WHEN IS A FUNCTION NOT FLAT? 1. Introduction. {e 1 0, x = 0. f(x) =

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1 WHEN IS A FUNCTION NOT FLAT? YIFEI PAN AND MEI WANG Abstrct. In this pper we prove unique continution property for vector vlued functions of one vrible stisfying certin differentil inequlity. Key words: unique continution, Crlemn s method. The function is well known for its property. Introduction fx) = {e x, x, x = f k) ) =, k but f. Such function is clled flt t the origin. On the other hnd, if f is rel nlytic function with Tylor expnsion on n open intervl contining, then f k) ) =, k implies f. The unique continution problem in PDE is to find conditions such tht the solutions of PDE enjoy the sme property. There re lrge mount of literture in this re originted from the ides by Crlemn [], clled Crlemn s method. In this pper we consider the simplest cse of one vrible. The following is the min theorem. The results cn be generlized to complex or vector vlued function cse Theorem 5). The proofs re in the subsequent sections. Theorem. Let fx) C [, b]), [, b], nd ) f n) x) C n k= for some constnt C nd n. Then, x [, b] f k) ) =, k implies f on [, b]. Remrks. The conditions in Theorem re optiml in the following sense: the infinite order vnishing f k) ) =, k cn not be relxed e.g. fx) = x N ), nd the singulrity order in ) is shrp, s illustrted in the exmple given fter Corollry. From Theorem we obtin the following corollry.

2 YIFEI PAN AND MEI WANG Corollry. Let fx) C [, b]), [, b], nd ) hold for some constnt C with n. Then f implies the zero set {f )} [, b] is finite. An exmple. We use n exmple in Groflo nd Lin [] to show tht the order of vnishing in ) is shrp in the following sense. There exists function fx) C [, ]), >, f n) x) C n k= +ε for x [, ] nd f k) ) =, k for some constnt C nd ε >, but f on [, ]. For m >, ε >, the following eqution is considered in []: or equivlently, x u x) + mxu x) cx ε ux) =, x, ), ) u x) + m x u x) c ux) =, x, ), x+ε Bessel differentil eqution [4]. The generl Bessel differentil eqution tkes the form 3) z u z) + α)zu z) + { β γ z γ + α ν γ ) } uz) =, z C. It is well known tht for non-integer) ν Z, the solution for 3) is uz) = z α [ C J ν βz γ ) + C J ν βz γ ) ], where C, C re rbitrry complex numbers, nd J ν is the Bessel function of order ν, i.e., solution of eqution 3) with α =, γ =. Notice tht eqution ) with c > is eqution 3) with α = m, β = i c ε, γ = ε, ν = m. ε By choosing m >, ε, ) such tht C = C = πe iνπ sinνπ), ν = m Z, ε the solution of eqution ) cn be written s ) c 4) ux) = x m )/ K m )/ε ε x ε/, x, ) where K ν z) = π e iνπ J ν iz) J ν iz)), rg z π, π/) sinνπ) is the modified Bessel function of the third kind [4], with the symptotic property K ν x) π x / e x s x +.

3 WHEN IS A FUNCTION NOT FLAT? 3 Therefore in 4), the function ux) π ) / { c x m + ε4 exp } c ε ε x ε/ s x is nontrivil solution of ) vnishing t x = of infinite order. Hence fx) = u x ), x [, ],, ) is well defined nd f C [, ]). Tking derivtives of eqution ) n times, d n { n u x) + m x u x) c } x +ε ux) =, x, ), we obtin u n) x) + n x)u n ) x) + + x)ux) =, x, ). For given m, c nd ε, the coefficients ) j x) C o, j =,,, n, x, ) x n j+ε for some constnt C o >. By the property of ux) ner x =, we hve with f n) x) + n x)f n ) x) + + x)fx) =, x [, ] f k) ) =, k, for some constnt C >. Thus f k) ) =, k, j x) C nd f from the non-trivility of u. f n) x) C x n j+ε n k= ), j =,,, n. +ε, x [, ], Theorem leds to pplictions in ODE s stted in the following two propositions. Bsed on the bove exmple, the order of the singulrity of the coefficients in the ssumption of the propositions is shrp. Proposition 3. Let fx) C be solution of with Then y n) + n x)y n ) + + x)y =, x [, ], > ) k x) = O s x, k =,,, n. f k) ) =, k = f on [, ].

