Construction of separately continuous functions with given restriction

0.12 MB·

by  V. V. Mykhaylyuk

CONSTRUCTION OF SEPARATELY CONTINUOUS FUNCTIONS
WITH GIVEN RESTRICTION
6
1
V.V.MYKHAYLYUK
0
2

n Abstract. It is solved the problem on constructed of separately continuous
a functionsonproductoftwotopologicalspaceswithgivenrestriction. Inpartic-
J ular,itisshownthatforeverytopologicalspaceXandfirstBaireclassfunction
9 g :X →R there exists separately continuous function f :X×X →R such
2 thatf(x,x)=g(x)foreveryx∈X.

]
N
G
.
h 1. Introduction
t
a
It ia well-known (see [1]) that the diagonals of separately continuous functions
m
oftworealvariablesareexactly the firstBairefunctions. It is shownin[2]that for
[ every topological space X with the normal square X2 and G -diagonal and every
δ
1 function g : X → R of the first Baire class there exists a separately continuous
v function f : X ×X → R, for which f(x,x) = g(x), i.e. every the first Baire class
4 function on the diagonal can be extended to a separately continuous function on
8
all product. Analogous question for functions of n variables was considered in [3].
9
On other hand, in the investigations of separately continuous functions f :
7
0 X × Y → R defined on the product of topological spaces X and Y the fol-
. lowing two topologies naturally arise (see [4]): the separately continuous topol-
1
ogy σ (the weakest topology with respect to which all functions f are continu-
0
6 ous) and the cross-topology γ (it consists of all sets G for which all x-sections
1 Gx ={y ∈Y :(x,y)∈G} and y-sections G ={x∈X :(x,y)∈G} are open in Y
y
v: and X respectively). Since the diagonal ∆ = {(x,x) : x ∈ R} is a closed discrete
i set in (R2,σ) or in (R2,γ) and not every function defined on ∆ can be extended
X to a separately continuous function on R2, even for X = Y = R the topologies σ
r
and γ are not normal (moreover, γ is not regular [4,5]). Besides, every separately
a
continuous function f : X ×Y → R is a Baire class function for a wide class of
the products X ×Y, in particular, if at least one of the multipliers is matrizable
[6]. Thus, the following question naturally arises: for which sets E ⊆ X ×Y and
σ-continuous(γ-continuous)functiong :E →Rofthe firstBaireclassthere exists
a separately continuous function f : X ×Y → R for which the restriction f|
E
coincides with g?
In this paper we generalize an approach proposed in [2] and solve the problem
formulated above for sets E of some type in the product of topological spaces.
2000 Mathematics Subject Classification. Primary54C08,54C30, 54C05.
Keywordsandphrases. separatelycontinuousfunctions,firstBaireclassfunction,diagonalof
mapping.
1
2 V.V.MYKHAYLYUK
2. Notions and auxiliary statements
A set A⊆X has the extension property in a topological space X,ifeverycontin-
uousfunctiong :A→[0,1]canbeextendedtoacontinuousfunctionf :X →[0,1].
According to Tietze-Uryson theorem [7, p.116], every closed set in a normal space
has the extension property.
Lemma2.1. Letsets X andY have theextensionproperty inatopological spaces
1 1
X and Y respectively, e:X →Y be a homeomorphism, E ={(x,e(x)):x∈X }
1 1 1
andg :E →[−1,1]beacontinuousfunction. Thenthereexistcontinuousfunctions
f :X×Y →[−1,1]andh:X×Y →[−1,1],whichsatisfiesthefollowingconditions:
(i) f| =g;
E
(ii) E ⊆h−1(0);
(iii) for every x′,x′′ ∈ X and y′,y′′ ∈ Y if x′ = x′′ or y′ = y′′ then |f(x′,y′)−
f(x′′,y′′)|=|h(x′,y′)−h(x′′,y′′)|.
