aa r X i v : . [ m a t h . F A ] M a r FIBERS OF L ∞ ALGEBRA.
MAREK KOSIEK
Abstract.
It is shown that Gelfand transforms of elements f ∈ L ∞ ( µ ) are con-stant at almost every fiber Π − ( { x } ) of the spectrum of L ∞ ( µ ) in the followingsense: for each f ∈ L ∞ ( µ ) there is an open dense subset U = U ( f ) of this spec-trum having full measure and such that the Gelfand transform of f is constant onthe intersection Π − ( { x } ) ∩ U . The proof of the main result bases on topological and measure properties of thespectrum of L ∞ ( µ ), (see [1], [2] I.9). This result is related to certain techniques con-nected with studying abstract approach to A-measures problem and corona problem.In this note we consider C ( X ), the algebra of all complex-valued continuous func-tions on a compact space X . Moreover we assume( ∗ ) µ is a regular Borel probabilistic measure on X such that X is equal to theclosed support of µ The set L ∞ ( µ ) of equivalence classes [ f ] of [ µ ] essentially bounded measurable func-tions f on X is a commutative C*-algebra under standard operations.Let Y be the spectrum of L ∞ ( µ ). By Gelfand-Naimark theorem, L ∞ ( µ ) is iso-metrically isomorphic (by the Gelfand transform [ f ] → c [ f ]) to C ( Y ).Let y ∈ Y and define a functional Π y on C ( X ) as follows:(1) Π y ( f ) := c [ f ]( y ) for f ∈ C ( X ) . Since Π y ( f g ) = d [ f g ]( y ) = [ [ f ][ g ]( y ) = ( c [ f ] c [ g ])( y ) = c [ f ]( y ) c [ g ]( y ) = Π y ( f )Π y ( g ) , we conclude that Π y is a linear-multiplicative functional on C ( X ), so it can beidentified with some point in X . Using this identification we can write f (Π y ) =Π y ( f ) for f ∈ C ( X ). Consider the mapping Π : y → Π y and observe that f ◦ Π = c [ f ]for f ∈ C ( X ). Hence f ◦ Π is a continuous function on Y for each f ∈ C ( X ).Consequently, since Gelfand topologies on X and Y are equal to the restrictions ofthe weak-star topologies to X and Y respectively, we have the following Lemma 1.
Projection
Π : Y → X is continuous. Let us consider the sequence of mappings(2) C ( X ) ∋ f → [ f ] → c [ f ] ∈ C ( Y ) . Mathematics Subject Classification.
Primary: 46J10; Secondary: 46E30, 28A20.
Key words and phrases. function algebra, measure, L ∞ algebra, fiber. By the assumption ( ∗ ), the first mapping is an isometry into L ∞ ( µ ). The last oneis the Gelfand transform: L ∞ ( µ ) → C ( Y ) which is also an isometry. Hence, by (1)we have for f ∈ C ( X )(3) sup x ∈ X | f ( x ) | = k f k = k c [ f ] k = sup x ∈ Π( Y ) | f ( x ) | . By Lemma 1, the set Π( Y ) is compact, and hence closed in X which by (3) impliesthat Π( Y ) contains Shilov boundary of C ( X ). Consequently Π( Y ) must be equalto X . So we have Proposition 2. (1)
Up to the isometric equivalence given by (2), C ( X ) can be considered as aclosed subalgebra of L ∞ ( µ ) . (2) Each element x ∈ X as a linear-multiplicative functional on C ( X ) has alinear-multiplicative extension y : [ f ] → c [ f ]( y ) to the whole L ∞ ( µ ) . From now on we will not distinguish in writing Borel, [ µ ] essentially boundedfunctions on X from their equivalence classes in L ∞ ( µ ). By the above consideration,if f ∈ C ( X ) then b f is constant on each fiber Π − ( { x } ) for x ∈ X .Since we identify L ∞ ( µ ) with C ( Y ), Riesz Representation Theorem gives a regularpositive Borel measure ˜ µ on Y ”representing µ ” i.e. such that k ˜ µ k = k µ k and(4) Z f dµ = Z ˆ f d ˜ µ for f ∈ L ∞ ( µ ) . For any Borel E ⊂ X its characteristic function c χ E as an idempotent in C ( Y ) isof the form χ U E , thus assigning a closed-open set U E in Y to any measurable E ⊂ X .Applying (4) to χ E we get for any Borel subset E of X the equality(5) µ ( E ) = ˜ µ ( U E ) . Moreover (Lemma 9.1 and Corollary 9.2 of [2]) we have
Lemma 3.
