On some properties of the dierence. spectrum. D. L. Salinger and J. D. Stegeman

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1 On some properties of the dierence spectrum D. L. Salinger and J. D. Stegeman Abstract. We consider problems arising from trying to lift the non-synthesis spectrum of a closed set in a quotient of a locally compact, non-metrizable, abelian group. In particular, we get contrasting results for the dierence spectra of linear combinations of elements of a regular Banach function algebra, depending on whether the Gelfand space is metrizable or not Mathematics Subject Classication: 43A45, 46J10 1. Conventions and Notation Let A be a commative, semisimple, regular Banach algebra with Gelfand space X. We shall identify A with its Gelfand representation, that is, we shall think of A as a Banach algebra of continuous functions (vanishing at innity) on the locally compact space X. We shall assume that the functions in A with compact supports are dense in A. Our standard example is the Fourier algebra A = A(G) =FL 1 (;), where G is a locally compact abelian group and ; is its dual group. In practice we often take G compact (hence ; discrete). As our emphasis is on local phenomena, this is not really a loss of generality. We now introduce notation that will be used througho. Consider a topological space X and a point x 2 X. We use the symbols = x, 2 x, x, ::: to denote equality, belonging, inclusion, :::, locally at x. Thus, for functions f and g dened on X, wesaythatf and g are locally equal at x, and we write f = x g, if there is a neighbourhood U of x such thatfj U = gj U similarly, two subsets E and F of X are called locally equal at x (which we denote by E = x F )ife \ U = F \ U for some neighbourhood U of x. We observe that E = x X if (and only if) x is an interior point ofe, whilee = x ifx 62 E, and6= x E 6= x X if x Note that the set of points where two functions, or two subsets, are equal, is always open. Suppose next that is a collection of functions dened on X and that f is a D.L.Salinger was supported by the British Council { DAAD link between Leeds and Saarbrucken under ARC project 713. We acknowledge this support with thanks. 1

2 2 D.L.Salinger and J.D.Stegeman function dened on X. Wesay that f belongs to, locally at x, and we write f 2 x, if there is a function g 2 such that f = x g. The set (f ) : = fx 2 X j f 62 x g will be called the spectrum of f with respect to. The spectrum is always a closed set. In particular, the spectrum of f with respect to O :=f0g is the support of f (in the usual sense) thus suppf =(f O) =fx 2 X j f 6= x 0g : Further, if 1 and 2 are two collections of functions dened on X, then we say that 1 is contained in 2, locally at x, and we write 1 x 2, if for each f 2 1 we have f 2 x 2. The two collections are called locally equal at x if 1 x 2 and 2 x 1. If the space X is locally compact, then, in the same way, we dene symbols = 1, 2 1, 1, :::, to indicate validity oside some compact subset of X. For instance, f = 1 0 if and only if f has compact support. We shall have occasion to use a more general form of the above notation. For an arbitrary subset Y of X we shall use symbols = Y, 2 Y, Y, :::, to denote validity of the relation concerned in a neighbourhood of Y, and we shallsay that the relation holds locally at Y. It is evident that if a relation holds locally at some subset Y of X, then, a fortiori, the same relation holds at any subset of Y. In the opposite direction we have the following: if Y X and if f = x g for all x 2 Y,thenf = Y g, and similarly for local equality oftwo sets E and F. However, if f 2 Y and f 2 Z (f a function, a collection of functions), it does not follow that f 2 Y [Z, nor does 1 Y 2 and 1 Z 2 imply 1 Y [Z 2.For instance, suppose that X consists of two points. We can then identify the real-valued functions on X with R 2. Now take 1 = f(1 0) (0 1)g and 2 = f(0 0) (1 0) (0 1) (1 1)g. Then 1 = x 2 for all x 2 X, yet 1 = X 2 does not hold. Finally, we shall use 6= x, 62 x, 6 x, :::, to denote that the relation in question does not hold, locally at x, and similarly for a subset Y instead of a point x. 2. The dierence spectrum Now let A be a Banach algebra as described in Section 1. In [2] we gave the following denition: the dierence spectrum of two (not necessarily closed) ideals I 1 and I 2 in A is the set (I 1 I 2 )ofallx2x where I 1 and I 2 are locally dierent thus (I 1 I 2 )=fx 2 X j I 1 6= x I 2 g : If the locally compact space X is not compact, we also dene the extended dierence spectrum 1 (I 1 I 2 ) to be the set 1 (I 1 I 2 )=fx 2 X [ f1g j I 1 6= x I 2 g :

