4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad which has the property that every fiite subcollectio is liearly idepedet. Ay elemet x X ca therefore be writte as some fiite liear combiatio of x γ. However, eve a separable ifiite-dimesioal Baach space would require a ucoutable Hamel basis. Moreover, the proof of the existece of Hamel bases for arbitrary ifiite-dimesioal spaces requires the Axiom of Choice (i fact, i ca be show that the statemet Every vector space has a Hamel basis is equivalet to the Axiom of Choice). Hece for most Baach spaces there is o costructive method of producig a Hamel basis. Example 4.1. [Gol66, p. 101]. We will use the existece of Hamel bases to show that if X is a ifiite-dimesioal Baach space, the there exist liear fuctioals o X which are ot cotiuous. Let {x γ } γ Γ be a Hamel basis for a ifiite-dimesioal Baach space X, ormalized so that x γ = 1 for every γ Γ. Let Γ 0 = {γ 1, γ 2,...} be ay coutable subsequece of Γ. Defie µ: X C by settig µ(γ ) = for ad µ(γ) = 0 for γ Γ\Γ 0, ad the extedig µ liearly to X. The this µ is a liear fuctioal o X, but it is ot bouded. More useful tha a Hamel basis is a coutable sequece {x } such that every elemet x X ca be writte as some uique ifiite liear combiatio x = c x. This leads to the followig defiitio. Defiitio 4.2. (a) A sequece {x } i a Baach space X is a basis for X if x X, uique scalars a (x) such that x = a (x) x. (4.1) (b) A basis {x } is a ucoditioal basis if the series i (4.1) coverges ucoditioally for each x X. (c) A basis {x } is a absolutely coverget basis if the series i (4.1) coverges absolutely for each x X. (d) A basis {x } is a bouded basis if {x } is orm-bouded both above ad below, i.e., if 0 < if x sup x <. (e) A basis {x } is a ormalized basis if {x } is ormalized, i.e, if x = 1 for every. Absolutely coverget bases are studied i detail i Chapter 5. Ucoditioal bases are studied i detail i Chapter 9. ote that if {x } is a basis, the the fact that each x X ca be writte uiquely as x = a (x) x implies that x 0 for every. As a cosequece, {x / x } is a ormalized basis for X.
40 4. BASES I BAACH SPACES If X possesses a basis {x } the X must be separable, sice the set of all fiite liear combiatios c x with ratioal c (or ratioal real ad imagiary parts if the c are complex) forms a coutable, dese subset of X. The questio of whether every separable Baach space possesses a basis was a logstadig problem kow as the Basis Problem. It was show by Eflo [Ef73] that there do exist separable, reflexive Baach spaces which do ot possess ay bases. otatio 4.3. ote that the coefficiets a (x) defied i (4.1) are liear fuctios of x. Moreover, they are uiquely determied by the basis, i.e., the basis {x } determies a uique collectio of liear fuctioals a : X F. We therefore call {a } the associated sequece of coefficiet fuctioals. Sice these fuctioals are uiquely determied, we ofte do ot declare them explicitly. Whe we do eed to refer explicitly to both the basis ad the associated coefficiet fuctioals, we will write ({x }, {a }) is a basis to mea that {x } is a basis with associated coefficiet fuctioals {a }. We show i Theorem 4.11 that the coefficiet fuctioals for ay basis must be cotiuous, i.e., {a } X. Further, ote that sice x m = a (x) x ad x m = δ m x are two expasios of x m, we must have a m (x ) = δ m for every m ad. We therefore say that the sequeces {x } X ad {a } X are biorthogoal, ad we ofte say that {a } is the biorthogoal system associated with {x }. Geeral biorthogoal systems are cosidered i more detail i Chapter 7. I particular, we show there that the fact that {x } is a basis implies that {a } is the uique sequece i X that is biorthogoal to {x }. Example 4.4. Fix 1 p <, ad cosider the space X = l p defied i Example 1.6. Defie sequeces e = (δ m ) m=1 = (0,..., 0, 1, 0,... ), where the 1 is i the th positio. The {e } is a basis for l p, ofte called the stadard basis for l p. ote that {e } is its ow sequece of coefficiet fuctioals. O the other had, {e } is ot a basis for l, ad ideed l has o bases whatsoever sice it is ot separable. Usig the l orm, the sequece {e } is a basis for the space c 0 defied i Example 1.6(c). We are primarily iterested i bases for which the coefficiet fuctioals {a } are cotiuous. We therefore give such bases a special ame. Defiitio 4.5. A basis ({x }, {a }) is a Schauder basis if each coefficiet fuctioal a is cotiuous. I this case, each a is a elemet of the dual space, i.e., a X for every. We shall see i Theorem 4.11 that every basis is a Schauder basis, i.e., the coefficiet fuctioals a are always cotiuous. First, however, we require some defiitios ad miscellaeous facts. I particular, the followig operators play a key role i aalyzig bases. otatio 4.6. The partial sum operators, or the atural projectios, associated with the basis ({x }, {a }) are the mappigs S : X X defied by S x = a (x) x.
