Example on p. 157

Transcription

1 Example Le where BV [, 1] = Example on p. 157 { g : [, 1] C g() =, g() = g( + ) [, 1), var (g) = sup g( j+1 ) g( j ) he supremum is aken over all he pariions of [, 1] (1) : = < 1 < < n = 1. j=1 } var (g) <, (i) For g BV [, 1], if we define g var = var (g), hen he space (BV [, 1], var ) is a Banach space. (This is rue even if one does no assume ha g() = g() = g( + ) for [, 1).) (ii) Moreover, BV [, 1] = C[, 1], namely, BV[,1] is he dual space of C[, 1], where C[, 1] is he space of coninuous funcions on [, 1]. Proof. By he dual heory, (i) is a direc resul of (ii). We refer o Prop 5 below for a direc proof. Now we urn o prove (ii). We firs verify ha BV [, 1] C[, 1]. (This is also rue even if one does no assume ha g() = g() = g( + ) for [, 1).) To his end, le g BV [, 1]. For φ C[, 1], he Sieljes inegral is defined by (2) φ() dg() = n 1 lim j=1 φ( j)[g( j+1 ) g( j )], where is he same as in (1), j [ j, j+1 ] for j n 1 If we define = max 1 j n j j 1. f, φ = hen i is easy o see ha f C[, 1] f C[,1] φ() dg(), dg() = var (g) = g var, which is desired. (The exisence of (2) refer o p. 347 of Real variable funcions by Minqiang Zhou. And we remark again ha we do no use he condiions g() = g() = g( + ) for [, 1) here.) 1

2 To prove C[, 1] BV [, 1], le f C[, 1]. We will prove ha here exiss a unique g BV [, 1] such ha for all φ C[, 1], (3) f, φ = φ() dg() (4) g var f C[,1]. We firs prove he uniqueness of g. Assume ha g 1 BV [, 1] also saisfies (3) (4) wih g replaced g 1. For any fixed s (, 1], le φ n () C[, 1] such ha φ n C[,1] 1 φ n χ (,s] as n in he sense of C[, 1]. By he righ coninuiy of g, he Lebesgue-Sieljes (L-S) inegral (L S) χ (,s]() dg() exiss. From his, he Lebesgue dominaed convergence heorem he fac ha if Riemann-Sieljes (R-S) inegral exiss, hen he corresponding L-S inegral exiss he wo inegrals are equal, we deduce ha (5) lim (R S) φ n () dg() (L S) χ (,s] () dg() 1 = lim (L S) [φ n () χ (,s] ()] dg() (L S) =. lim φ n() χ (,s] () dg() Moreover, by he definiion of L-S inegral ogeher wih he righ coninuiy of g g() =, we have (L S) which ogeher wih (5) gives ha g(s) = (L S) χ (,s] () dg() = g(s), = lim (R S) = lim (R S) = (L S) = g 1 (s). 2 χ (,1] () dg() φ n () dg() φ n () dg 1 () χ (,1] () dg 1 ()

3 Thus g = g 1. (We should remark ha g() = g() = g( + ) for [, 1) is used for he uniqueness of g.) To prove he exisence of g, le { } X = αχ {} + φ α R, φ C[, 1] f (αχ {} + φ) = f(φ) for all α R, φ C[, 1]. Then X is a linear subspace of L [, 1], by he Hahn-Banach heorem (Theorem 2.4.4, p. 111), here exiss f L [, 1] such ha for φ C[, 1], f, φ = f, φ, f(χ {} ) =, f L [,1] = f X = f C[,1], which can be seen from he definiion of operaor norms αχ {} + φ L [,1] = φ C[,1]. For < 1, since χ (,] L [, 1], we define v() = v() = f, χ (,]. To verify ha (4) for v, namely, v var f C[,1], le be any pariion as in (1). For k =, 1,, n 1, se ω k = v( k+1 ) v( k ), λ k = ω k ω k, if ω k, oherwise, Then, we have h () = λ k χ (k, k+1 ](). k= ω k = λ k ω k k= k= n 1 = λ k [v( k+1 ) v( k )] k= n 1 = λ k [ f, χ (,k+1 ] f, χ (,k ] ] k= 3