Proof. Consider the continuous function ϕ : X → [−1,1] and ψ : Y → [−1,1],
1 1
which defined by: ϕ(x) = g(x,e(x)), ψ(y) = g(e−1(y),y). Since X and Y have
1 1
the extension property in X and Y respectively, there exist continuous functions
ϕ˜ : X → [−1,1] and ψ˜ : Y → [−1,1] such that ϕ˜| = ϕ and ψ˜| = ψ. Put
X1 Y1
f(x,y) = ϕ˜(x)+ψ˜(y) and h(x,y) = ϕ˜(x)−ψ˜(y). Clearly that f and h are continuous
2 2
on X ×Y and valued in [−1,1]. Moreover, for every point p=(x,y)∈E we have
ϕ˜(x) = ϕ(x) = g(p) = ψ(y) = ψ˜(y). Therefore f| = g and h| = 0, i.e. the
E E
conditions (i) and (ii) are hold.
Let x′,x′′ ∈X and y ∈Y. Then
ϕ˜(x′)−ϕ˜(x′′)
′ ′′ ′ ′′
f(x,y)−f(x ,y)= =h(x,y)−h(x ,y).
2
If x∈X and y′,y′′ ∈Y, then
ψ˜(y′)−ψ˜(y′′)
′ ′′ ′′ ′
f(x,y )−f(x,y )= =h(x,y )−h(x,y )/
2
Thus, the condition (iii) is holds and lemma is proved. (cid:3)
In the case if the set E satisfies a compactness-type condition we will use the
following proposition.
Lemma2.2. LetX beatopological space, E beapseudocompact set inX, (f )∞
n n=1
be a sequence of continuous functions f : X → R which pointwise converges on
n
the set E. Then there exists a functionally closed set F ⊆X such that E ⊆F and
the sequence (f )∞ pointwise converges on the set F.
n n=1
Proof. Consider the diagonal mapping
f =n∆∈Nfn :X →RN, f(x)=(fn(x))n∈N.
Since the set E is pseudocompact and f is continuous, the set f(E) is a pseudo-
compact set in the metrizable space RN. Therefore f(E) is closed and the set
F = f−1(f(E)) is functionally closed. It remains to verify that the sequence
(f )∞ pointwise converges on F. Let x ∈ F. Then there exists an x ∈ E such
n n=1 1
that f(x) = f(x ), i.e. f (x) = f (x ) for every n ∈ N. Since the sequence
1 n n 1
(f (x ))∞ is convergent,the sequence (f (x))∞ is convergenttoo. (cid:3)
n 1 n=1 n n=1
SEPARATELY CONTINUOUS FUNCTIONS WITH GIVEN RESTRICTION 3
The following proposition we will use in a final stage of the construction of
separately continuous functions with the given restriction.
Lemma 2.3. Let X be a topological space, F be a functionally closed set in X,
(h )∞ be a sequence of continuous functions h : X → R such that F ⊆ h−1(0)
n n=1 n n
for every n ∈ N G = X \F. Then there exists a locally finite partition of the
unit (ϕ )∞ on G such that the supports G = suppϕ ={x ∈ G : ϕ (x) > 0} of
n n=0 n n n
functions ϕ satisfy the conditions:
n
(a) G ∩F =Ø for every n=0,1,2,...;
n
(b) G ⊆h−1 (−1, 1) for every n=1,2,....
n n n n
Proof. Let h : X →(cid:0)[0,1] be(cid:1)a continuous function such that F = h−1(0). For
0 0
every n∈N we put
n n
1 1 1 1
A = h−1 (− , ) , B = h−1 [− , ] ,
n k n n n k n n
k=0 (cid:18) (cid:19) k=0 (cid:18) (cid:19)
\ \
G = A \ B and, moreover, G = G \ B . Clearly that all sets G are
n n n+2 0 2 n
functionally open and G ⊆ h−1 (−1, 1) for every n ∈ N, i.e. the condition (b)
n n n n
holds. Note that
(cid:0) (cid:1)
∞ ∞ ∞
A = B = h−1(0)=F.