4. BASES I BAACH SPACES 41 The partial sum operators are clearly liear. We will show i Corollary 4.8 that if {x } is a basis the each partial sum operator S is a bouded mappig of X ito itself. The the fact that all bases are Schauder bases will follow from the cotiuity of the partial sum operators (Theorem 4.11). The ext propositio will be a key tool i this aalysis. It states that if {x } is a basis, the it is possible to edow the space Y of all sequeces (c ) such that c x coverges with a orm so that it becomes a Baach space isomorphic to X. I geeral, however, it is difficult or impossible to explicitly describe the space Y. Oe exceptio was discussed i Example 2.5: if {e } is a orthoormal basis for a Hilbert space H, the c e coverges if ad oly if (c ) l 2. Recall that a topological isomorphism betwee Baach spaces X ad Y is a liear bijectio S: X Y that is cotiuous. By the Iverse Mappig Theorem (Theorem 1.44), every topological isomorphism has a cotiuous iverse S 1 : Y X. Propositio 4.7. [Si70, p. 18]. Let {x } be a sequece i a Baach space X, ad assume that x 0 for every. Defie Y = { (c ) : c x coverges i X }, ad set The the followig statemets hold. (a) Y is a Baach space. (c ) Y = sup c x. (b) If {x } is a basis for X the Y is topologically isomorphic to X via the mappig (c ) c x. Proof. (a) It is clear that Y is a liear space. If (c ) Y the c x = lim c x coverges. Sice coverget sequeces are bouded, we therefore have (c ) Y < for each (c ) Y. Thus Y is well-defied. It is easy to see that (c ) + (d ) Y (c ) Y + (d ) Y ad a (c ) Y = a (c ) Y for every scalar a, so Y is at least a semiorm o Y. Suppose that (c ) Y = 0. The c x = 0 for every. I particular, c 1 x 1 = 0, so we must have c 1 = 0 sice we have assumed x 1 0, But the c 2 x 2 = 2 c x = 0, so c 2 = 0, etc. Hece Y is a orm o Y. It remais oly to show that Y is complete i this orm. Let A = (c ) be ay collectio of sequeces from Y which form a Cauchy sequece with respect to the orm Y. The for fixed, we have c M c x = (c M c ) x (c M k c k ) x 1 k + (c M k c k ) x k 2 A M A Y. k=1 Sice {A } is Cauchy ad x 0, we coclude that (c ) =1 is a Cauchy sequece of scalars, so must coverge to some scalar c as. Choose ow ay ε > 0. The sice {A } is Cauchy i Y, there exists a iteger 0 > 0 such that L M, 0, A M A Y = sup (c M c ) x < ε. (4.2) L k=1
42 4. BASES I BAACH SPACES Fix 0 ad ay L > 0, ad set y M = L (cm c ) x. The y M < ε for each M 0 by (4.2). However, y M y = L (c c ) x, so we must have y ε. Thus, we have show that L 0, sup (c c ) x ε. (4.3) L Further, (c 0 ) Y, so c 0 x coverges by defiitio. Hece, there is a M 0 > 0 such that > M M 0, c 0 x < ε. Therefore, if > M M 0, 0 the c x = (c c 0 ) x =M+1 (c c 0 ε + ε + ε = 3ε. =M+1 M ) x M + (c c 0 ) x + (c c 0 =M+1 ) x + c 0 =M+1 Therefore c x coverges i X, so A = (c ) Y. Fially, by (4.3), we kow that A A i the orm of Y, so Y is complete. (b) Suppose ow that {x } is a basis for X. Defie the map T : Y X by T (c ) = c x. This mappig is well-defied by the defiitio of Y. It is clearly liear, ad it is bijective because {x } is a basis. Fially, if (c ) Y the T (c ) = c x = lim c x sup c x = (c ) Y. Therefore T is bouded, hece is a topological isomorphism of Y oto X. A immediate cosequece of Propositio 4.7 is that the partial sum operators S are bouded. Corollary 4.8. Let ({x }, {a }) be a basis for a Baach space X. The: (a) sup S x < for each x X, (b) C = sup S <, ad (c) x = sup S x forms a orm o X equivalet to the iitial orm for X, ad satisfies C. Proof. (a) Let Y be as i Propositio 4.7. The T : X Y defied by T (c ) = c x is a topological isomorphism of X oto Y. Suppose that x X. The we have by defiitio that x = a (x) x ad that the scalars a (x) are uique, so we must have T 1 x = (a (x)). Hece sup S x = sup a (x) x = (a (x)) Y = T 1 x Y T 1 x <. (4.4) (b) From (4.4), we see that sup S T 1 <. x c 0 x
