A new theorem on the existence of the Riemnn-Stieltjes integrl nd n improved version of the Loéve-Young inequlity Rfł M. Łochowski rxiv:1403.5413v4 [mth.ca] 2 Dec 2015 Deprtment of Mthemtics nd Mthemticl Economics, Wrsw School of Economics ul. Mdlińskiego 6/8, 02-513 Wrszw, Polnd E-mil : rlocho314@gmil.com Abstrct Using the notion of truncted vrition we obtin new theorem on the existence nd estimtion of the Riemnn-Stieltjes integrl. As specil cse of this theorem we obtin n improved version of the Loéve-Young inequlity for the Riemnn-Stieltjes integrls driven by irregulr signls. Using this result we strenghten some results of Terry Lyons on the existence of solutions of integrl equtions driven by modertely irregulr signls. Keywords: the Loéve-Young inequlity, the Riemnn-Stieltjes integrl, irregulr signls 1 Introduction The purpose of this pper is to investigte the top-down structure of the Riemnn-Stieltjes integrl nd to stte some generl condition gurnteeing the existence of this integrl, expressed in terms of the functionl clled truncted vrition. b The simplest (nd rther not surprising) cse where the Riemnn-Stieltjes integrl fdg (RSI in short) exists, is the sitution when the integrnd nd the integrtor hve no common points of discontinuity, the former is bounded nd the ltter hs finite totl vrition. We will prove generl theorem (Theorem 1) encompssing this sitution s well s more interesting cse when the RSI exists, nmely when the integrnd nd the integrtor hve possibly unbounded vrition, but they hve finite p-vrition nd q-vrition respectively, with p > 1, q > 1 nd p 1 + q 1 > 1. The ltter result is due to Young ([10, p. 264, Theorem on Stieltjes integrbility]). For f : [;b] R nd p > 0, the p-vrition, which we will denote by V p (f;[;b]), is defined s V p (f,[;b]) sup n sup t 1 <t 2 < <t n b n 1 f (t i ) f (t i 1 ) p. The forementioned Theorem 1 provides lso n upper bound for the difference ˆ b fdg f()[g(b) g()]. 1
As specil cse of this bound we will obtin the following, improved version of the clssicl Loéve-Young inequlity: ˆ b fdg f ()[g(b) g()] C p,q(v p (f,[;b])) 1 1/q f 1+p/q p osc,[;b] (V q (g,[;b])) 1/q, (1) wherec p,q is some constnt depending onpndq only nd f osc,[;b] : sup s<t b f(t) f(s). The originl Loéve-Young estimte, published in 1936 in [10], reds s: ˆ b fdg f()[g(b) g()] ζ( p 1 +q 1) (V p (f,[;b])) 1/p (V q (g,[;b])) 1/q, where ζ(r) + k1 k r is the fmous Riemnn zet function. We sy tht our bound is improved version of the Loéve-Young inequlity since the constnt C p,q (lthough possibly greter thn ζ(p 1 +q 1 )) is irrelevnt in pplictions while the fct tht the term (V p (f,[;b])) 1/p in the originl Loéve-Young estimte is replced in our estimte by (V p (f,[;b])) 1 1/q f 1+p/q p osc,[;b] (notice tht 1/p > 1 1/q nd lwys (V p (f,[;b])) 1/p (1 1/q) f 1+p/q p osc,[;b] ) mkes it possible to obtin stronger results. For exmple, Proposition 1 obtined with the help of (1) is genuine improvement upon erlier known results of this type. Let us comment shortly how the results on the existence of the RSI were obtined so fr. The originl Young s proof utlilized elementry but clever induction rgument for finite sequences. Another proof of the Young theorem my be found in [4, Chpt. 6], where integrl estimtes bsed on control function nd the Lóeve-Young inequlity re used. This