More metrcs on cartesan products If (X, d ) are metrc spaces for 1 n, then n Secton II4 of the lecture notes we defned three metrcs on X whose underlyng topologes are the product topology The purpose of ths note s to explan how one can nterpolate a contnuous famly of metrcs between these examples; for each such metrc, the underlyng topology wll be the product topology Throughout ths dscusson p 1 wll denote a fxed real number Let x, y X, express them n terms of coordnates as (x 1,, x n ) and (y 1,, y n ) respectvely, and defne d p from ( X ) ( X ) to R as follows: d p (x, y) = d (x, y ) p ) 1/p The cases where p = 1 or 2 were consdered n the lecture notes It follows mmedately that d p satsfes all the propertes for a metrc except perhaps the fundamentally mportant Trangle Inequalty The latter s n fact a consequence of the followng basc result: Mnkowsk s Inequalty Suppose that we have u, v R n and we wrte these vectors n coordnates as (u 1,, u n ) and (v 1,, v n ) respectvely Then we have u + v p ) 1/p u p ) 1/p + v p ) 1/p Here are some references for a proof of Mnkowsk s Inequalty: W Rudn, Real and Complex Analyss (Thrd Edton Mc-Graw-Hll Seres n Hgher Mathematcs) McGraw-Hll, Boston-etc, 1987 ISBN: 0-07-054234-1 http://wwwplanetmathorg/encyclopeda/mkowskinequaltyhtml The ncorrect spellng Mkowsk needed to reach the planetmath lnk should be noted; the latter also gves further lnks to the proof of the nequalty, the statement and proof of the closely related Hölder Inequalty, and a statement and proof of the Young Inequalty whch can be used to prove Hólder s Inequalty; n fact, one generally begns by provng Hölder s Inequalty (ether as n the planetmath lnks or by some other means) and then derves Mnkowsk s nequalty from Hölder s Inequalty Hölder s Inequalty q > 1 such that Then we have Suppose that we have u, v R n as above wth p > 1, and that we choose ) u v 1 q + 1 p = 1 u p ) 1/p v q ) 1/q The planetmath references also contan a sequence of lnks to Hölder s nequalty and related facts whch can be used bo gve a self-contaned proof of the two gven nequaltes and some other basc results 1
Snce each of the metrcs d p for p = 1, 2, defnes the product topology, t s natural to speculate that the same holds for all choces of p, and n fact ths s true PROPOSITION For each p 1, the topology determned by the metrc d p s the product topology Furthermore, the dentty map from ( X, d α ) to ( X, d β ) s unformly contnuous for all choces of α, β such that 1 α, β Proof It suffces to prove the asserton n the second sentence, and the latter reduces to the specal case where one of α, β s ; f we know the result n such cases, we can retreve the general case usng the unform contnuty of the dentty mappngs ( ) ( ) ( ) X, d p X, d X, d r and the fact that a composte of unformly contnuous maps s unformly contnuous The unform contnuty statements are drect consequences of the followng nequaltes for nonnegatve real numbers u for 1 n: ( max { u } u p ) 1/p n max { u } One can then apply the argument n the notes to show that the dentty maps ( ) ( ) ( ) X, d X, d p X, d are unformly contnuous (and n fact the δ correspondng to a gven ε can be read off explctly from the nequaltes!), and of course all compostes of maps from ths dagram are also unformly contnuous The lmtng case The followng result s the motvaton for settng d equal to the maxmum dstance between coordnates: PROPOSITION In the settng above we have d = lm p d p Proof Ths reduces mmedately to provng the followng result: If u R n as above then max { u } = lm p ( u p ) 1/p Let M denote the expresson on the left hand sde, and for each p > 1 let Y p denote the value of the expresson whose lmt we wsh to fnd Clearly M Y p for all p because M s obtaned by deletng all but one summand from Y p However, snce u M for all, we also have Y p (n M p ) 1/p = M n 1/p Now the lmt of the rght hand sde as p s equal to M, and 2
thus we have sandwched Y p between two expressons, one of whch s equal to M and the other of whch has a lmt equal to M It follows that the lmt of Y p s also equal to M, whch s exactly the clam n the proposton If one graphs the set of all ponts n R 2 whose p-dstance from the orgn s equal to 1 for varous values of p 1, the result s a collecton of closed curves centered at the orgn such that the area enclosed by the curve ncreases wth p and the lmt of these curves s the boundary of the square whose vertces are the elements of the set {±1} {±1} Refned estmates It s straghtforward to check that the sold unt dsk n R 2 wth respect to the d 1 metrc s the sold square regon whose vertces are (±1, 0) and (0, ±1), whle the sold unt dsks n R 2 wth respect to the d 2 and d metrcs are (respectvely) the usual round unt dsk and the sold square [ 1, 1] [ 1, 1] In partcular, f α < β then the unt dsk wth respect to d β strctly contans the unt dsk wth respect to d α Lkewse, the correspondng unt dsks n R 3 are the sold octahedral regon wth vertces (±1, 0, 0), (0, ±1, 0) and (0, 0, ±1), and the sold cube [ 1, 1] 3, wth each set properly contaned n the next There are analogous statements n all hgher (fnte) dmensons We shall generalze these observatons to arbtrary metrcs d α and d β where 1 α < β THEOREM Let α and β satsfy 1 α < β, and suppose that n 2 Then the sold unt dsk n R n wth respect to the metrc d β strctly contans the analogous dsk wth respect to the metrc d α Of