4. BASES I BAACH SPACES 43 (c) It is easy to see that has the properties of at least a semiorm. ow, give x X we have ad x = sup x = S x sup S x = C x lim S x sup S x = x. It follows from these two statemets that is i fact a orm, ad is equivalet to. The umber C appearig i Corollary 4.8 is importat eough to be digified with a ame of its ow. Defiitio 4.9. If ({x }, {a }) is a basis for a Baach space X, the its basis costat is the fiite umber C = sup S. The basis costat satisfies C 1. If the basis costat is C = 1, the the basis is said to be mootoe. The basis costat does deped o the orm. Uless otherwise specified, the basis costat is always take with respect to the origial orm o X. Chagig to a equivalet orm for X will ot chage the fact that {x } is a basis, but it ca chage the basis costat for {x }. For example, we show ow that the basis costat i the orm is always 1. Propositio 4.10. Every basis is mootoe with respect to the equivalet orm defied i Corollary 4.8(c). Proof. ote first that the compositio of the partial sum operators S M ad S satisfies the rule S M S = { SM, if M, S, if M. Therefore, Hece, S x = sup M sup It follows from this that sup S = 1. S M S x = sup { S 1 x,..., S x }. S x = sup S x = x. ow we ca prove our mai result: the coefficiet fuctioals for every basis are cotiuous! Theorem 4.11. [Si70, p. 20]. Every basis ({x }, {a }) for a Baach space X is a Schauder basis for X. I fact, the coefficiets fuctioals a are cotiuous liear fuctioals o X which satisfy where C is the basis costat for ({x }, {a }). 1 a x 2C, (4.5)
44 4. BASES I BAACH SPACES Proof. Sice each a is a liear fuctioal o X, we eed oly show that each a is bouded ad that (4.5) is satisfied. Give x X, we compute a (x) x = a (x) x = 1 a k (x) x k a k (x) x k 1 a k (x) x k + a k (x) x k = S x + S 1 x 2C x. Sice each x is ozero, we coclude that a 2C/ x <. The fial iequality follows from computig 1 = a (x ) a x. Sice the coefficiet fuctioals a are therefore elemets of X, we use the otatios a (x) = x, a iterchageably. I fact, from this poit oward our preferred otatio is x, a. We ed this chapter with several useful results cocerig the ivariace of bases uder topological isomorphisms. Lemma 4.12. [You80, p. 30]. Bases are preserved by topological isomorphisms. That is, if {x } is a basis for a Baach space X ad S: X Y is a topological isomorphism, the {Sx } is a basis for Y. Proof. If y is ay elemet of Y the S 1 y X, so there are uique scalars (c ) such that S 1 y = c x. Sice S is cotiuous, this implies y = S(S 1 y) = c Sx. Suppose that y = b Sx was aother represetatio of y. The the fact that S 1 is also cotiuous implies that S 1 y = b x, ad hece that b = c for each. Thus {Sx } is a basis for Y. This lemma motivates the followig defiitio. Defiitio 4.13. Let X ad Y be Baach spaces. A basis {x } for X is equivalet to a basis {y } for Y if there exists a topological isomorphism S: X Y such that Sx = y for all. If X = Y the we write {x } {y } to mea that {x } ad {y } are equivalet bases for X. It is clear that is a equivalece relatio o the set of all bases of a Baach space X. ote that we could defie, more geerally, that a basis {x } for X is equivalet to a sequece {y } i Y if there exists a topological isomorphism S: X Y such that Sx = y. However, by Lemma 4.12, it follows immediately that such a sequece must be a basis for Y. Pelczyski ad Siger showed i 1964 that there exist ucoutably may oequivalet ormal-
ized coditioal bases i every ifiite dimesioal Baach space which has a basis. We show below i Corollary 4.15 that all orthoormal bases i a Hilbert space are equivalet. More geerally, we show i Chapter 11 that all bouded ucoditioal bases i a Hilbert space are equivalet (ad hece must be equivalet to orthoormal bases). Lidestrauss ad Pelczyski showed i 1968 that a o-hilbert space H i which all bouded ucoditioal bases are equivalet must be isomorphic either to the sequece space c 0 or to the sequece space l 1. We ca ow give a characterizatio of equivalet bases. Theorem 4.14. [You80, p. 30]. Let X ad Y be Baach spaces. Let {x } be a basis for X ad let {y } be a basis for Y. The the followig two statemets