pproch is further pplied in the rough-pth theory setting. Further generlistions of Young s theorem re possible, with p-vrition replced by more generl ϕ-vrition: V ϕ (f,[;b]) sup n sup t 1 <t 2 < <t n b n 1 ϕ( f (t i ) f (t i 1 ) ), where ϕ : [0;+ ) [0;+ ) is Young function, i.e. convex, strictly incresing function strting from 0 (see for exmple [11], [3] nd for survey bout nother results of this type see the recent books [2, Chpt. 3], [1, Sect. 4.4]). However, s fr s we know, Theorem 1 is new result on the existence of the RSI. The proof of Theorem 1 utilizes simple properties of the truncted vrition nd multiple ppliction of the summtion by prts. Similrly, no version of the Loéve-Young inequlity s estimte (1), s fr s we know, hs ppered so fr (see detiled historicl notes on the the Loéve-Young inequlity in [2, pp. 212-214]). We conjecture tht using Theorem 1 one my lso obtin vrition of the Loéve-Young inequlity for ϕ-vrition (see [2, Theorem 3.89, Corollry 3.90] or [1, Theorem 4.40]). We intend to del with this conjecture in the future. After hving obtined these results we were ble, following Lyons [7], nd Lyons, Crun nd Lévy [8], to solve few types of integrl equtions driven by modertely irregulr signls. By modertely irregulr signls we men continuous signls with finite p-vrition, where p (1; 2). It is well known tht for higher degrees of irregulrity, corresponding to p 2, one needs, constructing pproximtions of integrl equtions, to consider terms of new type (like Lévy s re). We believe tht the tuncted vrition pproch for such pths is lso possible nd this will be topic of our further reserch. Let us comment shortly on the orgnistion of the pper. In the next section we prove generl theorem on the existence of the Riemnn-Stieltjes integrl, expressed in terms of the truncted vrition functionls nd derive from it the stronger version of the Loéve-Young inequlity. Next, in Section 3, we del with the pplictions of this result to few types of integrl equtions driven by modertely irregulr signls. 2
2 A theorem on the existence of the Riemnn-Stieltjes integrl b In this section we will prove generl theorem on the existence of the RSI fdg formulted in terms of the truncted vrition. We will ssume tht both - integrnd f : [;b] R nd integrtor g : [;b] R re regulted functions. Let us stte the definition of the truncted vrition nd recll the definition of regulted function. For f : [;b] R its truncted vrition with the trunction prmeter δ 0 will be denoted by TV δ (f,[;b]). It my be simply defined s the gretest lower bound for the totl vrition of ny function g : [;b] R, uniformly pproximting f with the ccurcy δ/2, TV δ (f,[;b]) : inf { TV(g,[;b]) : f g,[;b] δ/2 }. f g,[;b] denotes here sup t b f(t) g(t) nd the totl vrition TV(g,[;b]) is defined s n 1 TV(g,[;b]) : sup sup g(t i ) g(t i 1 ). n t 1 <t 2 < <t n b It ppers tht the truncted vrition TV δ (f,[;b]) is finite for ny δ > 0 iff f is regulted (cf. [6, Fct 2.2]) nd then for ny δ > 0 the following equlity holds TV δ (f,[;b]) sup n sup t 1 <t 2 < <t n b n 1 mx{ f (t i ) f (t i 1 ) δ,0} (2) (cf. [5, Theorem 4]). Let us recll tht function h : [;b] R is regulted if there exist one-sided finite limits lim t + h(t) nd lim t b h(t), nd for ny t (;b) nd there exist one-sided finite limits lim t x h(t) nd lim t x+ h(t). We will lso need the