4 = f, h f L [,1] h L [,1] f L [,1] = f C[,1], which ells us ha v is a bounded variaion funcion saisfies (4). To verify (3) for v, namely, for φ C[, 1], f, φ = (R S) φ() dv(), for any ɛ >, ake a pariion as in (1) such ha for all, [ j, j+1 ], j =, 1,, n 1, φ() φ( ɛ ) 2( f C[,1] + 1), (R S) φ() dv() φ( j )[v( j+1 ) v( j )] < ɛ 2. j= ((R S) φ() dv() exiss since φ is coninuous v is a bounded variaion funcion.) Se φ = φ( j )χ (j, j+1 ] + φ()χ {}, we hen have j= f, φ (R S) φ() dv() f, φ f, φ + f, φ f L [,1] φ φ L [,1] + φ( j )(v( j+1 ) v( j )) < ɛ, j= φ() dv() φ() dv() which gives (3). Then by Prop 4 below, here exiss g BV [, 1] saisfies (3) (4), which complees he proof of Example We finishes he proof of example by he following several proposiions. 4

5 Prop 1. Le v be a bounded variaion funcion on [, 1]. If se g() = v(+) for (, 1), hen g(+) = v(+) g( ) = v( ), g(+) = v(+) g(1 ) = v(1 ). Proof. For any ɛ >, here exiss δ > such ha for any s (, + δ), v(s) v( + ) < ɛ/2, here exiss δ 1 (, δ + s) such ha for s 1 (s, s + δ 1 ), by < s < s 1 < + δ, Thus v(s 1 ) v(s + ) < ɛ/2 v(s 1 ) v( + ) < ɛ/2. g(s) v( + ) g(s) v(s 1 ) + v(s 1 ) v( + ) < ɛ, which implies ha g( + ) = g(). On he oher h, for any ɛ >, here exiss a δ > such ha for s ( δ, ), v(s) v( ) < ɛ/2, here exiss a δ 1 (, s) such ha for any s 1 (s, s + δ 1 ), by δ < s < s 1 <, Thus v(s 1 ) v(s + ) < ɛ/2 v(s 1 ) v( ) < ɛ/2. g(s) v( ) g(s) v(s 1 ) + v(s 1 ) v( ) < ɛ, which indicaes g( ) = v( ). This complees he proof of Proposiion 1. Prop 2. Le v w be bounded variaion funcion on [, 1]. If v() = w() a.e. [, 1] for any φ C[, 1], (R S) φ() dv() (R S) φ() dw() exis, hen (R S) φ() dv() = (R S) φ() dw(). Proof. This is a simple corollary of he definiion of R-S inegrals. Prop 3. Le v be a bounded variaion funcion. Then φ() dv() = 5

6 for all φ C[, 1] if only if v() = v(1), for any (, 1), v( ) = v() = v( + ). Proof. Sufficiency. Assume ha v() = v(1) v( ) = v() = v( + ) for any (, 1). Since v() is differenial for almos all [, 1], herefore coninuous for almos all [, 1], we hen have v() = v() for almos all [, 1], which ogeher wih Proposiion 2 indicaes ha φ() dv() = φ() dv(a) = for all φ C[, 1]. Necessiy. Assume ha φ() dv() = for all φ C[, 1]. Taking φ() = 1 for [, 1], we have v(1) v() = φ() dv() =, which indicaes v() = v(1). I is easy o see ha for [, 1), as h +, 1 h +h v() d v( + ) 1 v() d v( ). h h For [, 1) h [, 1 ), se 1, s φ(s) = 1 s h, < s + h, + h < s 1 By we have = φ(s) dv(s) = v() v() + +h +h φ(s) dv(s) = v() v(). On he oher h, applying inegral by pars, we obain +h +h φ(s) dv(s) = φ()v() = v() + 1 h 6 +h +h v(s) ds. φ(s) dv(s), v(s) dφ(s)