are equivalet. (a) {x } is equivalet to {y }. (b) c x coverges i X if ad oly if c y coverges i Y. Proof. (a) (b). Suppose that {x } is equivalet to {y }. The there is a topological isomorphism S: X Y such that Sx = y for every. Sice S is cotiuous, the covergece of c x i X therefore implies the covergece of c Sx i Y. Similarly, S 1 is cotiuous, so the covergece of c y i Y implies the covergece of c S 1 y i X. Therefore (b) holds. (b) (a). Suppose that (b) holds. Let {a } X be the coefficiet fuctioals for the basis {x }, ad let {b } Y be the coefficiet fuctioals for the basis {y }. Suppose that x X is give. The x = x, a x coverges i X, so Sx = x, a y coverges i Y. Clearly S defied i this way is liear. The fact that the expasio x = x, a x is uique esures that S is well-defied. Further, if Sx = 0 the 0 y = 0 = Sx = x, a y, ad therefore x, a = 0 for every sice {y } is a basis. This implies x = x, a x = 0, so we coclude that S is ijective. ext, if y is ay elemet of y, the y = y, b y coverges i Y, so x = y, b x coverges i X. Sice x = x, a x ad {x } is a basis, this forces y, b = x, a for every. Hece Sx = y ad therefore S is surjective. Thus S is a bijectio of X oto Y. It remais oly to show that S is cotiuous. For each, defie T : X Y by T x = x, a y. Sice each fuctioal a is cotiuous, we coclude that each T is cotiuous. I fact, T x = x, a y x, a y x a y. Sice T x Sx, we coclude that Sx sup T x < for each idividual x X. By the Uiform Boudedess Priciple (Theorem 1.42), it follows that sup T <. However, S sup T, so S is a bouded mappig. Corollary 4.15. All orthoormal bases i a Hilbert space are equivalet. Proof. Suppose that {e } ad {f } are both orthoormal bases for a Hilbert space H. The, by Theorem 1.19(a), c e coverges c 2 < c f coverges. Hece {e } {f } by Theorem 4.14. 45
46 5. ABSOLUTELY COVERGET BASES I BAACH SPACES It is ofte desirable to have a basis {x } such that the series x = x, a x has some special covergece properties. I this sectio we study those bases which have the property that this series is always absolutely coverget. We will see that this is a highly restrictive coditio, which implies that X is isomorphic to l 1. I Chapter 9 we will study those bases for which the series x = x, a x is always ucoditioally coverget. Defiitio 5.1. A basis ({x }, {a }) for a Baach space X is absolutely coverget if the series x = x, a x coverges absolutely i X for each x X. That is, we require that x X, x, a x <. Theorem 5.2. [Mar69, p. 42]. If a Baach space X possesses a absolutely coverget basis the X is topologically isomorphic to l 1. Proof. Suppose that ({x }, {a }) is a absolutely coverget basis for X. Defie the mappig T : X l 1 by T x = ( x, a x ). Certaily T is a well-defied, ijective, ad liear map. Suppose that y X, that y y X, ad that T y (c ) l 1. The lim y, a x c = Sice the coefficiet fuctioals a are cotiuous, we have by (5.1) that y, a x = lim T y (c ) l 1 = 0. (5.1) lim y, a x = c. Therefore T y = (c ), so T is a closed mappig. We coclude from the Closed Graph Theorem (Theorem 1.46) that T is cotiuous. ow choose ay (c ) l 1. The ( c x / x ) l 1, so x = c x x X. However, T x = (c ), so T is surjective. Therefore T is a topological isomorphism of X oto l 1. I particular, it follows from the Iverse Mappig Theorem (Theorem 1.44) that T 1 is cotiuous. Alteratively, we ca see this directly from the calculatio x = x, a x x, a x = T x l 1. Example 5.3. Let H be a separable, ifiite-dimesioal Hilbert space, ad let {e } be ay orthoormal basis for H. We saw i Example 2.5 that c e coverges if ad oly if (c ) l 2, ad that i this case the covergece is ucoditioal. O the other had, sice e = 1, we see that c e coverges absolutely if ad oly if (c ) l 1. Sice l 1 is a proper subset of l 2, this implies that {e } is ot a absolutely coverget basis for H. Moreover, sice H is topologically isomorphic to l 2, ad sice l 2 is ot topologically isomorphic to l 1, it follows from Theorem 5.2 that H does ot possess ay absolutely coverget bases.