following result (cf. [5, Theorem 4]): for ny regulted function f : [;b] R nd δ > 0 there exists regulted function f δ : [;b] R such tht f f δ,[;b] δ/2 nd TV 0( f δ,[;b] ) TV δ (f,[;b]). From formul (2), it directly follows tht the truncted vrition is superdditive functionl of the intervl, i.e. for ny d (;b) TV δ (f,[;b]) TV δ (f,[;d])+tv δ (f,[d;b]). Moreover, we lso hve the following esy estimte of the truncted vrition of function f perturbed by some other function h : TV δ (f +h,[;b]) TV δ (f,[;b])+tv 0 (h,[;b]), (3) which stems directly from the inequlity: for s < t b, mx{ f (t)+h(t) {f (s)+h(s)} δ,0} mx{ f (t) f (s) δ,0}+ h(t) h(s). Theorem 1 Let f,g : [;b] R be two regulted functions which hve no common points of discontinuity. Let η 0 η 1... nd θ 0 θ 1... be two sequences of non-negtive numbers, such tht η k 0, θ k 0 s k +. Define η 1 : sup t b f (t) f () nd S : 2 k η k 1 TV θ k (g,[;b])+ 2 k θ k TV η k (f,[;b]). 3
b If S < + then the Riemnn-Stieltjes integrl fdg exists nd one hs the following estimte ˆ b fdg f ()[g(b) g()] S. (4) Remrk 1 The ssumption tht f nd g hs no common points of discontinuity is necessry b for the existence of the RSI fdg. When more generl integrls re considered (e.g. the Moore-Pollrd integrl, c.f. [10, p. 263]), we my weken this ssumption nd ssume tht f nd g hve no common one-sided discontinuities. The proof of Theorem 1 will be bsed on the following lemm. Lemm 1 Let f,g : [;b] R be two regulted functions. Let c t 0 < t 1 <... < t n d be ny prtition of the intervl [c;d] [;b] nd let ξ 0 c nd ξ 1,...,ξ n be such tht t i 1 ξ i t i for i 1,2,...,n. Then for δ 1 : sup c t d f (t) f (c), δ 0 δ 1... δ r 0 nd ε 0 ε 1... ε r 0 the following estimte holds f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] r r 2 k δ k 1 TV ε k (g,[c;d])+ 2 k ε k TV δ k (f,[c;d])+nδ r ε r. Proof. Denote ε ε 0, by summtion by prts, we hve the following equlity f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] [f (ξ i ) f (c)][g ε (t i ) g ε (t i 1 )] + [f (ξ i ) f (c)][g(t i ) g ε (t i ) {g(t i 1 ) g ε (t i 1 )}] [f (ξ i ) f (c)][g ε (t i ) g ε (t i 1 )] + [g(d) g ε (d) {g(t i 1 ) g ε (t i 1 )}][f (ξ i ) f (ξ i 1 )], (5) where g ε : [c;d] R is regulted nd such tht Similrly, for δ δ 0 we my write g g ε,[c;d] 1 2 ε nd TV0 (g ε,[c;d]) TV ε (g,[c;d]). [g(d) g ε (d) {g(t i 1 ) g ε (t i 1 )}][f (ξ i ) f (ξ i 1 )] [g(d) g ε (d) {g(t i 1 ) g ε (t i 1 )}] [ f δ (ξ i ) f δ (ξ i 1 ) ] (6) + [ f (ξi ) f δ (ξ i ) { f (c) f δ (c) }] [{g(t i ) g ε (t i )} {g(t i 1 ) g ε (t i 1 )}], 4
where f δ : [c;d] R is regulted nd such tht f f δ,[c;d] 1 2 δ nd TV0( f δ,[c;d] ) TV δ (f,[c;d]). Since TV 0 (g ε,[c;d]) TV ε (g,[c;d]), TV 0( f δ,[c;b] ) TV δ (f,[c;d]), g g ε,[c;d] ε/2 nd f f δ,[c;d] δ/2, from (5) nd (6) we hve the following estimte f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] sup f (t) f (c) TV ε (g,[c;d])+ε TV δ (f,[c;d])+nδε. (7) c t d Denote g 1 : g g ε, f 1 : f f δ on [c;d]. By (5) nd (6), insted of the lst summnd nδε in (7) we my write the estimte [ f (ξi ) f δ (ξ i ) { f (c) f δ (c) }] [{g(t i ) g ε (t i )} {g(t i 1 ) g ε (t i 1 )}] [f 1 (ξ i ) f 1 (c)][g 1 (t i ) g 1 (t i 1 )] sup f 1 (t) f 1 (c) TV ε 1 (g 1,[c;d])+ε 1 TV δ 1 (f 1,[c;d])+nδ 1 ε 1 c t d δ TV ε 1 (g 1,[c;d])+ε 1 TV δ 1 (f 1,[c;d])+nδ 1 ε 1, (8) where the lst but one inequlity in (8) follows