course, f n = 1 then all the analogous dsks are the same Proof As usual, let α and β denote the α- and β-norms on R n The frst step s to show that f β > α and x α 1 then we also have x β 1 Snce x p only depends upon the absolute values of the coordnates of x, t suffces to consder the case where all the coordnates of x are nonnegatve Furthermore, snce x α = x β f x s a multple of a unt vector, t wll suffce to prove the result when at least two of the coordnates of x are nonzero, n whch case t follows that all the (absolute values of the) coordnates are all strctly less than 1 CLAIM: If x s as above and x α r 1 then for all β > α we have x β < r To prove the clam, consder the functon where p 1 For all p 1 we have N x (p) = ( x p 1 + + xp k ) (1/p) N x(p) = 1 p ( n =1 x p ) 1 p p x j 0 (log e x j ) x p j and the rght hand sde s negatve because () the values x p are all nonnegatve but less than 1, () at least two of the numbers x are postve, so that the assocated logarthmc coeffcents are negatve and the terms x p are postve Therefore by the Mean Value Theorem we know that N x s a strctly decreasng functon for p 1 Ths mmedately proves the clam f β < In the remanng case where β = we know that x = lm p N x (p) and snce N x s strctly 3
decreasng t follow sthat the lmt value s strctly less than x p for all p such that 1 p Ths completes the proof of the clam In partcular, the precedng dscusson shows that f x α 1 then x β 1 so that the unt dsk wth respect to d α s contaned n the unt dsk wth respect to d β To prove the statement about strct contanment, let x be the vector whose frst two coordnates are 2 α and whose remanng coordnates are zero, so that x α = 1 If we let b = x β, then b > 0 and by the precedng dscusson we know that b < 1 The basc propertes of norms now mply that b 1 x β = 1 whle b 1 x α = b 1 > 1, and therefore t follows that b x les n the unt dsk wth respect to d β but not n the unt dsk wth respect to d α, provng that the unt dsk wth respect to the frst metrc strctly contans the unt dsk wth respect to the second A fgure llustratng the frst quadrant portons of some unt dsks wth respect to d p metrcs appears n the fle dpuntdskspdf The precedng results for the d p metrcs on R n generalze mmedately to other products It wll be convenent to ntroduce a the followng property for metrc spaces Defnton Let ε > 0 A metrc space (X, d) s sad to be ε-weakly saturated at x X f for all δ [0, ε] there s a pont y X such that d(x, y) = δ Clearly a normed vector space determnes a weakly saturated metrc wth respect to every pont, but a set wth the dscrete metrc does not In practce, many nterestng examples of spaces satsfy weak saturaton condtons For example, f the underlyng topologcal space X s connected n the sense of Unt III and contans more than one pont, then for each x X one can fnd some ε(x) > 0 such that x s ε(x)-weakly saturated at x; a proof s gven at the end of ths document THEOREM Suppose that we are gven metrc spaces (X, d ) for 1 n, let 1 α < β, and let d α and d β be the assocated product metrcs on X Then for all x and y n the product we have the relaton d α (x, y) r 1 = d β (x, y) r Furthermore, f x = (x 1,, x n ) and for each the metrc for X s 1-weakly saturated at x, then there are ponts x, y X such that d β (x, y) = 1 but d α (x, y) > 1 Proof The dsplayed relaton follows mmedately from the precedng theorem To prove the second, t suffces to fnd a pont y = (y 1,, y n ) such that d 1 (x 1, y 1 ) = d 2 (x 2, y 2 ) = 2 β and y = x for all 3 (ths s an empty condton f n = 2) Then the argument n the prevous theorem mples that the β-dstance from x to y s 1 but the α-dstance s strctly greater than 1 Saturaton and connectedness We shall conclude by gvng smple condtons under whch a metrc satsfes saturaton hypotheses at each pont Ths dscusson nvolves the concept of connectedness, whch s ntroduced n Unt III The man facts about connected spaces that we shall need are () a subset C of the real lne s connected f and only f for all x, y C such that x < y and all z such that x < z < y we have z C, () f f : X Y s contnuous and X s connected then f[x] s also connected Gven a metrc space (X, d), we shall also need and use the basc contnuty propertes of functons defned n terms of d PROPOSITION Let (X, d) be a connected metrc space consstng of more than one pont () For each x X there s some ε x > 0 such that X s ε x -weakly saturated at x 4
() If the metrc d s unbounded (n other words, ts mage s not a bounded subset of R), then for all x X and all ε > 0 the metrc space (X, d) s ε-weakly saturated at x Proof () Let x X be arbtrary, and let y X be such that y x Take ε x = d(x, y) > 0, and let h : X R be the (contnuous) functon h(u) = d(x, u) Then we know that h[x] s a connected subset of R Snce h(x) = 0 and h(y) = ε x, by connectedness we know that h[x] must contan the entre nterval 0, ε x ] () By the precedng argument t suffces to show that f ε > 0 and x X, then there s some pont y X such that d(x, y) ε Suppose ths s false for some partcular x X and ε > 0, so that d(x, y) < ε for all y The unboundedness asserton for the metrc mples that there are ponts u, v X such that d(u, v) > 2ε By prevously derved consequences of the Trangle Inequalty we then have d(x, u) d(u, v) d(x, v) 2ε ε = ε whch contradcts our hypothess that d(x, w) < ε for all w X Ths yelds the statement at the begnnng of the paragraph 5