4 4 YIFEI PAN AND MEI WANG Proposition 4. Let fx), gx) C be solutions of with Then y n) + n x)y n ) + + x)y = bx), x [, ], > ) k x) = O s x, k =,,, n. f k) ) = g k) ), k = f g on [, ]. The result in Theorem cn be generlized to complex or vector vlued functions s stted in the following theorem. Theorem 5. Let [, b], fx) = ux) + ivx), where ux), vx) C [, b]), or fx) = u x),, u d x)) for some d, where u x), u x),, u d x) C [, b]). Assume 5) f n) x) C n k= for some constnt C nd n. Then, x [, b] f k) ) =, k implies f on [, b]. Remrks. Theorem 5 hs interesting geometric menings. A d-dimensionl curve u t),, u d t)), d, stisfying 5) my hve t most finite order singulrity t t =, which mens there must be t lest one component u i with u k) i ) for some k.. Proof of Theorem nd its corollry Severl lemms re needed for the proof of Theorem. The bsic ide of the following lemm ws considered in [3]. Lemm 6. Let vx) C [, b]). Assume v k) ) =, k. Then for α, 6) Proof. d [vx)] x α+ 4 α + ) [v x)] ) x α+) v x) = α + )x α+) v x) + x α+) vx)v x) Since v k) ) = for k [α/] +, we hve [vx)] Ox k+) ) = ox α+ ), thus d ) x α+) v x) = [vb)] b α+ lim [vx)] x + x α+ = [vb)] b α+. x α

5 Therefore, α + ) v x) xα+ = WHEN IS A FUNCTION NOT FLAT? 5 vx)v x) x α+ { α ) + / vx) α + v x) + xα+ x α+)/ α + } { α + v x)) by pplying b + b to the lst inequlity. Consequently, α + which is equivlent to 6). v x) xα+ α + x α v x)), x α ) } / v x) x α/ Lemm 7. Let ux) C [, b]). Assume u k) ) =, k. Then for β, n, [ u k) x) ] x β+n k) 4 b [u n) x)] β + ) x β, for k =,, n. Proof. Applying Lemm 6 to vx) = u n j) x), α = β + j, β, j =,,, n, we obtin [ u n j) x) ] 4 b [u n j) x)] x β+j+ β + j + ) x β+j, j =,,, n, or equivlently, [ u k) x) ] x β+n k) 4 β + n k ) + ) m=k [u k+) x)], xβ+n k ) k =,,, n. Applying the bove inequlity repetedly, we hve [ u k) x) ] { n } xβ+n k) 4 4 b [u n) x)] β + n m) + ) β + ) x β 4 β + ) [u n) x)], becuse β, β + n m ) + for m =,..., n. x β Lemm 8. Let ux) C [, b]). Assume u k) ) =, k. Then for β, n, b n x β k= ) u k) x) x n k 4n β + ) x β [un) x)].