by the sme resoning for f 1 nd g 1 s inequlity (5) for f nd g. Repeting these rguments, by induction we get f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] r r δ k 1 TV ε k (g k,[c;d])+ ε k TV δ k (f k,[c;d])+nδ r ε r, (9) where δ 1 : sup c t c f (t) f (), g 0 g, f 0 f nd for k 1,2,...,r, g k : g k 1 g ε k 1 k 1, f k : f k 1 f δ k 1 k 1 re defined similrly s g 1 nd f 1. Since ε k ε k 1 for k 1,2,...,r, by (3) nd the fct tht the function δ TV δ (h,[c;d]) is non-incresing, we estimte ( TV ε k (g k,[c;d]) TV ε k gk 1 g ε k 1 k 1,[c;d]) TV ε k (g k 1,[c;d])+TV 0( g ε k 1 k 1,[c;d]) TV ε k (g k 1,[c;d])+TV ε k 1 (g k 1,[c;d]) Hence, by recursion, for k 1,2,...,r, Similrly, for k 1,2,...,r, we hve 2TV ε k (g k 1,[c;d]). TV ε k (g k,[c;d]) 2 k TV ε k (g,[c;d]). TV δ k (f k,[c;d]) 2 k TV δ k (f,[c;d]). By (9) nd lst two estimtes we get the desired estimte. 5
Remrk 2 Notice tht strting in (5) from the summtion by prts, then splitting the difference f (ξ i ) f (ξ i 1 ) : f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] [g(d) g(t i 1 )][f (ξ i ) f (ξ i 1 )] [g(d) g(t i 1 )] [ f δ (ξ i ) f δ (ξ i 1 ) ] + [g(d) g(t i 1 )] [ f (ξ i ) f δ (ξ i ) { f (ξ i 1 ) f δ (ξ i 1 ) }] nd proceeding similrly s in the proof of Lemm 1 we get the symmetric estimte f (ξ i )[g(t i ) g(t i 1 )] f (c)[g(d) g(c)] r r 2 k ε k 1 TV δ k (f,[c;d])+ 2 k δ k TV ε k (g,[c;d])+nδ r ε r, (10) where ε 1 sup c t d g(d) g(t). Remrk 3 Setting in Lemm 1, n 1 for ny ξ [c;d] we get the estimte (f (ξ) f (c))[g(d) g(c)] r 2 k δ k 1 TV ε k (f,[c;d])+ r 2 k ε k TV δ k (g,[c;d])+nδ r ε r. (11) nd similrly, setting in Remrk 2, n 1 we get similr estimte, where the right side of (11) is replced by the right side of (10). Now we proceed to the proof of Theorem 1. Proof. It is enough to prove tht for ny two prtitions π { 0 < 1 <... < l b}, ρ { b 0 < b 1 <... < b m b} nd ν i [ i 1 ; i ], ξ j [b j 1 ;b j ], i 1,2,...,l, j 1,2,...,m, the difference l m f (ν i )[g( i ) g( i 1 )] f (ξ j )[g(b j ) g(b j 1 )] is s smll s we plese, provided tht the meshes of the prtitions π nd ρ, defined s mesh(π) : mx,2,...,l ( i i 1 ), mesh(ρ) : mx j1,2,...,m (b j b j 1 ) respectively, re sufficiently smll. Define σ π ρ { s 0 < s 1 <... < s n b} 6 j1
nd for i 1,2,...,l consider f (ν i)[g( i ) g( i 1 )] We estimte f (ν i)[g( i ) g( i 1 )] k:s k 1,s k [ i 1 ; i ] k:s k 1,s k [ i 1 ; i ] f (s k 1 )[g(s k ) g(s k 1 )]. f (s k 1 )[g(s k ) g(s k 1 )] f (ν i )[g( i ) g( i 1 )] f ( i 1 )[g( i ) g( i 1 )] + f (s k 1 )[g(s k ) g(s k 1 )] f ( i 1 )[g( i ) g( i 1 )]. k:s k 1,s k [ i 1 ; i ] Recll the definition of S. If there exists N 0,1,2,... such tht η N 0 or θ N 0 then TV(f,[;b]) or TV(g,[;b]) is finite, moreover, both functions f nd g re bounded (since they b re regulted), hence the integrl fdg exists. Thus we my nd will ssume tht η N > 0 nd θ N > 0 for ll N 0,1,2,... Choose N 1,2,... By the ssumption tht f nd g hve no common points of discontinuity, for sufficiently smll mesh(π), for i 1,2,...,l we hve or sup f (s) f ( i 1 ) η N 1 (12) i 1 s i sup i 1 s i g( i ) g(s) θ N 1. (13) To see this, ssume tht for every h > 0, there exist [ h ;b h ] [;b] such tht b h h h nd sup x,y [h ;b h ] f (y) f (y) > η N 1 nd sup x,y [h ;b h ] g(x) g(y) > θ N 1. We choose convergent subsequence of the sequence ( 1/n +b 1/n ) /2, n 1,2,..., nd we see tht the