The family { U E : E ⊂ Y, E measurable } form a basis for the topologyof Y . If U is an open non-empty subset of Y , then ˜ µ ( U ) > . Lemma 4. If E, F are Borel subsets of X , and E ⊂ F then c χ E c χ F and U E ⊂ U F .Proof. If E ⊂ F then χ E = χ E · χ F . Hence c χ E = c χ E · c χ F which means that c χ E c χ F .Since χ U E = c χ E and χ U F = c χ F , we have U E ⊂ U F . (cid:3) Lemma 5. If E ⊂ X is open then Π − ( E ) ⊂ U E and χ Π − ( E ) c χ E . If E ⊂ X isclosed then Π − ( E ) ⊃ U E and χ Π − ( E ) > c χ E .Proof. Let E be open in X and x ∈ E . Then there is a continuous function f : X → [0 ,
1] such that f ( x ) = 1 and f χ E . Hence ˆ f is equal 1 on Π − ( { x } ) and f = f · χ E , which implies ˆ f = ˆ f · c χ E = ˆ f · χ U E . Consequently d χ U E is equal 1 onΠ − ( { x } ) which means that Π − ( { x } ) ⊂ U E . Since x was an arbitrary point of E ,we have Π − ( E ) ⊂ U E . Then also χ Π − ( E ) χ U E = c χ E .If E is closed then X \ E is open and χ E · χ X \ E = 0, χ E + χ X \ E = 1. Consequently χ U E χ U X \ E = c χ E · [ χ X \ E = 0 and χ U E + χ U X \ E = c χ E + [ χ X \ E = 1. It means that IBERS OF L ∞ ALGEBRA. 3 U E ∩ U X \ E = ∅ and U E ∪ U X \ E = Y which implies the desired statement for closedsets. (cid:3) Remark 6.
Till now the regularity of µ has not been used. Lemma 7. If E is a Borel subset of X then (6) µ ( E ) = ˜ µ (Π − ( E )) = ˜ µ ( U E ) . If E ⊂ X is open then Π − ( E ) = U E . If E ⊂ X is closed then int (Π − ( E )) = U E .Proof. By the regularity of µ , for any ε > K ⊂ X and anopen set V ⊂ X such that K ⊂ E ⊂ V and µ ( V \ K ) < ε . Also there exists f ∈ C ( X )such that χ K f χ V . By the continuity of f we have χ Π − ( K ) ˆ f χ Π − ( V ) .(Proposition 2 and the consideration following it). Hence | µ ( E ) − R f dµ | < ε and | ˜ µ (Π − ( E )) − R ˆ f d ˜ µ | < ε which by (4) and free choice of ε gives µ ( E ) = ˜ µ (Π − ( E )).The second equality in (6) we get by (5).If E is closed then U E ⊂ Π − ( E ) by Lemma 5. So U E ⊂ int(Π − ( E )) andint(Π − ( E )) \ U E is open since U E is closed-open. Consequently int(Π − ( E )) = U E by Lemma 3.The assertion for open sets follows from the equalities int(Π − ( E )) = Y \ Π − ( X \ E )and U E = Y \ U X \ E . (cid:3) Theorem. If µ is a measure satisfying ( ∗ ), Y is the spectrum of L ∞ ( µ ) , and h ∈ L ∞ ( µ ) , then there exists an open dense subset U of Y with ˜ µ ( U ) = ˜ µ ( Y ) such that ˆ h is constant on Π − ( { x } ) ∩ U for all x ∈ X .Proof. Let h ∈ L ∞ ( µ ), and let ε >