3 On some properties of the dierence spectrum 3 A partition-of-unity argument gives I 1 = I 2 if (and only if) 1 (I 1 I 2 )=([1, p.19]). When I 1 and I 2 are closed, we already have I 1 = I 2 if (I 1 I 2 ) = (since the functions in A with compact supports are dense, by assumption). In a completely analogous way the following result is proved. (Note that the assumption that I be an ideal is essential, in view of the remarks towards the end of Section 1.) Proposition 2.1. Let A and X be asbefore, and let Y be acompact subset of X. Let I be an ideal in A, and let f be a function, dened onx. If f 2 x I for all x 2 Y, then f 2 Y I. Let I be an ideal in A. Then the hull of I (also called the zero setor the cospectrum of I) is the set hull(i) =(I A)=fx 2 X j I 6= x Ag. Let E be a closed subset of X. The kernel of E is the ideal I E = ff 2 A j fj E =0g : It is the largest ideal with hull E. The smallest ideal with hull E is the ideal j E = ff 2 A j f = E[f1g 0g : The closure of j E is denoted by J E it is the smallest closed ideal with hull E. As is well-known, E is called a set of (spectral) synthesis if J E = I E otherwise E is called a set of non-synthesis. We denote the dierence spectrum (J E I E )by E, and we call this set the spectrum of non-synthesis of E. Note that and that E = if and only if E is a set of synthesis. We also use the following terminology: the spectrum and the cospectrum of an ideal I are dened as the dierence spectra of I and A and of I and O = f0g, respectively thus cosp(i) =(I A)=fx 2 X j I 6= x Ag spec(i) =(O I) =fx 2 X j O 6= x Ig In [1, p.125] the spectrum of an ideal is dened as the closure of the above set spec(i) our denition above seems more natural. For instance, the spectrum of j E is the open set X n E. In the special case that X is metrizable, the following holds (cf. [2, p. 532]): if I 1 is closed, then fx 2 X j I 1 x I 2 ) is open. As a consequence we have the following result. Proposition 2.2. Let A and X be as before. Suppose that X is metrizable. Then: (i) if I 1, I 2 are closed ideals of A, then (I 1 I 2 ) is closed (ii) the cospectrum (the hull) of any ideal is closed (iii) the spectrum of any closed ideal is closed. Proof. (I 1 I 2 ) is the complement of the intersection of two open sets. (I A) and (O I) are the complements of fx 2 X j A x Ig and fx 2 X j I x Og, respectively.