n n n
n=1 n=1 n=0
\ \ \
Since A ⊆B ⊆A for every n∈N,
n+1 n+1 n
A \A ⊆A \B ⊆A \A =(A \A ) (A \A ).
n n+1 n n+2 n n+2 n n+1 n+1 n+2
[
Therefore
∞ ∞ ∞ ∞
G = (A \B )= (A \A )=A \( A )=A \F.
n n n+2 n n+1 1 n 1
n=1 n=1 n=1 n=1
[ [ [ \
∞
Thus, G =(G\B )∪(A \F)=G.
n 2 1
n=0
WesShowthatthefamily(Gn :n=0,1,...)islocallyfiniteonG. Letx∈G,i.e.
h (x) 6= 0. We choose n ∈ N such that 1 < |h (x)|. Then x 6∈ B and the set
0 0 n0 0 n0
G\B is a neighborhoodof x. On other hand, G ⊆A ⊆B for every n≥n .
n0 n n n0 0
ThereforeG ∩(G\B )=Øforeveryn≥n . Thus,thefamily(G :n=0,1,...)
n n0 0 n
is locally finite at the point x.
Since the sets G are functionally open, there exist continuous functions ψ :
n n
X → [0,1] such that G = ψ−1((0,1]). The function ψ : G → [0,+∞) which
n n
∞
definedby ψ(x)= ψ (x) is continuous,moreover,suppψ =G. Foreveryx∈G
n
n=0
X
and n = 0,1,... we put ϕ (x) = ψn(x). The functions ϕ are continuous and
n ψ(x) n
formed a locally finite partition of the unit on G, moreover G =suppϕ .
n n
Itremainstoverifythecondition(a). SinceG ⊆X\B ⊆X\A andthe
n n+2 n+2
setX\A is closed, G ⊆X\A , i.e. G ∩A =Ø. Moreover,F ⊆A ,
n+2 n n+2 n n+2 n+2
therefore G ∩F =Ø for every n=0,1,.... (cid:3)
n
4 V.V.MYKHAYLYUK
3. Main results
Theorem3.1. LetsetsX andY havetheextensionpropertyin topological spaces
1 1
X and Y respectively, e:X →Y be a homeonorphism, E ={(x,e(x)):x∈X },
1 1 1
g :E →R be the first Baire class function and at least one of the following condi-
tions: E is pseudocompact, E is functionally closed in X ×Y, X is functionally
1
closed in X, Y is functionally closed in Y holds. Then there exists a separately
1
continuous function f :X ×Y →R such that f| =g.
E
Proof. Wetakeasequenceofcontinuousfunctionsg :E →[−n,n]whichpointwise
n
convergestothefunctionganduseLemma2.1. Weobtainasequenceofcontinuous
functionsf :X×Y →[−n,n]andh :X×Y →[−n,n]whichsatisfythefollowing
n n
conditions (i)-(iii).
We showthatthe setE is containedinsomefunctionallyclosedsetF onwhich
1
thesequence(f )∞ pointwiseconverges. IfE isfunctionallyclosed,thenF =E.
n n=1 1
It follows from Lemma 2.2 the existence of such set F for pseudocompact set E.
1
Itremains to verify this inthe case whenX orY is functionally closedinX or Y
1 1
respectively. Let X is functionally closed in X. Now we put
1
∞
F =(X ×Y)∩ h−1(0) .
1 1 n
!
n=1
\
It follows from the property (ii) that E is contained in a functionally closed set
F . We take a point (x,y) fromthe set F . Using the condition (iii)of Lemma 2.1
1 1
we obtain |f (x,y)−f (x,e(x))| =|h (x,y)−h (x,e(x))| =0. Hence, f (x,y)=
n n n n n
f (x,e(x)). Since the sequence (f )∞ pointwise converges on E, the sequence
n n n=1
∞
(f (x,y)) converges,because (x,e(x))∈E.
n n=1
Now we use Lemma 2.3 to the functionally closed set
∞
F =F ∩ h−1(0)
1 n
!
n=1
\
in the space X ×Y and to the sequence of continuous functions h and obtain a
n
locally finite partition of the unit (ϕ )∞ on G=(X×Y)\F, which satisfies the
n n=0
conditions (a) and (b).