7 (see p. 43 of Real variable funcions wrien by Senlin Xu). Thus v() = 1 h +h v(s) dv(s), which converges o v( + ) as h converges o +, namely, v( + ) = v(). Similarly, we have v( ) = v(1) for (, 1]. This complees he proof of Proposiion 3. Prop 4. Le v be bounded variaion funcion, g() =, g(1) = v(1) v() g() = v( + ) v() for < < 1. Then for all φ C[, 1], (6) φ() dg() = φ() dv(), (7) var (g) var (v). Proof. To prove (6), by Prop 3, i suffices o verify ha for (, 1), g( ) v( ) = g( + ) v( + ) = g() v(). In fac, we have g() v() = v() = g(1) v(1). For (, 1), by Prop 1, we have g( + ) = v( + ) v() g( ) = v( ) v(), hus g( + ) v( + ) = v() = g( ) v( ) = g() v(). This gives (6). To verify (7), le f() = v(+) for [, 1) f(1) = v(1). Then g() = f() v() var (g) = var (f) var (v). In fac, obviously var (g) = var (f). If v is increasing on [, 1], hen var (v) = v(1) v(), moreover, f is increasing var (f) = f(1) f() = v(1) v( + ) v(1) v() = var (v). For v BV [, 1], by leing v 1 () = 1 var 2 (v) + 1v() v 2 2 = 1 var 2 (v) 1 v(), we 2 hen have v = v 1 v 2, v 1 v 2 are increasing funcions, where var (v) denoes he 7

8 variaion of he funcion v on he inerval [, ]. (See p. 248 of Real variable funcion wrien by Minqiang Zhou.) From his, i follows ha var (v 1 ) + var (v 2 ) = v 1 (1) v 1 () + v 2 (1) v 2 () = 1 2 var (v) (v(1) v()) var (v) 1 (v(1) v()) 2 = var (v). By le f 1 () = v 1 ( + ) f 2 () = v 2 ( + ) for [, 1], hen f() = f 1 () + f 2 () hus var (f) = var (f 1 + f 2 ) var (f 1 ) + var (f 2 ) var (v 1 ) + var (v 2 ) = var (v), ha is, var (f) var (g), which complees he proof of Prop 4. Prop 5. (BV [, 1], var ) is a Banach space. Proof. I suffices o verify he compleeness. Le {f l } l N be a Cauchy series in BV [, 1]. For (, 1], considering he pariion = < 1 = 1, by f l () = f m () =, we have (8) f l () f m () [f l () f m ()] [f l () f m ()] which ells ha lim l f l () exiss. Se var (f l f m ), l, m, f() = lim f l () for [, 1]. Obviously, f() =. We claim ha f BV [, 1]. In fac, for any pariion in (1), we have [f m ( j+1 ) f( j+1 )] [f m ( j ) f( j )] j=1 lim [f m ( j+1 ) f l ( j+1 )] [f m ( j ) f l ( j )] l j=1 lim l var (f l f m ), 8

9 which gives ha hus var (f f m ) lim l var (f l f m ) (9) lim var (f f m) lim var (f l f m ) =. m m,l From his, we deduce ha f BV [, 1] var (f) var (f f m ) + var (f m ) <. (We remark ha if he righ coninuiy of funcions in BV [, 1] is removed, hen he above argumen already indicaes ha BV [, 1] is a Banach space, oherwise we need he following discussion.) For [, 1), considering he pariion = < 1 = 1, by f l () = f() =, we have f l () f() [f l () f()] [f l () f()] var (f l f). From his (9), for all ɛ >, here exiss N N such ha l N for all s [, 1], f l (s) f(s) ɛ/3. Fix l, since f l ( + ) = f l (), here exiss a δ >, which dependen on l, such ha for s (, + δ), f l () f l ( + s) ɛ/3, herefore, f() f( + s) f() f l () + f l () f l ( + s) + f l ( + s) f( + s) ɛ, which gives ha f() = f( + ). This ogeher wih f BV [, 1] (9) indicaes ha BV [, 1] is complee hus complees he proof of Prop 5. 9