limit of this sequence is point of discontinuity for both f nd g, which is contrdiction with the ssumption tht f nd g hve no common points of discontinuity. Let I be the set of ll indices i 1,2,...,l for which (12) holds. Now, for i I, set δ j 1 : η N+j 1, ε j : θ N+j, j 0,1,2,..., nd define S i : 2 j η j 1 TV θ j (g,[ i 1 ; i ])+ 2 j θ j TV η j (f,[ i 1 ; i ]). By Lemm 1 we estimte k:s k 1,s k [ i 1 ; i ] f (s k 1 )[g(s k ) g(s k 1 )] f ( i 1 )[g( i ) g( i 1 )] 2 j δ j 1 TV ε j (g,[ i 1 ; i ])+ 2 j η N+j 1 TV θ N+j (g,[ i 1 ; i ])+ 2 N S i. 7 2 j ε j TV δ j (f,[ i 1 ; i ]) 2 j θ N+j TV η N+j (f,[ i 1 ; i ])
Similrly, Hence f (ν i )[g( i ) g( i 1 )] f ( i 1 )[g( i ) g( i 1 )] 2 N S i. f (ν i)[g( i ) g( i 1 )] 2 1 N S i. k:s k 1,s k [ i 1 ; i ] f (s k 1 )[g(s k ) g(s k 1 )] (14) The truncted vrition is superdditive functionl of the intervl, from which we hve TV θ j (g,[ i 1 ; i ]) TV θ j (g,[;b]), i I TV η j (f,[ i 1 ; i ]) TV η j (f,[;b]). i I By (14) nd lst two inequlities, summing over i I we get the estimte f (ν i)[g( i ) g( i 1 )] f (s k 1 )[g(s k ) g(s k 1 )] i I k:s k 1,s k [ i 1 ; i ] 2 1 N S i 2 1 N S. (15) i I Now, let J be the set of ll indices, for which (13) holds. For i 1,2,...,l define T i : 2 j θ j TV η j (f,[ i 1 ;])+ 2 j η j TV θ j+1 (g,[ i 1 ; i ]). For i J, by the summtion by prts nd then by Lemm 1 we get f ( i)[g( i ) g( i 1 )] f (s k 1 )[g(s k ) g(s k 1 )] k:s k 1,s k [ i 1 ; i ] g(s k )[f (s k ) f (s k 1 )] g( i 1 )[f ( i ) f ( i 1 )] k:s k 1,s k [ i 1 ; i ] 2 j θ N+j 1 TV η N+j (f,[ i 1 ; i ])+ 2 j η N+j TV θ N+j (g,[ i 1 ; i ]). 2 1 N T i 2 1 N S i. Similrly, by Lemm 1, f ( i )[g( i ) g( i 1 )] f (ν i )[g( i ) g( i 1 )] g( i 1 )[f (ν i ) f ( i 1 )]+g( i )[f ( i ) f (ν i )] g( i 1 )[f ( i ) f ( i 1 )] 2 1 N S i. 8
From lst two inequlities we get f (ν i)[g( i ) g( i 1 )] k:s k 1,s k [ i 1 ; i ] f (s k 1 )[g(s k ) g(s k 1 )] f ( i )[g( i ) g( i 1 )] f (ν i )[g( i ) g( i 1 )] + f ( i)[g( i ) g( i 1 )] f (s k 1 )[g(s k ) g(s k 1 )] 2 2 N S i. k:s k 1,s k [ i 1 ; i ] Summing over i J nd using the superdditivity of the truncted vrition s function of the intervl, we get the estimte f (ν i)[g( i ) g( i 1 )] f (s k 1 )[g(s k ) g(s k 1 )] i J 2 2 N i J Finlly, from (15) nd (16) we get f (ν i)[g(b) g()] k:s k 1,s k [ i 1 ; i ] S i 2 2 N S. (16) f (s k 1 )[g(s k ) g(s k 1 )] 6 2 N S. k1 Similr estimte holds for m f (ξ j )[g(b j ) g(b j 1 )] ij f (s k 1 )[g(s k ) g(s k 1 )], provided tht mesh(ρ) is sufficiently smll. Hence l m f (ν i )[g( i ) g( i 1 )] f (ξ j )[g(b j ) g(b j 1 )] 12 2 N S. ij k1 provided tht mesh(π) nd mesh(ρ) re sufficiently smll. Since N my be rbitrry lrge, we get the convergence of the pproximting sums to n universl limit, which is the Riemnn- Stieltjes integrl. The estimte (4) follows directly from the proved convergence of pproximting sums to the Riemnn-Stieltjes integrl nd Lemm 1. Using Remrk 2 nd resoning similrly s in the proof of Theorem 1, we get the symmetric result. Theorem 2 Let f,g : [;b] R be two regulted functions which hve no common points of discontinuity. Let η 0 η 1... nd θ 0 θ 1... be two sequences of non-negtive numbers, such tht η k 0, θ k 0 s k +. Define θ 1 : sup t b g(b) g(t) nd S : 2 k θ k 1 TV η k (f,[;b])+ 2 k η k TV θ k (g,[;b]). If S b < + then the Riemnn-Stieltjes integrl fdg exists nd one hs the following estimte ˆ b fdg f ()[g(b) g()] S. (17) 9
