LIPSCHITZ RETRACTIONS IN HADAMARD SPACES VIA GRADIENT FLOW SEMIGROUPS

Transcription

1 LIPSCHITZ RETRACTIONS IN HADAMARD SPACES VIA GRADIENT FLOW SEMIGROUPS MIROSLAV BAČÁK AND LEONID V KOVALEV arxiv: v1 [mahfa] 7 Mar 016 Absrac Le Xn for n N be he se of all subses of a meric space Xd of cardinaliy a mos n The se Xn equipped wih he Hausdorff meric is called a finie subse space In his paper we are concerned wih he exisence of Lipschiz reracions r: Xn Xn 1 for n I is nown ha such reracions do no exis if X is he one-dimensional sphere On he oher hand L Kovalev has recenly esablished heir exisence in case X is a Hilber space and he also posed a quesion as o wheher or no such Lipschiz reracions exis for X being a Hadamard space In he presen paper we answer his quesion in he posiive 1 Inroducion Le Xd be a meric space For each n N we denoe he se of all subses of X wih cardinaliy a mosnbyxnthesexnequippedwihhehausdorffmericd H iscalledafinie subse space Unlie Caresian powers X n or he space of unordered n-uples X n /S n finie subse spaces admi canonical isomeric embeddings ι: Xn Xn+1 Following [6 7 9] we are ineresed in Lipschiz reracions r: Xn Xn 1 L Kovalev proved heir exisence for X being a Euclidean space [6] and X being a Hilber space [7] On he oher hand a resul of J Mosovoy [9] yields ha in general here is no coninuous mapping r: Xn Xn 1 wih r ι = id if X is he one-dimensional sphere S 1 ; here id sands for he ideniy operaor on Xn I is herefore naural o as wheher Lipschiz reracions r: Xn Xn 1 exis if X is a nonposiively curved meric space Indeed his quesion appears explicily in [7 Quesion 33] Our main resul Theorem 3 provides he posiive soluion o his problem The soluion for Hilber spaces from [7] is based on he exisence of gradien flow rajecories in a finie dimensional subspace which is assured by he classical ODE heory In he presen paper we also define he desired reracions via gradien flows of cerain convex funcionals on he n-h power of a Hadamard space and mae use of he Lie-Troer-Kao formula proved recenly in [3 11] Finally noe ha given a Hadamard space Hd we obain Lipschiz reracions r: Hn Hn 1 wih Lipschiz consan max4n 3 +1n +n 1 whereas he resul from [7] for X being a Hilber space has Lipschiz consan maxn 1n 3 Preliminaries We firs recall some basic facs abou Hadamard spaces as well as more recen resuls which shall be used in our proof For furher deails we refer he reader o [] Le Hd be a Hadamard space ha is a complee meric space wih geodesics saisfying 1 dzx 1 dzx 0 +dzx 1 1 dx 0 x 1 Dae: Ocober Mahemaics Subjec Classificaion Primary: 53C3 Secondary: 47H0 54E40 58D07 Key words and phrases Finie subse space gradien flow Hadamard space Lie-Troer-Kao formula Lipschiz reracion L V Kovalev was suppored by he Naional Science Foundaion gran DMS