4 4 D.L.Salinger and J.D.Stegeman Now consider the case where X is non-metrizable. In [2] we have proved the following: Theorem 2.3. Let G be a non-metrizable compact abelian group (thus ; is a discrete uncountable group). (i) There is a set of non-synthesis E such that its spectrum of non-synthesis E is not closed. (ii) The set (O I) (the spectrum of I) is closed whenever I is a closed ideal. From Section 1 we recall that the spectrum (f I) of an individual function f 2 A with respect to an arbitrary ideal I of A is closed, even if X is non-metrizable. 3. Linear combinations of functions Let A and X be as before. Consider a compact non-synthesis subset E of X. For f 2 I E we shall denote the spectrum (f J E )off with respect to J E,by f, or by E f, if it is necessary to specify the set E. For two functions f g 2 I E it is obvious that f+g f [ g. It is also clear that equality need not hold in general: it may happen that part of the `bad behaviour' of f is neralized by that of g. If this happens, we can avoid it by adding to f, not g itself, b a suitable multiple of g. The multiple has to be chosen in such a way that no `neralization of bad behaviour' occurs at some other part of the spectrum. The following lemma shows that suitable multiples are always available when X is metrizable. Lemma 3.1. Suppose that the Gelfand space X of the Banach algebra A is metrizable. Let E be acompact set of non-synthesis in X. Consider two functions f g 2 I E nj E. Then there areatmostcountably many 2 C such that f+g 6= f [ g. Proof. Clearly f+g f [ g,any 2 C. Consider the sets U, 2 C, dened by U =( f [ g ) n f+g. Suppose that x 2 U \ U. Then f + g 2 x J E and f + g 2 x J E, hence (subtract!) ( ; )g 2 x J E and, similarly, ( ; )f 2 x J E. If 6=, this implies that x 62 f [ g, which gives a contradiction. We conclude that the sets U ( 2 C ) are disjoint. It now follows from the metrizability ofx that there can be at most countably many 's for which U 6=. Indeed, we can x a metric d giving the topology on X. The sets U are relativelyopenin f [ g, hence for a point x 2 U there exists > 0 such that, for y 2 f [ g,wehave d(x y) < ) y 2 U. If there were uncountably many non-empty U, there would be a >0such that for innitely many distinct values of, giving an innite set of points x with no accumulation point in the compact set f [ g, which is impossible. Our aim is to prove that in the case of a metrizable Gelfand space the result of the lemma can be extended to innite linear combinations. B rst we need another lemma.

5 On some properties of the dierence spectrum 5 Lemma 3.2. Let A and X be as before. Suppose that X is metrizable. Let E be acompact subset of X of non-synthesis with spectrum of non-synthesis E. Then there exists a sequence of functions ( k ) k1 in A such that for every x 2 E, every neighbourhood U of x and every n 1 there isak>nsuch that k = x 1 and k = y 0 for every y 62 U. Proof. Since E E is metrizable and compact, we may select a countable collection of open sets U n, n 2 N, inx, such that U n \ E is a basis of open sets in the relative topology on E,each U n is compact and for each x 2 U n, there is an m such that x 2 U m U m U n. It follows from the regularity ofa that, given any pair U m U n such that U m U n, there exists a function m n 2 A such that m n =1onU m and m n = 0 oside U n. The collection of such m n is countable and we relable it as k, k 2 N, in such away thateach m n occurs innitely often. The sequence k now has the required properties. Theorem 3.3. Suppose that the Gelfand space X of the Banach function algebra A is metrizable. Let E be a compact non-synthesis subset of X. Let functions f j 2 P I E n J E (j 1) be given. Then there exist j 2 C (j 1) such that 1 g := j=1 jf j exists and 1[ g = j=1 fj : Proof. As in the proof of Lemma 3.1, one inclusion is obvious. As the set g is closed, it therefore suces to show that for every n 1 one has fn g. We can further assume, witho loss of generality, that jjf j jj = 1 for all j 1. Take a sequence ( j ) j1 as in Lemma 3.2. We shall choose the numbers k P k (k 1) inductively, andwe shall denote the partial sums j=1 jf j by g k.take 1 = 1. Suppose that for some k 2 the numbers 1 ::: k;1 have been chosen. Take k 2 C nf0g such that k[ (i) gk = j=1 fj, (ii) j k jj k;1 j=2, (iii) if g k;1 k 62 J E,thenj k j < dist(g k;1 k J E ), 2jj k jj where, for f 2 A and A, we dene dist(f ) = inffjjf ; gjj j g 2 g. It follows from Lemma 3.1 that (i) can be satised for arbitrarily small values of. Therefore a choice as above isalways possible. Now takeany x 2 fn and suppose that x 62 g. From (i) it follows that x 2 gn.write U = X n g. By Lemma 3.2, there is a k>nsuch that k = x 1 and k = y 0 for every y 62 U. If y 2 U then y 62 g, hence g 2 y J E, hence also g k 2 y J E. If y 62 U then g k = y 0 2 J E. It follows that gk =, hence g k 2 J E.

