f (u + v) = f (2x)
which contradicts (36).
Hence f is continuous at 2x. Using the arguments given in the proof of Theorem 4 this
implies the continuity of f on X .
{ 13 {
If O 2 IRn is an open, bounded and connected set which has a connected boundary
@O, then it is possible to choose C = @O in Lemma 3 and we obtain
2O @O + @O:
Furthermore we conclude by Theorem 6 that f (satisfying the functional equation (35))
must be continuous on IRn if its restriction to @O is continuous. Thus Theorem 6 is in
fact a generalization of Theorem 5 in the nite dimensional case.
It is however possible to derive an even more general result.
Lemma 30 . Let O IRn be a bounded, open and connected set. Then
2O @O + @O and therefore (@O + @O) 6= ;:
Corollary 1. Let X = IRn with n 2, Y , f , g be given as in Theorem 5. Suppose
moreover that there exist a bounded, open and connected set O IRn such that the
restriction f j@O is contiuous. Then f is continuous on IRn .
Once the assertion in Lemma 30 is established, Corollary 1 follows from Lemma 30 in
exactly the same way as Theorem 6 follows from Lemma 3. Therefore we focus on the
proof of Lemma 30 .
Proof of Lemma 30 . Let O IRn be open, bounded and connected. The complement
IRn n O contains exactly one unbounded, open, connected component which we denote
by O1 . Let C = @O1 . Then C is compact and C @O. (c.f. [D, p.356, Th. 1.2] for
these facts.) C is separating the space IRn in the sense that IRn n C is not connected.
Especially O and O1 lie in dierent components of IRn n C (Suppose they lie in the
same component, then there exists a path in this component which connects some
point in O with some other point in O1 . This path however must intersect @O1 = C
contradicting the fact that lies entirely in IRn n C .)
It can be proved by means of Zorn's Lemma that there exists a minimal compact set
(with respect to inclusion) C C such that O and O1 lie in dierent components of
IRn n C .
The set C is connected. This is proved as follows. Assume that C = C1 [ C2 with
C1 \ C2 = ;, Ci 6= ; and Ci closed for i = 1; 2 and let p 2 O and q 2 O1 be given.
Suppose that p and q belong to the same component of IRn n C1 and that they also
belong to the same component of IRn n C2 . Let p : IRn n fpg ! S n,1 be dened as
, p and : IRn n fqg ! S n,1 be analoguosly given by (x) = x , q . It
p (x) = jxx ,
q
q
pj
jx , qj
is known [D, p.359, Prop. 4.1] that two points p and q belong to the same component of
IRn n A, where A is some compact set, if and only if the restrictions p jA and q jA are
homotopic. Hence there exist homotopies i : [0; 1] Ci ! S n,1 with i (0; x) = p (x)
and i (1; x) = q (x) for x 2 Ci and i = 1; 2. But now it is easy to construct a homotopy
between p and q on C = C1 [ C2 . We dene : [0; 1] C ! S n,1 by
if x 2 C1
1 (t; x)
(t; x) := 2(t; x) if x 2 C2 .
{ 14 {
It is easily seen that is continuous and satises (0; x) = p (x) and (1; x) = q (x)
for all x 2 C . Using again [D, p.359, Prop. 4.1], we conclude that p and q belong to
the same component of IRn n C . This however is a contradiction to the fact that C separates O from O1 . Therefore p and q are separated by at least one of the Ci -s say
by C1 . Since O and O1 are connected and O \ C1 = O1 \ C1 = ;, it follows that O
and O1 are subsets of dierent components of IRn n C1 , which cannot be true since C is a minimal set fullling this requirement. Thus C is connected.
By O we denote the component of IRn n C which includes O as a subset. Then we
have @O C [D, p356, Th. 1.2 (3)] and O is bounded since it is separated from
the (unique) unbounded component of IRn n C (which includes O1 ) by C . Therefore
Lemma 3 can be applied to O and C respectively and we nally get
2O 2O C + C @O + @O
which completes the proof.
In dimension n = 2, this result can be generalized.
Lemma 4. Let C IR2 be a compact, connected set, containing three noncollinear
points a; b and c. Then
(C + C ) 6= ;:
Proof.
: C 6= ;, then obviously (C + C ) 6= ;.
2
2. case: If IR n C has a bounded component O , then @O C and Lemma 3 implies
that (C + C ) 6= ;.
2
3. case: Assume that C = ; and IR n C is connected. Under these assumptions C
cannot contain all edges of the triangle, whose vertices are a; b and c. Without loss of
generality we suppose that there exists a point x 2 [ a; b] n C . Here [ a; b] denotes the
closed line-segment joining a and b. We denote by L the straight line passing through
x, perpendicular to [ a; b] . Then there exist two points x1 ; x2 2 L n fxg, such that
[ x1; x2] \ C = ; and [ x1; x2] \ [ a; b] = fxg, since IR2 n C is open.