6. SOME TYPES OF LIEAR IDEPEDECE OF SEQUECES 47 6. SOME TYPES OF LIEAR IDEPEDECE OF SEQUECES I a ifiite-dimesioal Baach space, there are several possible types of liear idepedece of sequeces. We list three of these i the followig defiitio. We will cosider miimal sequeces i particular i more detail i Chapter 7. Defiitio 6.1. A sequece {x } i a Baach space X is: (a) fiitely idepedet if c x = 0 implies c 1 = = c = 0, (b) ω-idepedet if c x coverges ad equals 0 oly whe c = 0 for every, (c) miimal if x m / spa{x } m for every m. Theorem 6.2. Let {x } be a sequece i a Baach space X. The: (a) {x } is a basis = {x } is miimal ad complete. (b) {x } is miimal = {x } is ω-idepedet. (c) {x } is ω-idepedet = {x } is fiitely idepedet. Proof. (a) Assume that ({x }, {a }) is a basis for a Baach space X. The {x } is certaily complete, so we eed oly show that it is miimal. Fix m, ad defie E = spa{x } m. The, sice {x } ad {a } are biorthogoal, we have x, a m = 0 for every x E. Sice a m is cotiuous, this implies x, a m = 0 for every x Ē = spa{x } m. However, we kow that x m, a m = 1, so we coclude that x m / Ē. Hece {x } is miimal. (b) Suppose that {x } is miimal ad that c x coverges ad equals 0. Let m be such that c m 0. The x m = 1 c m m c x spa{x } m, a cotradictio. (c) Clear. oe of the implicatios i Theorem 6.2 are reversible, as the followig examples show. Example 6.3. [Si70, p. 24]. Miimal ad complete = / basis. Defie C(T) = { f C(R) : f(t + 1) = f(t) }, the space of all cotiuous, 1-periodic fuctios. The C(T) is a Baach space uder the uiform orm L. Cosider the fuctios e (t) = e 2πit for Z. ot oly are these fuctios elemets of C(T), but they defie cotiuous liear fuctioals o C(T) via the ier product f, e = 1 0 f(t) e 2πit dt. Further, {e } Z is its ow biorthogoal system sice e m, e = δ m. Lemma 7.2 below therefore implies that {e } Z is miimal i C(T). The Weierstrass Approximatio Theorem [Kat68, p. 15] states that if f C(T) the f = c e L < ε for some scalars c. Hece spa{e } Z is dese i C(T), ad therefore {e } Z is complete i C(T). Alteratively, we ca demostrate the completeess as follows. Suppose that f C(T) satisfies f, e = 0 for every. Sice C(T) L 2 (T) ad sice {e } Z is a orthoormal basis for L 2 (T), this implies that f is the zero fuctio i the space
48 6. SOME TYPES OF LIEAR IDEPEDECE OF SEQUECES L 2 (T), hece is zero almost everywhere. Sice f is cotiuous, it follows that f(t) = 0 for all t. Hece {e } Z is complete both C(T) ad L 2 (T) by Corollary 1.41. Thus, {e } Z is both miimal ad complete i C(T). Further, if f = c e coverges i C(T), the it is easy to see from the orthoormality of the e that c = f, e. However, it is kow that there exist cotiuous fuctios f C(T) whose Fourier series f = f, e e do ot coverge uiformly [Kat68, p. 51]. Therefore, {e } Z caot be a basis for C(T). Example 6.4. [Si70, p. 24]. ω-idepedet = / miimal. Let X be a Baach space such that there exists a sequece {x } that is both miimal ad complete i X but is ot a basis for X (for example, we could use X = C(T) ad x (t) = e (t) = e 2πit as i Example 6.3). Sice {x } is miimal, it follows from Lemma 7.2 that there exists a sequece {a } X that is biorthogoal to {x }. Sice {x } is ot a basis, there must exist some y X such that the series y, a x does ot coverge i X. Cosider the sequece {y} {x }. This ew sequece is certaily complete, ad sice y spa{x }, it caot be miimal. However, we will show that {y} {x } is ω-idepedet. Assume that cy + c x = 0, i.e., the summatio coverges ad equals zero. If c 0 the we would have y = 1 c c x. The