6 6 YIFEI PAN AND MEI WANG Proof. Apply Lemm 7 to k =,, n nd sum up both sides. Lemm 9. Let fx) C [, b]), [, b]. Assume f k) ) =, k. Then for β, n, ) b n f k) x) 4n b x β x n k β + ) x β [f n) x)]. k= Proof. The function ux) = f x) stisfies the ssumptions for Lemm 8 on [, ], so ) n f k) x) 4n x β x n k β + ) x β [f n) x)]. k= Substitute the vrible x by x. Since the terms in the sum re of even powers nd both sides hve the sme x β term, the bove inequlity cn be written s ) n f k) x) 4n x β x n k β + ) x β [f n) x)]. k= From Lemm 8 the desired inequlity is lredy true for fx) on [, b]. Combining the results on [, ] nd [, b], Lemm 9 follows. The following is the proof of Theorem. Proof. fx) stisfies the ssumptions in Lemm 9 on [, b], so for ny β, β + ) ) b n f k) x) b 7) 4n x β x n k x β [f n) x)]. From ), k= ) n f n) x) C for some C >, thus ) b n 8) x β f n) x) C x β. Combining 7) nd 8) we hve b n f k) x) x β x n k k= k= k= ) 4nC β + ) x β n k= ).

7 WHEN IS A FUNCTION NOT FLAT? 7 If f on [, b], we would hve > on some sub-intervl of [, b] thus the integrls >, which would imply 4nC, β = Contrdiction. β + ) Therefore fx) on [, b]. This completes the proof of Theorem. The proof of Corollry follows immeditely. Proof. Under the ssumption of Corollry, f implies f k) ) = β for some k bsed on the result of Theorem. If β >, then f C yields f k) x) > β/ >, x [ δ, δ ] [, b] for some δ >. Consequently the k-fold integrl fx) = x x f k) y)dy dy > x k β/ >, < x < δ. The cse β < implies fx) < on n open intervl contining. For ny x [, b] such tht fx ) =, the condition f implies fx), < x x < δ x, x [, b] for some δ x >. By the compctness of [, b], the zero set {f )} is t most finite in [, b]. This completes the proof of Corollry. 3. Proof of Theorem 5 Proof. First consider the complex vlued cse fx) = ux) + ivx), ux), vx) C [, b]). Then f k) ) =, k implies u k) ) =, v k) ) = k. Therefore both ux) nd vx) stisfy Lemm 9 on [, b], so for ny β, β + ) ) b n u k) x) b 4n x β x n k k= x β [un) x)], β + ) ) b n v k) x) b 4n x β x n k x β [vn) x)]. Since k= f k) x) = u k) x) + v k) x), k, we hve inequlity 7) for fx) : β + ) n f k) x) ) 4n x β x n k k= x β [f n) x)].

8 8 YIFEI PAN AND MEI WANG The rest of the proof is the sme s in tht for Theorem. From the ssumption in Theorem 5 we hve ) n f n) x) C for some C >, thus x β f n) x) C k= x β Combining the bove we hve n f k) x) ) 4nC x β x n k β + ) k= n k= x β ). n k= ). If f on [, b], we would hve > on some sub-intervl of [, b] thus the integrls >, which would imply 4nC, β = Contrdiction. β + ) Therefore fx) on [, b]. The proof for vector vlued cse is immedite, becuse f x) k) = u k) x) u k) + + d x), k holds. This completes the proof of Theorem 5. References [] T. Crlemn 938) Sur un problème d unicité pour les systèmes d équtions ux dérivées prtielles à deux vribles indépendntes, Arkiv Mt., Astron. Fysik, 6B7), -9. [] N. Groflo nd F. Lin 986) Monotonicity properties of vritionl integrls, A p weights nd unique continution, Indin Univ. Mth. J. 35), [3] Y. Pn 99) Unique continution of Schrödinger opertors with singulr potentils, Communictions in Prtil Differentil Eqution 7, [4] G. N. Wtson 944) A Tretise on the Theory of Bessel Functions, nd ed. Cmbridge University press. Yifei Pn, Deprtment of Mthemticl Sciences, Indin University - Purdue University Fort Wyne, Fort Wyne, IN School of Mthemtics nd Informtics, Jingxi Norml University, Nnchng, Chin E-mil ddress: pn@ipfw.edu Mei Wng, Deprtment of Sttistics, University of Chicgo, Chicgo, IL 6637 E-mil ddress: meiwng@glton.uchicgo.edu