We set C~ := C [ [ a; b] and we shall show now that IR2 n C~ has at least one bounded
component. Assume to the contrary that IR2 n C~ is connected. Since x1 ; x2 2 IR2 n C~ ,
there exists a broken line Px ;x IR2 n C~ with endpoints x1 and x2 , which does not
cross itself. Here we use the fact that the open, connected set IR n C~ is also polygonaly
connected (in the sense described above, see [R, p150, Thm. 14.29]). We also may
assume that Px ;x \ [ x1 ; x2] = fx1 ; x2g. Hence J := Px ;x [ [ x1 ; x2] is a closed
polygonal Jordan curve. Moreover a and b lie in dierent components of IR2 n J , since
the line connecting a and b intersects the Jordan curve at exactly one point x. On the
other hand, a and b are elements of the connected set C , and therefore C must have a
nonempty intersection with the boundary of the bounded component of IR2 n J , which
is J . However we assumed that [ x1; x2] \ C = ; and Px ;x \ C = ;, in contradiction
to J \ C 6= ;.
Let O be one of the (open) bounded components of IR2 n C~ and let w 2 @O be a
point with maximal distance > 0 from a; b, the line through a; b. Moreover let L be
the line parallel to a; b, with distance =2 from a; b which lies between a; b an w. We
now consider the nonempty and open intersection of the open halfplane H , dened by
1. case
1
1
2
2
1
1
{ 15 {
2
2
@H = L and w 2 H, with O. We put O := H \ O and take any z 2 O . Then,
by Lemma 3, there exist points u; v 2 @O C [ [ a; b] such that z = u+2 v . It is easy to
see that u; v 62 [ a; b] , because u 2 [ a; b] a; b would imply that the distance between
v 2 @O and a; b is twice the distance between z and a; b, which is greater that , in
contradiction to the maximality of . Therefore we have u; v 2 C and we can write
2O C + C , i.e. (C + C ) 6= ;.
With Lemma 4, we can generalize a result by Kuczma [K, p.219 (x5 plane curves) Thm.
2].
Theorem 7. Let D IR2 be open and convex and let f : D ! IR be a Jensen convex
function, i.e.
f x +2 y f (x) +2 f (y) for all x; y 2 D:
Furthermore let C D be a connected compact set containing three noncollinear
points, such that f is bounded from above on C . Then f is continuous on D.
Proof. Let f (z) M for all z 2 C , then
f x +2 y f (x) +2 f (y) M +2 M = M
for all x; y 2 C , or, what is the same, f (u) M for all u 2 12 (C + C ). By Lemma 4,
we have ( 21 (C + C )) 6= ;, and therefore, by the Theorem of Bernstein-Doetsch, we get
continuity of f on D.
By the same method we get the following generalization of Theorem 6 in the case of
dimension 2.
Theorem 60 . Let f : IR2 ! Y fulll the equation
f (x + y) = g(f (x); f (y))
with Y a topological space and g : Y Y ! Y a continuous function. If C IR2 is a
compact connected set, containing three noncollinear points, and if f jC is continuous,
then f is continuous on IR2 .
Proof. Imitate the proof of Theorem 6 by using Lemma 4.
Conjecture: Let n 2 and let C IRn be a compact connected set which contains
n +1 anely independent points, then the n-fold sum C + + C has nonempty interior.
References
[D] Dugundji, J., Topology, Allyn and Bacon, Boston, 1966.
[G1] Ger, R., Note on convex functions bounded on regular hypersurfaces, Demonstratio Mathematica, vol. 4, (1973), 97{103.
[G2] Ger, R., Thin sets and convex functions, Bull. del l' Acad. Polonaise des sci. Serie
des science math., astr. et phys. vol. 21 No 5 (1973), 413{416.
{ 16 {
[KP] Krantz, S. G. and Parks, H. R., A Primer on Real Analytic Functions, Basler
Lehrbucher, vol. 4, Birkhauser, Basel, 1992.
[K] Kuzcma, M., An Introduction to the Theory of Functional Equations and Inequalities, Panstowe Wydawnictwo Naukowe, Uniwersitet Slaski, Warszawa { Krakow
{ Katowice, 1985.
[N] Narasimhan, R., Analysis on Real and Complex Manifolds, Adv. Studies in Pure
Math., Masson, Paris, 1968.
[R] Rinow, W., Lehrbuch der Topologie, Hochschulbucher fur Mathematik, vol. 79,
VEB Deutscher Verlag der Wissenschaften, Berlin, 1975.
[Z] Zorzitto, F., Homogeneous summands of exponentials, Publ. Math. Debrecen
45 / 1-2 (1994), 177{182.
Authors' address:
Wolfgang Ring, Peter Schopf, Jens Schwaiger
Institut fur Mathematik, Karl-Franzens-Universitat Graz
Heinrichstrae 36, A-8010 Graz, Austria
{ 17 {