biorthogoality of {x } ad {a } the implies that y, a = c /c. But the y, a x coverges, which is a cotradictio. Therefore, we must have c = 0, ad therefore c x = 0. However, {x } is miimal, ad therefore is ω-idepedet, so this implies that every c is zero. Thus {y} {x } is ω-idepedet ad complete, but ot miimal. Alteratively, we ca give a Hilbert space example of a complete ω-idepedet sequece that is ot miimal [VD97]. Let {e } be ay orthoormal basis for ay separable Hilbert space H, ad defie f 1 = e 1 ad f = e 1 + e / for 2. The {f } is certaily complete sice spa{f } = spa{e }. However, f 1 f = e / = 1/ 0. Therefore f 1 spa{f } 2, so {f } is ot miimal. To see that {f } is ω-idepedet, suppose that c f coverges ad equals zero. The ( c f = c )e 1 + c e 0. =2 Therefore, ( c )e 1 + 2 c e =2 = 2 c + c 2 0. =2 This implies immediately that c = 0 for each 2, ad therefore c 1 = 0 as well. Example 6.5. [Si70, p. 25]. Fiitely idepedet = / ω-idepedet. Let ({x }, {a }) be a basis for a Baach space X, ad let x X be ay elemet such that x, a 0 for every. For example, we could take x = x 2 x. ote that x caot equal ay x because x, a m = 0 whe m. Cosider the the ew sequece {x} {x }. This is certaily complete, ad x + x, a x = 0, so it is ot ω-idepedet. However, we will show that it is fiitely idepedet. Suppose that cx + c x = 0. Substitutig the fact that
49 x = x, a x, it follows that ( ) c x, a + c x + =+1 c x, a x = 0. However, {x } is a basis, so this is oly possible if c x, a +c = 0 for = 1,..., ad c x, a = 0 for >. Sice o x, a is zero we therefore must have c = 0. But the c 1 = = c = 0, so {x} {x } is fiitely idepedet.
50 7. BIORTHOGOAL SYSTEMS I BAACH SPACES A basis {x } ad its associated coefficiet fuctioals {a } are a example of biorthogoal sequeces. We study the properties of geeral biorthogoal systems i this chapter. Defiitio 7.1. Give a Baach space X ad give sequeces {x } X ad {a } X, we say that {a } is biorthogoal to {x }, or that ({x }, {a }) is a biorthogoal system, if x m, a = δ m for every m,. We associate with each biorthogoal system ({x }, {a }) the partial sum operators S : X X defied by S x = x, a x. We show ow that the existece of sequece biorthogoal to {x } is equivalet to the statemet that {x } is miimal. Lemma 7.2. [You80, p. 28], [Si70, p. 53]. Let X be a Baach space, ad let {x } X. The: (a) {a } X biorthogoal to {x } {x } is miimal. (b) uique {a } X biorthogoal to {x } {x } is miimal ad complete. Proof. (a). Suppose that {a } X is biorthogoal to {x }. Fix ay m, ad choose z spa{x } m, say z = j=1 c j x j. The z, a m = j=1 c j x j, a m = 0 sice x j x m for all j. Sice a m is cotiuous, we the have z, a m = 0 for all z spa{x } m. However x m, a m = 1, so we must have x m / spa{x } m. Therefore {x } is miimal.. Suppose that {x } is miimal. Fix m, ad defie E = spa{x } m. This is a closed subspace of X which does ot cotai x m. Therefore, by the Hah Baach Theorem (Corollary 1.40) there is a fuctioal a m X such that x m, a m = 1 ad x, a m = 0 for x E. Repeatig this for all m we obtai a sequece {a } that is biorthogoal to {x }. (b). Suppose there is a uique sequece {a } X that is biorthogoal to {x }. We kow that {x } is miimal by part (a), so it remais oly to show that {x } is complete. Suppose that x X is a cotiuous liear fuctioal such that x, x = 0 for every. The x m, x + a = x m, x + x m, a = 0 + δ m = δ m. Thus {x + a } is also biorthogoal to {x }. By our uiqueess assumptio, we must have x = 0. The Hah Baach Theorem (Corollary 1.41) therefore implies that spa{x } = X, so {x } is complete.