From Theorem 1, Theorem 2 nd Remrk 3 we lso hve Corollry 1 Let f, g : [; b] R be two regulted functions which hve no common points of discontinuity, ξ [;b] nd S nd S be s in Theorem 1 nd Theorem 2 respectively. If S < + or S b < + then the Riemnn-Stieltjes integrl fdg exists nd one hs the following estimte ˆ b fdg f (ξ)[g(b) g()] 2min{S, S}. 2.1 Young s Theorem nd the Loéve-Young inequlity Let for p > 0, V p ([;b]) denote the fmily of functions f : [;b] R with finite p vrition. Note tht if f V p ([;b]) then f is regulted. The dditionl reltion we will use, is the following one: if f V p ([;b]) for some p 1, then for every δ > 0, TV δ (f,[;b]) V p (f,[;b])δ 1 p. (18) As fr s we know, the first result of this kind, nmely, TV δ (f,[;b]) C f δ 1 p for continuous function f V p ([;b]) nd some constnt C f < + depending on f, ws proven in [9, Sect. 6]. In [9], TV ε (f,[;b]) is clled ε vrition nd is denoted by V f (ε). However, being equipped with formul (2) we see tht reltion (18) follows immeditely from the inequlity: for ny s < t b, f (t) f (s) p mx{ f (t) f (s) δ,0}, δ p 1 which is n obvious consequence of the estimte: { δ p 1 0 if δ x mx{ x δ,0} x p 1 mx{ x δ,0} if 0 < δ < x x p for ny δ > 0 nd ny rel x. Let us denote f p vr,[;b] : (V p (f,[;b])) 1/p (19) nd recl tht f osc,[;b] sup s<t b f (t) f (s). Now we re redy to stte Corollry stemming from Theorem 1, which ws one of the min results of [10]. The second prt of this Corollry is n improved version of the Loéve-Young inequlity. Corollry 2 Let f,g : [;b] R be two functions with no common points of discontinuity. If f V p ([;b]) nd g V q ([;b]), where p > 1, q > 1, p 1 +q 1 > 1, then the Riemnn Stieltjes b fdg exists. Moreover, there exist constnt C p,q, depending on p nd q only, such tht ˆ b fdg f ()[g(b) g()] C p,q f p p/q p vr,[;b] f 1+p/q p osc,[;b] g q vr,[;b]. Proof. By Theorem 1 it is enough to prove tht for some positive sequences η 0 η 1... nd θ 0 θ 1..., such tht η k 0, θ k 0 s k + nd η 1 sup t b f (t) f () one hs S : 2 k η k 1 TV θ k (g,[;b])+ C p,q f p p/q p vr,[;b] f 1+p/q p osc,[;b] g q vr,[;b]. 2 k θ k TV η k (f,[;b]), 10
The proof will follow from the proper choice of (η k ) nd (θ k ). Since p 1 + q 1 > 1, we hve (q 1)(p 1) < 1. We choose ( ) α (q 1)(p 1);1, β sup f (t) f (), γ > 0 t b nd for k 0,1,..., define η k 1 β 2 (α2 /[(q 1)(p 1)]) k +1, By (18) we estimte θ k γ 2 (α2 /[(q 1)(p 1)]) k α/(q 1). η k 1 TV θ k (g,[;b]) β 2 /[(q 1)(p 1)]) (α2 k +1 ( V q (g,[;b]) γ 2 /[(q 1)(p 1)]) (α2 k α/(q 1) ) 1 q nd similrly 2 (1 α)(α2 /[(q 1)(p 1)]) k +1 V q (g,[;b])βγ 1 q, Hence S θ k TV η k (f,[;b]) γ 2 /[(q 1)(p 1)]) (α2 k α/(q 1) V p (f,[;b]) (β 2 /[(q 1)(p 1)]) ) (α2 k+1 1 p +1 2 /[(q 1)(p 1)]) (1 α)(α2 k α/(q 1)+1 p V p (f,[;b])β 1 p γ. 2 k η k 1 TV θ k (g,[;b])+ 2 k θ k TV η k (f,[;b]) ( + ) 2 k 2 /[(q 1)(p 1)]) (1 α)(α2 k +1 V q (g,[;b])βγ 1 q ( + ) + 2 k 2 /[(q 1)(p 1)]) (1 α)(α2 k α/(q 1)+1 p V p (f,[;b])β 1 p γ. Since α < 1 nd α 2 /[(q 1)(p 1)] > 1, we esily infer tht S < +, from which we get tht b the integrl fdg exists. Moreover, denoting { + C p,q mx 2 /[(q 1)(p 1)]) k+2 (1 α)(α2 k, 2 /[(q 1)(p 1)]) k+2 (1 α)(α2 k α/(q 1) p we get S 1 2 C p,q( V q (g,[;b])βγ 1 q +V p (f,[;b])β 1 p γ ). Setting in this expression γ (V q (g,[;b])/v p (f,[;b])) 1/q β p/q we obtin } S C p,q (V q (g,[;b])) 1/q (V p (f,[;b])) 1 1/q β 1+p/q p C p,q g q vr,[;b] f p p/q p vr,[;b] f 1+p/q p osc,[;b]. 11