Let f ≡0 on X ×Y. We consider the function
0
∞
ϕ (x,y)f (x,y), if (x,y)∈G,
n n
f(x,y)=
 n=0
X
 lim fn(x,y), if (x,y)∈F.
n→∞
Since (ϕn)∞n=0 is a locallyfinite partition of the unit on the set G and all functions
f are continuous,the function f is correctly defined and continuous on the set G.
n
Note that F ⊆ F . Therefore the sequence (f )∞ pointwise converges on F and
1 n n=0
the function f correctlydefined on F. Moreover,since E ⊆h−1(0) for everyn and
n
E ⊆F , E ⊆F and f| = lim f | = lim g =g.
1 E n E n
n→∞ n→∞
It remains to verify that the function f is separately continuousat points of the
set F. Let p = (x ,y ) ∈ F ε > 0. We choose n ∈ N such that 1 < ε and
0 0 0 0 n0 2
|f (p )−f(p )|< ε foreveryn≥n . Itfollowsfromthecondition(a)thattheset
n 0 0 2 0
n0
W =X ×Y \ G ,
n
n=0
(cid:0)[ (cid:1)
SEPARATELY CONTINUOUS FUNCTIONS WITH GIVEN RESTRICTION 5
where G = suppϕ , is an open neighborhood of p in X ×Y. We take a neigh-
n n 0
borhood U of x in X such that U ×{y } ⊆ W. Let x ∈ U. If p = (x,y ) ∈ F,
0 0 0
then h (p) = 0 for every n ∈ N. Then according to the condition (iii), we obtain
n
|f (p )−f (p)|=|h (p )−h (p)|=0,i.e. f (p )=f (p). Thereforef(p )=f(p).
n 0 n n 0 n n 0 n 0
If p6∈F, then
∞ ∞
f(p)= ϕ (p)f (p)= ϕ (p)f (p),
n n n n
nX=0 nX=n0
because p∈W. Then
∞ ∞
|f(p )−f(p)|= ϕ (p) f(p )−f (p ) + ϕ (p)f (p )−
0 n 0 n 0 n n 0
(cid:12)nX=n0 (cid:0) (cid:1) nX=n0
(cid:12)
∞ ∞
− ϕ (p)f (p) ≤ ϕ (p)|f(p )−f (p )|+
n n n 0 n 0
nX=n0 (cid:12) nX=n0
∞ (cid:12) ∞
ε
+ ϕ (p)|f (p )−f (p)|< ϕ (p)· +
n n 0 n n
2
nX=n0 nX=n0
∞ ∞
ε
+ ϕ (p)|h (p )−h (p)|= + ϕ (p)|h (p)|.
n n 0 n n n
2
nX=n0 nX=n0
It follows from the property (b) of sets G that if ϕ (p) 6= 0, then |h (p)| < 1.
n n n n
Thus,
∞ ∞ ∞
1 1 1 ε
ϕ (p)|h (p)|≤ ϕ (p)· ≤ ϕ (p)= < .
n n n n
n n n 2
0 0
nX=n0 nX=n0 nX=n0
Hence, |f(p )−f(p)|<ε. Thus, f is continuous at p with respect to x.
0 0
The continuity of f at p with respect to y can be proved analogously. Thus, f
0
is separately continuous and the theorem is proved. (cid:3)
In the case of X = Y = X = Y we obtain the following theorem which
1 1
generalized the result from [2].
Theorem 3.2. Let X be a topological space and g : X → R be a function of the
firstBaire class. Then thereexistsa separately continuousfunction f :X×X →R
such that f(x,x)=g(x) for every x∈X.
