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# Review Sheet for MATH 527A at UA

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Date Created: 02/06/15

NORMED LINEAR SPACES AND DUALS SHANKAR VENKATARAMANI Review Dr Flaschka7s notes for the de nition of linear space and norm 1 LINEAR MAPS Given two vector spaces V and W over a eld F a map f V a W is a linear map if fau vafu fv7 W756 UWEV In order to de ne boundednedd for such linear rnaps we need additional structure on the vector spaces V and W In particular we can de ne the notion of a bounded linear map if V and W are both norrned linear spaces Remark 1 A linear map f V a W is bounded if 30 lt 00 such that HfiHW S OHDllvVZ E V Note that not every linear map is bounded if V or W is in nite dimensional in contrast to the situation for maps between nite dimensional norrned linear spaces For example consider the derivative operator as a linear map between 010 1 R and C01R where we use the sup norm for both the vector spaces lf fnz sinn7rzn is a sequence of functions in the domain 01 then llfnlloo 171 a 0 so that fn a 0 However 7139 for all n so that fl 74gt 0 showing that the map f gt gt f is unbounded In class we have used the symbol 8V W to denote all bounded linear rnaps between V and W This agrees with the de nitions used in the book by Hunter and Nachtergaele henceforth The set 8V W inherits the structure of a linear space from W lf W is a vector space over F and fg E 8V Woz E F we de ne af g to be the map 1 gt gt afo go for all i E V It is easy to check that of g is also a bounded linear map In fact 8V W is a norrned linear space where the induced norm is de ned by WWW sup W Hvuweo llvllv The fact that this operation de nes a norm is proved for instance in Prop 2340 on Pg l 76 of the class notes Remark 2 There are other equivalent de nitions of this norm A very useful eccercise is to show that sup Mam eRl HfvllwOHvHv WV Hvuweo llvllV sup HfWHW HEW1 sup Hlelw llvllv l 1 2 SHANKAR VENKATARAMANI A particular case of this general construction is of prticular interest7 the case where W R Assume that V is a normed linear space over R Then the space V 3V7 R is called the dual of V7 and it consists of all bounded7 real valued7 linear functionals on V For example7 if V is C07ll7R with the sup norm7 then the maps f gt gt f05 and f gt fol z2fxdx are indeed elements of the dual space V However if V is the linear space CO7 17R with the L1 norm7 then the map f gt gt f05 is linear but unbounded7 and hence does not belong to the dual V 2 SEMINORMS V is a vector space over reals A seminorm is a non negative real valued function p V a 07007 that satis es all the requirements on a norm7 except the condition 131 0 gt v 0 ln particular7 p needs to satisfy the triangle inequality Remark 3 Note that every norm is also a serninorrn A common scenario in which seminorm7s arise is given by the following proposition Proposition 1 IfL V a R is a linear map then the function de ned by 131 lL1l is a serninorrn Proof Clearly 131 2 0 for all v E V AlSO71Oz1 lLavl laLvl la 131 Finally7 10111 112 lL 1 2N lL 1 L 2l S lL11l lL12l 1001 10 D Remark 4 Note that we don t need a map into R If L V a W is a linear map and p is a seminorm on w then 131 pL1 is also a serninorrn and this fact is proved in emactly the same manner as above Example 1 Show that 2 pm lf05l2 1 ftdt 0 is a serninorrn on CO7 We can de ne a linear map from C071 to R2 by 1 f H Lf lt 0 ftdt7f05 consequentb llLfH2 is a seminorm Proposition 2 V is a vector space over a eld F Given afarnily of serninorrns 130470 E A it follows that 191 suppaW 04614 is a serninorrn on a vector space V de ned by V 1 E Vl sup13a1 lt 00 046A Remark 5 You ve proved this in your homework NORMED LINEAR SPACES AND DUALS 3 Remark 6 The proposition has two distinct aspects which require proof Firstly we are constructing a space V which we must show is indeed a vector space After that we need to show that the function p which is clearly well de ned on V is a seminorm The proof for the second part has emactly the same ideas as the proof in the class notes that the induced norm is indeed a norm Proof We rst show that V is indeed a vector space Assume that 111 112 E V and c1 c2 are scalars in the underlying eld F Then ow1 yg E V because V Q V and V is a vector space For each oz 6 A we have Palt01111 0202 Spa01111 Palt02112 lcllpa 1 lczlPaWz Sl61lsuplp v1l lczlsuplmvz 96A 76A lCilP11 lczlPQz lt 00 Since this is true for each oz 6 A we see that sugpdcwl 62112 S lollpm lczlpvz lt 00 046 implying that c1111 egg 6 V This proves that V is a vector space Also the last equation implies that 100111 0202 S lCllP 1 lCleW for all 0102 6 V c1 c2 6 F Taking c1 c2 1 yields