Remrk 4 Let f, g, p, q nd C p,q be the sme s in Corollry 2. Using Theorem 2 insted of Theorem 1, we get the following, similr estimte ˆ b fdg f ()[g(b) g()] C p,q f p vr,[;b] g q q/p q vr,[;b] g 1+q/p q osc,[;b]. From Corollry 1 nd the obtined estimtes, we lso hve tht for ny ξ [;b] ˆ b fdg f (ξ)[g(b) g()] { 1+p/q p } f osc,[;b] 2C p,q f p vr,[;b] g q vr,[;b] min, g 1+q/p q osc,[;b]. f 1+p/q p p vr,[;b] g 1+q/p q q vr,[;b] Remrk 5 From Corollry 2, resoning in the similr wy s in [7, p. 456], we get the t following importnt estimte of the q-vrition of the function t fdg ˆ ( fdg C p,q f p p/q p vr,[;b] f 1+p/q p osc,[;b] + f,[;b] ) g q vr,[;b], q vr,[;b] ) (C p,q f p vr,[;b] + f,[;b] g q vr,[;b], where f, g, p, q nd C p,q re the sme s in Corollry 2. 3 Integrl equtions driven by modertely irregulr signls Let p (1;2). The preceding section provides us with tools to solve integrl equtions of the following form y(t) y 0 + ˆ t F(y(s))dx(s), (20) where x is continuous function from the spce V p ([;b]) nd F : R R is α-lipschitz. The functionl vr,p,[;b] : V p ([;b]) [0;+ ) defined s f vr,p,[;b] : f() + f p vr,[;b] is norm nd the spce V p ([;b]) equipped with this norm is Bnch spce. For our purposes it will be enough to work with the following definition of loclly or globlly α-lipschitz function when α (0;1]. For x (x 1,...,x n ) R n we denote x mx,...,n x i. Definition 1 Let F : R n R nd α (0;1]. For ny R > 0 we define its locl α-lipschitz prmeter KF α (R) s { } K (α) F (y) F (x) F (R) : sup y x α : x,y R n,x y, x R, y R nd its globl α-lipschitz prmeter K (α) F s K (α) F F will be clled loclly α-lipschitz if for every R > 0, K (α) F globlly α-lipschitz if K (α) F < +. : lim R + K F (R) < +. The function (R) < + nd it will be clled In the cse, when there is no mbiguity wht is the vlue of the prmeter α nd wht is the function F, we will write K F (R), K F, K(R) or even K. First we will consider the cse p 1 < α < 1. In this cse we hve the existence but no uniqueness result. We will obtin stronger result thn similr results [7, Lemm, p. 459] or [8, Theorem 1.20]. Nmely, we will prove tht there exists solution to (20) which is n element of the spce V p ([;b]), not only n element of the spce V q ([;b]) for rbitrry chosen q > p. This will be possible with the use of Remrk 5. 12
Proposition 1 Let p (1;2), y 0 R, x be continuous function from the spce V p ([;b]) nd F : R R be globlly α-lipschitz where p 1 < α < 1. Eqution (20) dmits solution y, which is n element of V p ([;b]). Moreover, y vr,p,[;b] R, where R > 0 stisfies the equlity R ( C p/α,p +2 ) K (α) F x p vr,[;b] Rα + y 0 + F(0) x p vr,[;b], with C p/α,p being the sme s in Corollry 2 nd Remrk 5. Now we proceed to the proof of Proposition 1. We will proceed in stndrd wy, but with the more ccurte estimte of Remrk 5 we will be ble to obtin the finiteness of vr,p,[;b] norm of the solution. Proof. Let f V p ([;b]). By [8, Lemm 1.18], F(f( )) V p/α ([;b]) nd since α/p+1/p > 1, we my pply Remrk 5 nd define the opertor T : V p ([;b]) V p ([;b]), Tf : y 0 + ˆ F(f(t))dx(t). Denote K K (α) F. Using Remrk 5 nd [8, Lemm 1.18] we estimte ˆ Tf vr,p,[;b] y 0 + F(f(t))dx(t) vr,p,[;b] ˆ y 0 + [F(f(t)) F(f())]dx(t) + F(f()) x p vr,[;b] p