7. BIORTHOGOAL SYSTEMS I BAACH SPACES 51. Suppose that {x } is both miimal ad complete. By part (a) we kow that there exists at least oe sequece {a } X that is biorthogoal to {x }, so we eed oly show that this sequece is uique. Suppose that {b } X is also biorthogoal to {x }. The x, a m b m = δ m δ m = 0 for every m ad. However, {x } is complete, so the Hah Baach Theorem (Corollary 1.41) implies that a m b m = 0 for every m. Thus {a } is uique. ext, we characterize the additioal properties that a miimal sequece must possess i order to be a basis. Theorem 7.3. [Si70, p. 25]. Let {x } be a sequece i a Baach space X. The the followig statemets are equivalet. (a) {x } is a basis for X. (b) There exists a biorthogoal sequece {a } X such that x X, x = x, a x. (c) {x } is complete ad there exists a biorthogoal sequece {a } X such that x X, sup S x <. (d) {x } is complete ad there exists a biorthogoal sequece {a } X such that sup S <. Proof. (a) (b). Follows immediately from the defiitio of basis ad the fact that every basis is a Schauder basis (Theorem 4.11). (b) (a). Assume that statemet (b) holds. We eed oly show that the represetatio x = x, a x is uique. However, each a m is cotiuous, so if x = c x, the x, a m = c x, a m = c δ m = c m. (b) (c). Assume that statemet (b) holds. The the fact that every x ca be writte x = x, a x implies that spa{x } is dese is X, hece that {x } is complete. Further, it implies that x = lim S x, i.e., that the sequece {S x} is coverget. Therefore statemet (c) holds sice all coverget sequeces are bouded. (c) (d). Each S is a bouded liear operator mappig X ito itself. Therefore, this implicatio follows immediately from the Uiform Boudedess Priciple (Theorem 1.42). (d) (b). Assume that statemet (d) holds, ad choose ay x spa{x }, say x = M c x. The, sice S is liear ad {x } ad {a } are biorthogoal, we have for each M that ( M ) S x = S c m x m = m=1 M c m S x m = m=1 M c m m=1 x m, a x = M c m x m = x. m=1
52 7. BIORTHOGOAL SYSTEMS I BAACH SPACES Therefore, we trivially have x = lim S x = x, a x whe x spa{x }. ow we will show that x = lim S x for arbitrary x X. Let C = sup S, ad let x be a arbitrary elemet of X. Sice {x } is complete, spa{x } is dese i X. Therefore, give ε > 0 we ca fid a elemet y spa{x } with x y < ε/(1 + C), say y = M m=1 c mx m. The for M we have x S x x y + y S y + S y S x x y + 0 + S x y (1 + C) x y < ε. Thus x = lim S x = x, a x for arbitrary x X, as desired. The ext two theorems give a characterizatio of miimal sequeces ad bases i terms of the size of fiite liear combiatios of the sequece elemets. Theorem 7.4. [Si70, p. 54]. Give a sequece {x } i a Baach space X with all x 0, the followig two statemets are equivalet. (a) {x } is miimal. (b) M, C M 1 such that M, c 0,..., c, M c x C M c x. Proof. (a) (b). Assume that {x } is miimal. The there exists a sequece {a } X that is biorthogoal to {x }. Let {S } be the partial sum operators associated with ({x }, {a }). Suppose that M, ad that c 0,..., c are ay scalars. The M c x = ( S ) M c x S M c x. Therefore statemet (b) follows with C M = S M. (b) (a). Assume that statemet (b) holds, ad let E = spa{x }. Give x = c x E ad M, we have M M 1 c M x M = c M x M c x + c x C M c x + C M 1 c x = (C M + C M 1 ) x.