2 M BAČÁK AND L V KOVALEV for each zx 0 x 1 H and [01] where x := 1 x 0 +x 1 is a unique poin on he geodesic [x 0 x 1 ] such ha dx 0 x = dx 0 x 1 An equivalen and more geomeric formulaion of inequaliy 1 is he following relaion beween a riangle wih verices p q r H and is comparison riangle wih verices pqr R where dpq = p q drq = r q and dpr = p r If x:= 1 p + q and y:= 1 sp+sr for some s [01] and we denoe heir comparison poins by x:= 1 p+q and y:= 1 sp+sr respecively inequaliy 1 implies ha dxy x y Here he symbol sands for he Euclidean norm on R Given wo geodesics [x 0 x 1 ] and [y 0 y 1 ] we have 3 dx y 1 dx 0 y 0 +dx 1 y 1 for each [01]; see [ 14] Given a funcion f: H ] denoe is domain by domf:= {x H: fx < } and he se of is minimizers by Minf:= {x H: fx = inff} We say ha a funcion f: H ] is convex if for each geodesic γ: [01] H he funcion f γ is convex Given a convex lower semiconinuous lsc for shor funcion is resolven wih parameer > 0 is given by 4 J x:= argmin y H and i saisfies he following imporan inequaliy [ fy+ 1 dxy ] x H 5 f J x+ 1 dxj x + 1 dj xy fy+ 1 dxy for every xy H Given x domf he gradien flow semigroup associaed o f is defined by x 6 S x:= lim J x domf for every [0 Lie in Hilber spaces he semigroup is comprised of nonexpansive operaors ha is 7 ds xs y dxy for each [0 and xy domf The above menioned heory of gradien flows in Hadamard spaces was firs sudied by J Jos [5] and U Mayer [8] A more recen resul from [1] esablished he asympoic behavior of a gradien flow I relies upon he noion of wea convergence in Hadamard spaces which was inroduced by J Jos in [4] Le us recall ha a bounded sequence x H converges wealy o a poin x H provided lim dp γ x x = 0 for each geodesic γ: [01] H wih x γ Here P γ sands for he meric projecion ono he image of γ Now we are ready o sae he heorem on asympoic behavior of a gradien flow Theorem 1 Le f: H ] be a convex lsc funcion which aains is infimum on H Then given x domf he associaed gradien flow semigroup wealy converges o a poin x Minf Proof See [1] or [ Thm 5116] The funcion values hen converge o he infimum of f Theorem Le f: H ] be a convex lsc funcion which aains is infimum on H Then given x domf we have f S x inff as Proof This can be seen from he proof of Theorem 1 in [1] or for insance lie in [ Prop 511]

3 LIPSCHITZ RETRACTIONS 3 The proof of our main heorem uses gradien flows of convex funcions o define a desired Lipschiz reracion These convex funcions have a special form namely hey are given as a finie sum of some elemenary convex funcions and heir gradien flow can be approximaed by he Lie-Troer-Kao formula We will now sae he necessary facs precisely Le N N and consider a funcion f: H ] of he form N 8 f:= f j j=1 where f j : H ] are convex lsc funcions for every j = 1N Le us denoe he resolven of he funcion f j by J [j] and he gradien flow semigroup of f by S Theorem 3 Lie-Troer-Kao formula Le f: H ] be of he form 8 Then we have x 9 S x = lim J [N] J [1] for every [0 and x domf Proof The original proof appeared in [11] For a simplified proof see [3] In fac he gradien flow we are going o use in he proof of Theorem 3 is no on H bu on is n-h power The n-h power of H denoed by H n is equipped wih he meric n 1 dxy:= dx j y j xy H n j=1 and is hen also a Hadamard space Noe ha we use he same symbol d for he original meric on H as well as for he meric on H n I is always clear from he argumens which one is mean Given a poin x = x 1 x n H n we shall denoe {x}:= {x 1 x n } a subse of H However given xy H n we shall denoe he Hausdorff disace beween {x} and {y} by d H xy insead of d H {x}{y} 3 The exisence of Lipschiz reracions The desired Lipschiz reracions r: Hn Hn 1 will be defined via a gradien flow of a convex funcional on H n Specifically we define his funcional as 10 Fx:= dx i x j x = x 1 x n H n 1 i<j n and show ha i is indeed convex and Lipschiz Lemma 31 The funcion F: H n R is convex and n 3 -Lipschiz Proof Convexiy follows from 3 For he Lipschiz propery we esimae Fx Fy dx i x j dy i y j dx i y i +dx j y j n 3 dxy 1 i<j n 1 i<j n where we wice used he riangle inequaliy and hen he Cauchy-Scharz inequaliy We are now ready o prove he main heorem Theorem 3 Le Hd be a Hadamard space Then for each ineger n here exiss a Lipschiz reracion r: Hn Hn 1 wih Lipschiz consan max4n 3 +1n +n 1