vr,[;b] ( ) y 0 + C p/α,p F(f( )) p/α vr,[;b] + F(f( )) F(f()),[;b] + F(f()) x p vr,[;b] ( (Cp/α,p y 0 + +1 ) ) F(f( )) + F(f()) p/α vr,[;b] x p vr,[;b] ( (Cp/α,p y 0 + +1 ) K f α p vr,[;b] ) x + F(f()) p vr,[;b]. (21) By the Lipschitz property, Denoting from (21) we get F(f()) K f() α + F(0) K f α vr,p,[;b] + F(0). A ( C p/α,p +2 ) K x p vr,[;b] nd B y 0 + F(0) x p vr,[;b], (22) Tf vr,p,[;b] A f α vr,p,[;b] +B. (23) For α < 1 let R be the lest positive solution of the inequlity R A R α + B (i.e. R { A R α + B). From (23) } we hve tht the opertor T mps the closed bll B(R) f V p ([;b]) : f vr,p,[;b] R to itself. 13
Now, for f,g V p ([;b]) we re going to investigte the difference Tf Tg. Agin, using Remrk 5 nd the Lipschitz property we estimte ˆ Tf Tg vr,p,[;b] [F(f(t)) F(g(t))]dx(t) p vr,[;b] ˆ [F(f(t)) F(g(t)) {F(f()) F(g())}] dx(t) + F(f()) F(g()) x p vr,[;b] p vr,[;b] ( C p/α,p +1 ) F(f( )) F(g( )) (p 1)/α p/α vr,[;b] F(f( )) F(g( )) (α+1 p)/α osc,[;b] x p vr,[;b] + F(f()) F(g()) x p vr,[;b] ( C p/α,p +1 ) ( ) (p 1)/α f K f α p vr,[;b] + g α p vr,[;b] g α+1 p osc,[;b] x p vr,[;b] +K f() g() α x p vr,[;b]. (24) From (24) we see tht T is continuous. Moreover, from the first inequlity in Remrk 4 nd the continuity of x we get tht functions belonging to the imge T(B(R)) re equicontinuous. Let U be the closure of the convex hull of T(B(R)) (in the topology induced by the norm vr,p,[;b] ). It is esy to see tht functions belonging to U re lso equicontinuous. Moreover, U B(R) (since T(B(R)) B(R) nd B(R) is convex) nd T(U) U (since T(U) T(B(R))). Now, let V T(U). From the equicontinuity of U, Arzel-Ascoli Theorem nd (24) we see tht the set V is compct in the topology induced by the norm vr,p,[;b]. Thus, by the fixed-point Theorem of Schuder, we get tht there exists point y U such tht Ty y. Now we will consider the cse α 1. Fct 1 Let p (1;2), y 0 R, x be continuous function from the spce V p ([;b]) nd F : R R be globlly 1-Lipschitz. Eqution (20) dmits solution y, which is n element of V p ([;b]). Proof. To prove the ssertion we my proceed in similr wy s in the proof of Proposition 1. The only thing we need is to ssure tht the inequlity R A R+B, whereandb re defined in disply (22), holds for sufficiently lrger. This my be chieved by splitting the intervl [;b] into smll intervls, such tht A < 1 on ech of these intervls, nd then solving eqution (20) on ech of these intervls with the initil condition being equl the terminl vlue of the solution on the preceding intervl. This is possible since for ny ε > 0 there exists δ > 0 such tht for ny [c;c+δ] [;b], x p vr,[c;c+δ] ε, which is the consequence of the fct tht the function [;b] t x p vr,[;t] is continuous nd we hve x p vr,[;c+δ] x p vr,[;c] + x p vr,[c;c+δ]. Acknowledgments I would like to thnk Rouf Ghomrsni for drwing my ttention to pper [9] nd Professor Terry Lyons for drwing my ttention to his pper [7]. Competing interests The uthor declres tht there re no competing interests regrding the publiction of this pper. References [1] Appel, J, Bnś, J nd Merentes, DNJ: Bounded vrition nd round. Series in Nonliner Anlysis nd Applictions 17. De Gruyter, Berlin (2013) [2] Dudley, RM, Norviš, R: Concrete Functionl Clculus. Springer Monogrphs in Mthemtics. Springer, New York (2010) 14
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