the triangle inequality The other requirements for p to be a seminorm are easily veri ed and this proves that p is indeed a seminorm D A natural question at this stage is under what circumstances is p a norm This is answered by the following proposition Proposition 3 Using the same notation as above p is a norm on V if for every 1 31 0 in V there is an oz 6 A such that paw gt 0 The proof is straightforward since if i 31 0 and i E V Q V we see that 191 2 19041 for all oz 6 A implying that pi gt 0 so that pz 0 gt z 0 The following examples illustrate this strategy for constructing norms Example 2 Let Lx x fori 12 n be n linear functions from R to R such that L ecctracts the ith component of a point x E R By an earlier result the functions px are seminorms Also the function supidl 2 n is a norm since for every x 31 0 in R at least one of its components is non zero which means at least one of the p is positive Example 3 Let V M2X2R be the set of all 2 gtlt 2 matrices with real entries For all x abT E R2 we can de ne a linear map from V to R2 by Q E V gt gt Qx E R2 Consequently it follows that for each x E R2 the mapping pxQ HQxHZ is a semi norm 4 SHANKAR VENKATARAMANI For instance7 if we take x 17 71T7 then the above construction yields the mapping 2 ZHllaibicidll2 07b2cid2 which is indeed a serninorrn Note that7 this is not a norm because the matrix 1 1 Q i s s is nonzero7 but p11Q 0 If we consider the set of all x 6 R27 it is easy to see that SUP HQXllz 00 XER2 for all Q 31 0 Thus it doesnt yield a useful norm on the space of matrices However7 if we restrict ourselves to the set x E R2 l HxHZ 17 we see that pQ sup pxQ sup MCtz llxll21 lxll21 is indeed a serninorrn that is de ned on all of ngg This is also a norrn7 because Q 31 0 implies either Q071T 31 0 or Q17 0T 31 07 whence it follows that pQ gt 0 Note that7 this norrn p is precisely the induced norm on the space of 2 gtlt 2 matrices considered as a space of linear rnaps between R2 with the Euclidean norm and itself 3 INDUCED NORMS AND DUALITY As before7 we de ne 3V7 W as the space of all bounded linear rnaps between V and W with respect to the induced norrn f 1 W HfllwW sup M Hvaeo llvllv In light of the above discussion7 we can also interpret the induced norrn as follows Every element 1 E V is a linear map from V7 W to W by f l7Wl gtf1EW Consequently7 the mapping pv V7 W a 0700 de ned by pvf HfvHW is a serninorrn on V7 We can now construct a serninorrn by taking the suprernurn of pv over an appropriate set of vectors 1 If we take the suprernurn over all 1 6 V7 it is easy to see that the suprernurn is in nity unless f 07 so this is not a useful construction Another idea is to take the suprernurn only over those 1 E V with HUHV 1 This will de ne a serninorrn7 but not necessarily on V7 Rather7 the serninorrn is only de ned on the set f E V7 W l Hs t lp1 llf llW lt 00 v This is precisely the space that we denoted by 3V7 W7 that is7 all linear maps with the property that there is a C lt 00 such that Hfvllw 0 V1 6 Vsuch that WV 1 NORMED LINEAR SPACES AND DUALS 5 This justi es among other things why 8V W is itself a linear vector space and is the set on which the induced norrn was originally de ned Note that the serninorrn de ned by W sup llfvllw Hvll1 is indeed a norm since fo 0 for all 1 implies that f 0 that is for every f 31 0 there is a i such that fo 31 0 and setting u gives an element u with 1 such that fu 31 0 The particular case where the range space of the linear maps is R gives the dual of V that is V 8V R Given a concrete space V the problem of nding the dual V is often solved in two steps rst nding the appropriate set of linear maps that de ne V and then nding the induced norm for the dual I will illustrate this below with a few examples Example 4 V R Find all the linear maps 15 V a R Remark 7 Note that we haven t de ned any norm on V The answer to this question is seemingly independent of the norm that we choose to impose on V Going back to the question posed in the example let e1e2 en be the standard basis for R Then for any x 12 xn E R we have x xlel xgez pnen Consequently if f R a R is a linear function we have fx z1fe1 z2fe2 pnfen alpl a2p2 anxn ax where a fei and a a1a2 an E R Conversely given any point b E R it is easy to see that 9X b X does indeed de ne a linear function g R a R Consequently we see that the set of linear functions from R to R is in a one to one correspondence with R and we can indeed represent all linear functionals on R as xgt gtaxfor sorneaE R As discussed in class an alternative and sometimes useful way to think of linear functions is in terms of its level sets If f R a R is a non trivial function ie not identically zero then the set f 