4. Functions on the product of compacts
Now we consider the case when X and Y satisfy compactness type conditions.
A set E in a product X ×Y is called horizontally and vertically onepointed if for
every x ∈ X and y ∈ Y the sets E ∩({x}×Y) and E ∩(X ×{y}) are at most
countableandhorizontally and vertically n-pointedifcorrespondingsetscontainat
most n elements.
Theorem 4.1. Let X and Y be compacts, E be a closed horizontally and vertically
onepointed set in X×Y and g :E →R be a function of the first Baire class. Then
there exists a separately continuous function f :X ×Y →R, for which f| =g.
E
6 V.V.MYKHAYLYUK
Proof. SincethesetE ishorizontallyandverticallyonepointed,theprojectionsthe
compactsetE totheaxisX andY arecontinuousinjectivemappings. Hence, E is
the graph of a homeomorphism e:X →Y , where X and Y are the projections
1 1 1 1
ofE onX andY respectively. Nowtheexistenceofdesiredfunctionf followsfrom
Theorem 3.1. (cid:3)
Theorem 4.2. Let X and Y be a locally compact spaces such that X ×Y be a
paracompact,E beaclosedhorizontallyandverticallyonepointedsetandg :E →R
be the first Baire class function. Then there exists a separately continuous function
f :X ×Y →R for which f| =g.
E
Proof. For every p = (x,y) ∈ X ×Y we choose open neighborhoods U and V
p p
of x and y in X and Y respectively such that the closure X = U and Y =
p p p
V are compacts and the set E = E ∩(X ×Y ) is horizontally and vertically
p p p p
onepointed. AccordingtoTheorem4.2thereexistsaseparatelycontinuousfunction
f :X ×Y →Rforwhichf | =g| . Since the spaceX×Y isaparacompact,
p p p p Ep Ep
there exists a partition of the unit (ϕ : i∈I) on X ×Y which is subordinated to
i
the open cover(W =U ×V :p∈X×Y) ofX×Y [7, p.447]. Foreveryi∈I we
p p p
choose p ∈X ×Y such that suppϕ⊆W and put
i i pi
f (x,y), if (x,y)∈W ,
g (x,y)= pi pi
i 0, if (x,y)6∈W .
(cid:26) pi
Notethatthefunctionsϕ ·g areseparatelycontinuousonX×Y and(ϕ g )| =
i i i i E
(ϕ | )g. Then the function f = ϕ g is the required. (cid:3)
i E i i
i∈I
X
5. Example
Finally we give a example which show the essentiality of conditions under the
set E in Theorem 4.1.
Let X =Y =[0,1],
2k−1 2k−1 1
E ={( , + ):k =1,...,2n−1,n∈N},
1 2n 2n 2n+1
1, x∈E ,
E = {(x,x) : x ∈ X}, E = E ∪E , g : E → R, g(x) = 1 Clearly
2 1 2 0, x∈E .
2
(cid:26)
that E is a closed horizontally and vertically 2-pointed set in X ×Y and g is a
function of the first Baire class. Since the set E is dense in E, the set D(g) of
1
discontinuitypointssetofthefunctiongcoincideswithE. Thereforetheprojections
of D(g) on the axis X and Y coincide with X and Y respectively. On other hand,
it is well-known that for every separately continuous function f : X ×Y → R the
setD(f)ofpointsofdiscontinuityoff iscontainedintheproductA×B ofmeagre
sets A ⊆ X and B ⊆ Y respectively. Thus, D(g) 6⊆ D(f) and the function f can
not be extension of the function g.
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SEPARATELY CONTINUOUS FUNCTIONS WITH GIVEN RESTRICTION 7
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Departmentof Mathematics,Chernivtsi National University, str. Kotsjubyn’skogo
2,Chernivtsi,58012Ukraine
E-mail address: [email protected]