4 4 M BAČÁK AND L V KOVALEV Proof We divide he proof ino several seps Sep 1 Le J and S be he resolven and gradien flow semigroup respecively associaed wih he funcion F from 10 Le us denoe Given x H n we define D:= {x = x 1 x n H n : x i = x j for some 1 i < j n} 11 δx:= min 1 i<j n dx ix j and Tx:= inf{ > 0: S x D} We will firs show ha 1 Tx 1 δx In order o be able o apply he Lie-Troer-Kao formula 9 denoe by J [ij] wih he funcion x dx i x j x = x 1 x n H n he resolven associaed where 1 i < j n Then 9 reads 13 S x = lim where R := J [n 1n] J [16] J [45] J [35] R J [5] x x domf J [15] J [34] J [4] Nex we claim ha for > 0 and i = 3n he following holds: i if dy 1 y hen dz 1 z dy 1 y ii if dy 1 y < hen dz 1 z dy 1 y + J [14] J [3] J [13] J [1] for every y = y 1 y n H n and z := J [i] J [1i] y We will now show boh i and ii by using comparison riangles To his end denoe u:= J [1i] y and consider he riangle wih verices y 1 y u i in H along wih is comparison riangle wih verices y 1 y u i R Then denoe he comparison poins of z 1 and z by z 1 and z respecively Nex observe ha i and ii hold rue if we replace all he poins involved by heir comparison poins and consider he Euclidean disance in R of course This can be seen by elemenary geomery argumens in R Finally applying gives i and ii Choose x H n and N Denoe := δx Wihou loss of generaliy one may assume dx 1x = δx Define now x l[ij] := J [ij] J [3] J [13] J [1] for each l = 1 and 1 i < j n and observe ha i and ii imply 14 d x [1n] 1 x [n] = δx R l 1x where he subscrip indices denoe he coordinaes in H n Indeed each applicaion of J [1] shorens he disance beween he firs wo coordinaes by addiive consan while he applicaion of any oher resolven han J [1] does no expand i or expands i by addiive consan a mos as we now

5 LIPSCHITZ RETRACTIONS 5 from i and ii More precisely we have d dx 1 x = δx d x 1[1] 1 x 1[1] = max 0δx δx δx δx max δx d x [1] 1 x [1] max 0δx δx d x 1[n 1n] 1 x 1[n 1n] d x [n 1n] 1 x [n 1n] max x 1[n 1n] 1 x 1[n 1n] d x [1] 1 x [1] δx max = 0 δx δx δx and hence 14 holds rue Passing o he limi in 14 and recalling 13 hen give d 1 = 0 S1 δx x or in oher words we have jus proved 1 Sep Le xy H n By 5 we have S1 δx x δx 1δx = δx 1 dj xy Fy F J x+ 1 dxy Consider now [0Tx] Fix N and employ he above inequaliy -imes o obain d J xy [ ] Fy F J x +dxy xy d J [ ] x Fy F J +d J xy xy d J [ Fy F J x ] +d J Summing up hese inequaliies dividing by and puing x:= y gives xx x d J F x F J d n 3 and afer aing limsup we obain Hence by virue of 1 ds xx n 3 15 d H S xx ds xx n 3 δxn 3 x 1xy J x