6 6 D.L.Salinger and J.D.Stegeman On the other hand, fn gk;1 (by (i)), so g k;1 k = x g k;1 =2 x J E, and hence g k;1 k =2 J E.Wehave jjg k ; g k;1 k jj = 1X 1X j f j k j=k j=k j j jjj k jj 2 j k jjj k jj < dist(g k;1 k J E ) from (ii) and (iii) above. It follows that g k 62 J E, a contradiction. We now give an example showing that the metrizability of the space X is crucial. Proposition 3.4. There exists a compact non-metrizable abelian group G (i.e., the dual group ; is discrete and uncountable), a closed set E G of non-synthesis, and functions f g 2 I E nj E such that f+g 6= f [ g for all 2 C with jj1. Proof. Denote the set f 2 C jjj 1g by D. We need the following three ingredients: (i) A function ' 2 A(T 2 ) with range(') =D: for example '(x y) =e 2ix cos 2y. (ii) A set of non-synthesis F Tand a function 2 I F n J F wemay assume that F = F = F. (iii) A non-metrizable compact abelian group H. We take H = Y (x y)2t 2 f0 1g = (" x y ) j " x y 2f0 1g ((x y) 2 T 2 ) : (the corners of an uncountable-dimensional cube). The zero element is0=(0) x y. The elements p x y =(" x 0 y 0)with" x y =1," x 0 y0 = 0 otherwise, constite the set C 1 of all those corners of the cube which are adjacent to 0. Take G = T 2 TH, and consider the subset e F = f(0 0)gF f0g of G. By the injection theorem for sets of synthesis ([1, p.149]) e F is a subset of non-synthesis of G. For each (x y) 2 T 2 wenow consider the set E x y = f(x y)gf fp x y gg. All E x y are translates of e F, hence they are sets of non-synthesis in G as well. Let E be the closure of the union of these sets. Then it is not hard to see that E = [ (x y)2t 2 f(x y)gf fp x y g 1 A ; [ T 2 F f0g : Now dene f g 2 I E by f(x y z t) =;'(x y) (z) and g(x y z t) = (z), where (x y z) 2 T 3 and t 2 H. It is easily seen that g = E and that f g, thus f [ g = g here= E. Moreover, we have (f + g)(x y z t) = ( ; '(x y)) (z). For (x y) such that'(x y) = the function f + g vanishes on f(x y)gtfp x y g, a set of synthesis. Therefore, f+ag \ E x y = (when '(x y) =), and hence f+ag 6= f [ g.

7 On some properties of the dierence spectrum 7 4. Lifting the spectrum of non-synthesis Let G be a compact abelian group and let H be a closed subgroup. We write for the canonical mapping : G! G=H. Let E be a closed subset of G=H. In [1, p.145] it is shown that E is a set of non-synthesis for G=H if and only if ;1 E is a set of non-synthesis for A(G). Does this result extend to the non-synthesis spectrum? It is not dicult to adapt Reiter's method to show that ;1 E ;1 E [2]. B trying to prove the opposite inclusion meets with diculties, which we now describe. (It was an attempt to circumvent those diculties which led to the results of the previous section.) First we discuss a natural way to write any function f 2 A(G) as a sum of functions in A(G), each ofwhich is the product of a character by a function that is constant on cosets of H. Lemma 4.1. For every f 2 A(G) there is a countable (possibly nite) set R ; such that each coset of H? contains at most one element of R and such that for each 2 R there is a function f 2 A(G=H) such that f = X 2R f : X X Proof. For f 2 A(G), we have f(x) = bf()hx i with jb f()j < 1. Consider the cosets of H? in ;. At most countably many of them contain at least one 2; 2; element for which f() b 6= 0. Take a set R ; such that each of these cosets contains exactly one element ofr (such a set R is called a setofrepresentatives for these cosets). X Then X X f(x) = hx i f( b + )h(x) i : 2R 2R 2H? 2H? b f( + )hx + i = X Identifying H? with the X dual of G=H, and dening, for arbitrary 2 ;, functions f 2 A(G=H) by f = f, as desired. 2H? b f( + ), we obtain f = X Now take f 2 I ;1 E. Consider any y 2 ;1 E. Wehave then, for arbitrary z 2 H, y + z 2 ;1 E as well. Writing f as in Lemma 4.1 we obtain, for arbitrary z 2 H, 0=f(y + z) = X 2R (hy if ((y))) hz i : 2R This is the Fourier series for the function z 7! f(y + z), which in this case is the zero function. Therefore all its Fourier coecients are zero. We conclude that the functions f in the above proof satisfy f ((y))=0forally 2 ;1 E: thus f 2 I E.