As x M 0, we therefore have 7. BIORTHOGOAL SYSTEMS I BAACH SPACES 53 c M (C M + C M 1 ) x. (7.1) x M I particular, x = 0 implies c 1 = = c = 0. Thus {x } is fiitely liearly idepedet. Sice E is the fiite liear spa of {x }, this implies that every elemet of E has a uique represetatio of the form x = c x. As a cosequece, we ca defie a scalar-valued mappig a m o the ( set E by a m c ) x = cm (where we set c m = 0 if m > ). By (7.1), we have a m (x) (C m + C m 1 ) x / x m for every x E, so a m is cotiuous o E. Sice E is dese i X, the Hah Baach Theorem (Corollary 1.38) implies that there is a cotiuous extesio of a m to all of X. This exteded a m is therefore a cotiuous liear fuctioal o X which is biorthogoal to {x }. Lemma 7.2 therefore implies that {x } is miimal. For a arbitrary miimal sequece, the costats C M is Theorem 7.4 eed ot be uiformly bouded. Compare this to the situatio for bases give i the followig result. Theorem 7.5. [LT77, p. 2]. Let {x } be a sequece i a Baach space X. The the followig statemets are equivalet. (a) {x } is a basis for X. (b) {x } is complete, x 0 for all, ad there exists C 1 such that M, c 1,..., c, M c x C c x. (7.2) I this case, the best costat C i (7.2) is the basis costat C = sup S. Proof. (a) (b). Suppose that {x } is a basis for X, ad let C = sup S be the basis costat. The {x } is complete ad x 0 for every. Fix M, ad choose ay c 1,..., c. The M c x = ( S ) M c x S M c x C c x. (b) (a). Suppose that statemet (b) holds. It the follows from Theorem 7.4 that {x } is miimal, so by Lemma 7.2 there exists a biorthogoal system {a } X. Let S deote the partial sum operators associated with ({x }, {a }). Sice {x } is complete, it suffices by Theorem 7.3 to show that sup S <. So, suppose that x = M c x spa{x }. The: M = S x = > M = S x = M c x C c x = C x, M c x = x.
54 7. BIORTHOGOAL SYSTEMS I BAACH SPACES As C 1 we therefore have S x C x for all wheever x spa{x }. However, each S is cotiuous ad spa{x } is dese i X, so this iequality must therefore hold for all x X. Thus sup S C <, as desired. This iequality also shows that the smallest possible value for C i (7.2) is C = sup S. The followig result is a applicatio of Theorem 7.5. Give a basis {x } for a Baach space X, it is ofte useful to have some boud o how much the elemets x ca be perturbed so that the resultig sequece remais a basis for X, or at least a basis for its closed liear spa. The followig result is classical, ad is a typical example of perturbatio theorems that apply to geeral bases. For specific types of bases i specific Baach spaces, it is ofte possible to derive sharper results. For a survey of results o basis perturbatios, we refer to [RH71]. Theorem 7.6. Let ({x }, {a }) be a basis for a Baach space X, with basis costat C. {y } X is such that R = a x y < 1, If the {y } is a basis for spa{y }, ad has basis costat C 1+R C. Moreover, i this case, the 1 R basis {x } for X ad the basis {y } for Y = spa{y } are equivalet i the sese of Defiitio 4.13. Proof. ote that, by defiitio, {y } is complete i Y = spa{y }. Further, if some y = 0 the we would have R a x 1 by (4.5), which cotradicts the fact that R < 1. Therefore, each y must be ozero. By Theorem 7.5, it therefore suffices to show that there exists a costat B such that M M, c 1,..., c, c y B c y. (7.3) Further, if (7.3) holds, the Theorem 7.5 also implies that the basis costat C for {y } satisfies C B. So, assume that M ad that c 1,..., c are give. Before showig the existece of the costat B, we will establish several useful iequalities. First, sice {x } ad {a } are biorthogoal, we have that K K K m, c m = c x, a m am c x. Therefore, for each K > 0 we have K c m (x m y m ) m=1 K m=1 K m=1 c m x m y m ( a m K ) c x x m y m
= K c x K m=1 a m x m y m K R c x. (7.4) 55 As a cosequece, M c y Further, (7.4) implies that M c x + M M c (y x ) (1 + R) c x. (7.5) c x c y + c (x y ) c y + R c x. Therefore, c y (1 R) c x. (7.6) Fially, sice {x } is a basis with basis costat C, Theorem 7.5 implies that M c x C c x. (7.7) Combiig (7.5), (7.6), ad (7.7), we obtai M M c y (1 + R) c x (1 + R)C c x 1 + R 1 R C c y. Hece (7.3) holds with B = 1+R 1 R C, ad therefore {y } is a basis for spa{y } with basis costat C B. Fially, calculatios similar to (7.5) ad (7.6) imply that (1 R) c x =M+1 =M+1 c y (1 + R) c x. =M+1 Hece c x coverges if ad oly if c y coverges. It therefore follows from Theorem 4.14 that {x } is equivalet to {y }.