0. By Lusin Theorem there is g ∈ C ( X ) with k g k k h k and a closed set Z ⊂ X such that µ ( X \ Z ) < ε while Z ⊂ { g = h } . ByLemma 7 we have U Z = int(Π − ( Z )) , ˜ µ ( U Z ) = µ ( Z ) > − ε. Since Z ⊂ { g = h } then χ Z · ( g − h ) = 0. Consequently χ U Z · (ˆ g − ˆ h ) = c χ Z · (ˆ g − ˆ h ) = 0which implies { ˆ g = ˆ h } ∩ U Z = ∅ . Put Z := Z and ε = 1 /
2. Repeating the previous construction we find a sequence { g n } ⊂ C ( X ) and a sequence { Z n } of closed subsets of X such that Z n ⊂ { g n = h } and µ ( X \ Z n ) < / n . Then˜ µ ( U Z n ) = µ ( Z n ) > − / n , { ˆ g n = ˆ h } ∩ U Z n = ∅ . The last equality implies that ˆ h is constant on each Π − ( { x } ) ∩ U Z n for all x ∈ X and n ∈ N . We define a sequence of open sets as follows: U := U Z , U n := U Z n \ Π − ( Z ∪ ... ∪ Z n − ) . By the above definition and Lemma 5, for k ∈ N we have Π − ( Z k ) ⊃ U Z k ⊃ U k ,hence Z k ⊃ Π( U Z k ) ⊃ Π( U k ), and consequently(7) Π( U n ) ∩ Π( U m ) = ∅ for n = m MAREK KOSIEK since Π( U n ) ∩ Z k = ∅ for k < n . By Lemma 7 we have ˜ µ (Π − ( Z n ) \ U Z n ) = 0 andhence(8) ˜ µ ( U n ) = ˜ µ ( U Z n \ Π − ( Z ∪ ... ∪ Z n − )) = ˜ µ (Π − ( Z n ) \ Π − ( Z ∪ ... ∪ Z n − ))= ˜ µ (Π − ( Z n \ ( Z ∪ ... ∪ Z n − )) = µ ( Z n \ ( Z ∪ ... ∪ Z n − )) . Put now Z ′ := Z and Z ′ n := Z n \ ( Z ∪ ... ∪ Z n − ) for n >
1. All the sets { Z ′ n } are pairwise disjoint and a direct calculation gives the equality Z ′ n ∪ Z ′ n − ⊃ Z n \ ( Z ∪ ... ∪ Z n − ) which by induction leads to the assertion Z ′ ∪ ... ∪ Z ′ n ⊃ Z n .Hence, by (8) and pairwise disjointness of { U n } and { Z ′ n } , we get˜ µ ( U ∪ ... ∪ U n ) = ˜ µ ( U ) + ... + ˜ µ ( U n ) = µ ( Z ′ ) + ... + µ ( Z ′ n )= µ ( Z ′ ∪ ... ∪ Z ′ n ) > µ ( Z n ) > − / n . Put U := P ∞ n =1 U n . Hence U is open, ˜ µ ( U ) = 1 = ˜ µ ( Y ), and consequently, byLemma 3, U is dense in Y . The function ˆ h is constant on each Π − ( { x } ) ∩ U n forall x ∈ X and n ∈ N and sets Π( U n ), n ∈ N are pairwise disjoint by (7). It meansthat each fiber Π − ( { x } ) intersects at most one of the sets U n . Hence ˆ h is constanton each Π − ( { x } ) ∩ U for all x ∈ X . (cid:3) Remark 8.
If the closed support of µ is not equal to X then L ∞ ( µ ) is isometricallyisomorphic to the algebra { f | supp( µ ) : f ∈ L ∞ ( µ ) } . In such a case Π − ( { x } ) = ∅ forall x outside of the closed support of µ . Assuming that each function is constant onempty set we conclude that the result of Theorem holds true also when the closedsupport of µ is a proper subset of X . References
1. J. Dixmier
Sur certain espaces consid´er´es par M. H. Stone , Summa Brasil. Math. 2 (1951),151-182.2. T. W. Gamelin
Uniform Algebras , Prentice Hall, Inc., Englewood Clifs, N.J. 1969.
Instytut Matematyki, Uniwersytet Jagiello´nski, Lojasiewicza 6, 30-348, Krak´ow,Poland
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