8 8 D.L.Salinger and J.D.Stegeman For w 62 E we know that, for all 2 R, there exists a neighbourhood U of w such that f 2 U J E. In the case where G=H is metrizable, we can choose U independently of and conclude the proof. We do not know how to do this for the general non-metrizable case. On the other hand, if we knew it failed in a specic case, then we would have an example where ;1 E 6= ;1 E, as the following result shows. Proposition 4.2. Let G 0 beacompact abelian group and let E be a closed subset of G 0. Let w 2 E be such that there isasequence (f n ) n1 in I E such that for all neighbourhoods U of w there is ann such that f n 62 U J E. element t 2 Tsuch that (w t) 2 EṪ Then there isan Proof. Suppose there exists no t 2 Tsuch that (w t) 2 EṪWemay assume that kf n k A(G0) =2 ;n (n 1). Let f 2 I ETbe dened by f(x t) = 1X n=1 f n (x)e int for x 2 G 0 and t 2 T = [; [. Then, for each t 2 Tthere exist a function g t 2 J ETand open neighbourhoods U t of w in G 0 and V t of t in Tsuch that f = g t on U t V t. Nowchoose a nite subcover V t1 :::V tn of T: we denote the intersection of the corresponding U tj by U, and we write g k instead of g tk. Let W 1 ::: W n be an open cover of T such that W k W k V tk, for k = 1 ::: n. Let k, for k =1 ::: n, be functions in A(T) such that k (t) 0 for all t, k (t) > 0 when t 2 W k and supp k V tk. Dene e k 2 A(G 0 T) by e k (x t) = k (t). Set g =( nx k=1 e k g k )( nx k=1 e k ) ;1 : ( P e k > 0onG 0 T,so,by Wiener's theorem, ( P n k=1 e k ) ;1 2 A(G 0 T)). Then g 2 J ETand g = f on U T. To nish the proof, we must show that this leads to a contradiction. Let 2 A(G 0 )besuch that supp U and such that there exists an open neighbourhood U 0 of w with = 1 on U 0. We may also assume that 0 1. Dene e 2 A(G0 T) by e (x t) = (x). Then g e = f e 2 JEṪ This means that, given ">0, there exists an open neighbourhood N of E T with a function h 2 A(G 0 T) which is identically zero on N and such that kh ; f ~ k A <. We can write h(x t) = P h n (x t)e int, where P kh n k A(G0) = khk A(G0T). By looking at the Fourier transforms, we see that kh n ; f n k A(G0) < ", for n 1. Let M be a neighbourhood of E such that M T N. For all t 2 T, and each xed x 2 M, wehave P h n (x)e int = 0. Hence, by the uniqueness theorem for Fourier series, each h n (x) =0onM. Since ">0 is arbitrary, we conclude that f n 2 J E and thus f n 2 U 0J E, for every n 1, contrary to the hypotheses.

9 On some properties of the dierence spectrum 9 To get an example where E does not lift, it thus suces to nd a compact nonmetrizable abelian group G 0, with E and w as in the proposition b satisfying, in addition, the condition w =2 E. However, can these two requirements be met simultaneously? As our construction in [2] suggests, a set E G for which E is not closed may have to be extremely thin (in some sense) at every point w 2 E n E, which might be an obstruction for the construction of a sequence (f n ) n1 as in Proposition 4.2. So far we have not been able to settle this problem. References [1] H, Reiter, Classical Harmonic Analysis and Locally Compact Groups. Oxford University Press [2] D.L. Salinger, J.D. Stegeman, Dierence spectra of ideals for non-metrizable groups. J. London Math. Soc. (2), 26, (1982).