6 6 M BAČÁK AND L V KOVALEV For fuure reference we also record ha he nonexpansiveness of he gradien flow semigroup7 implies 16 d H S xs y ds xs y dxy n 1 max 1 j n dx jy j Sep 3 Given x Hn we number is elemens {x 1 x n } and consider x := x 1 x n H n We may assume ha dx 1 x = δx Then we define rx:= { S Tx x } However we will wrie x insead of x in he sequel Le us now show ha r: Hn Hn 1 is a Lipschiz reracion Firs of all observe ha r is he ideniy on he canonical embedding of Hn 1 To prove he Lipschiz propery choose xy Hn and examine he following wo alernaives If δx+δy 4d H xy hen d H rxry d H rxx+d H xy+d H yry n 3 δx+dh xy+n 3 δy 4n 3 +1 d H xy where we used 15 o obain he second inequaliy If on he oher hand δx +δy > 4d H xy hen we may assume δx > d H xy wihou loss of generaliy The fac δx > d H xy hen implies ha we can renumber he poins {y 1 y n } in such a way ha 17 dx j y j d H xy for each j = 1n In he remainder of he proof we will use 17 only wihou referring o δx > d H xy We can hence wihou loss of generaliy assume Tx Ty on accoun ha he roles of x and y in 17 are inerchangeable Recall ha rx = S Tx x and pu z:= S Tx y Inequaliy 16 implies ha d H rxz n 1 d H xy Consequenly By 15 we have Finally one arrives a δz = δz δrx d H zrx n 1 dh xy d H zrz n 3 δz n d H xy d H rxry d H rxz+d H zrz+d H rzry n 1 dh xy+n d H xy+0 where he zero on he righ hand side is due o he semigroup propery of he gradien flow The proof is complee Remar 33 Asympoic behavior of he flow Given x H n denoe x:= max 1 i<j n dx i x j Using he same argumens as in he previous proof we can show ha for τ := 1 x one obains S τ x MinF Alernaively we can obain he asympoic behavior of he flow as follows By Theorem 1 given x H n he flow S x wealy converges o a poin x = x 1 x n MinF; and obviously x 1 = = x n Nex we show ha his convergence is in fac srong To his end we firs observe ha x 1 co{s x} [0 Indeed by virue of 6 and he semigroup propery i is sufficien o show ha {J x} co{x} This inclusion howeverfollowsdirecly by a projecion argumen Now use Theorem o conclude F S x 0 and herefore diamco{s x} 0 We hence have S x x Remar 34 Open quesions We end he paper by posing a few quesions many of which have appeared already in [7] The Lipschiz consan of r: Hn Hn 1 guaraneed by Theorem 3 is max4n 3 + 1n + n 1 Can one improve upon his consan? Can one show ha for every n N wih n here exis Lipschiz reracions r: Hn Hn 1 wih Lipschiz consans independen of n? Can one exend Theorem 3 ino spaces of nonposiive curvaure in he sense of Busemann? In paricular does an analog of Theorem 3 hold in sricly convex or uniformly convex Banach spaces?

7 LIPSCHITZ RETRACTIONS 7 Can one exend Theorem 3 ino p-uniformly convex spaces? Recall ha a geodesic meric space X d is called p-uniformly convex for p if here exiss K > 0 such ha dzx p 1 dzx 0 p +dzx 1 p K1 dx 0 x 1 p for each zx 0 x 1 H and [01] where x := 1 x 0 +x 1 This definiion was inroduced in [10 Definiion 3] as a generalizaion of p-uniform convexiy in Banach spaces References [1] M Bačá The proximal poin algorihm in meric spaces Israel J Mah pp [] Convex analysis and opimizaion in Hadamard spaces vol of De Gruyer Series in Nonlinear Analysis and Applicaions De Gruyer Berlin 014 [3] A new proof of he Lie Troer Kao formula in Hadamard spaces Commun Conemp Mah p pages [4] J Jos Equilibrium maps beween meric spaces Calc Var Parial Differenial Equaions 1994 pp [5] Nonlinear Dirichle forms in New direcions in Dirichle forms vol 8 of AMS/IP Sud Adv Mah Amer Mah Soc Providence RI 1998 pp 1 47 [6] L V Kovalev Symmeric producs of he line: embeddings and reracions Proc Amer Mah Soc pp [7] Lipschiz reracion of finie subses of Hilber spaces Bull Aus Mah Soc pp [8] U F Mayer Gradien flows on nonposiively curved meric spaces and harmonic maps Comm Anal Geom pp [9] J Mosovoy Laices in C and finie subses of a circle Amer Mah Monhly pp [10] A Naor and L Silberman Poincaré inequaliies embeddings and wild groups Compos Mah pp [11] I Sojovic Approximaion for convex funcionals on non-posiively curved spaces and he Troer-Kao produc formula Adv Calc Var 5 01 pp Max Planc Insiue for Mahemaics in he Sciences Inselsr Leipzig Germany address: baca@mismpgde 15 Carnegie Syracuse Universiy Syracuse NY 1344 USA address: lvovale@syredu