10 ie the kernel of the linear map f de nes a n 7 1 dimensional hypersurface in R Likewise the family of hypersurfaces f 1c c E R gives a foliation of R by parallel hypersurfaces that correspond to the level sets ofthe function f Given this family of hypersurfaces it is of course easy to recover the corresponding rnap z gt gt a K using the results a feii 12 n Remark 8 We have addressed the question of the representation of all linear functions on a vector space For V R it is easy to answer this qiestion but for other in nite dimensional spaces the answers can be surprisingly complicated Example 5 Find the dual of the R l with 1 S p 3 oo 6 SHANKAR VENKATARAMANI By the preceding argument7 if f is a linear map from R to R there is a vector 21 E R such that fv av Zeal111 i1 for all V E R Conversely7 every 21 E R does induce a bounded linear map from R l to R by x gt gt a x since Ho39lder7s inequality implies that laXl S llxllpllallq where p 1 1 1 We have thus identi ed the set of linear maps in the dual Now we have to calculate the induced norm For fx aX7 la Xl llxllp so that Hfllmdmed S Hallq Also7 given a 31 0 E R set S Hallq Signwi 0 17laz l llalloo 0 PLWVWMW signailailPp 1 1 lt p lt oo signai p 00 In each of these cases7 it is easy to check that7 with these choices la Xl Hxllp implying that Hfllmdmed 2 Hallq CHECK THESE CLAIMSE Combining the two inequalities7 we see that the induced norm is the lq norm7 so that the dual space is V an7 lg with 1 1 1 ip l Hallm Example 6 IfV R3 with a norm Hz1727z3lla lz1l x x determine the dual V5 The same argument as above shows that7 as a set7 V R3 we need to calculate the induced norm on R3 to determine the dual The unit ball in Ha is given by the set l1lwz z 1 For a17a27a3 E R3 the induced norm is Nah 027asllv SUP la i 02952 13ml lleq mgm 1 lt mp imadee ee lml lmgm 1 mwwm ee NORMED LINEAR SPACES AND DUALS 7 where we have used the Cauchy Schwarz inequality in the second line7 and lal lJmag 3 3 1 in obtaining the last line This suggests that the norm on the dual is 11017 127 13Nlmduced maxltlall7 0 a To show that this is indeed the case7 we need to show there is an x E R3 such that lalal agag agxgl 2 max lall7 a3 a HxHa From the above analysis7 it is clear that to get an equality in the 3rd line7 we need lxll 1 if lall 2 Ma a and lxll 0 otherwise Also7 for equality in the Cauchy Schwarz inequality7 we need 2 a27x3 a3 Combining all of these requirements7 we see that7 for the choice signa1 lallZ l02l2l03l2 1 0 otherwise 0 W 2 lazlz w 2 7 a2 otherwise 0 W 2 lazlz w 3 7 as otherwise we see that lalzl a22 a3xgl max lall7 a3 ag wag Combining this with the preceding inequality7 we obtain 1101702703Nlmduced max lam Example 7 De ne p R2 a 07 00 by 7 MHzl 96120 7 xxz y otherwise a Show thatp is norm on R2 b Determine the dual space State clearly any identi cations you make between spaces offunctions and points in R We will construct the norm out of the linear functions de ning the tangents to the unit ball If Ly is in the rst or the third quadrant and p7y 17 it is clear that z y i1 Further7 the lines z y i1 are tangent to the unit ball If x7 y is in the second or the fourth quadrant7 and 10x7 y 17 it follows that 2y2 1 so z i c0S97y i sin0 for 6 E Tr277T and the tangents to the unit ball at the point cost97 sin6 is given by zcos0 y sin0 1 Based on this intuition7 we claim pm a maXW at sup 196 0039 y sin9 l 967r27r 8 SHANKAR VENKATARAMANI For all my it is clear that lzcos0 ysin6l S x2 y2 by the Cauchy Schwarz inequality If x and y have the same sign we have lyl2 22zyy2 2 2y2 so that if xy gt 0 it follows that maxltl96 yl sup WOW ysin9l l96 yl Way sigh2m lf zy S 0 setting 00 arctanyx where the arctangent is de ned on a fundamental domain 07T we see that yz S 0 so that 00 E 7r27T Therefore sup lx cos0 y sin6l 2 lx cos60 y sin00l 2 y sigh2m Combining this with the result above from the Cauchy Schwarz inequality we see that supeewg lx cos0 y sin6l 2 yz Since xy 3 0 we also have lyl2 22zyy2 S 2y2 so that if xy 3 0 it follows that maxltl96 yl7 sup WOW ysin9l V962 112 pmy sigh2m Thus we have shown that Way sup lLanN where L0 is a suitable family of linear functions in particular the linear functions that determine the tangents to the unit ball By Proposition 2 it follows that p is a seminorm Also if my 31 00 it follows from the de nition that py 31 0 This proves that p is a norm We will now compute the dual space As before there is a one to one identi cation between V and R2 by for all a E R2 f a x and for 239 1 2 a fe Also if 10 is the induced norm on the dual for 1102 E R2 the induced norm is 1001702 SUP la 1le 121 1 3 min Na a3 sup x y2maxla1l lagl sup Pltmgty1 pmy1 where we have used the Holders inequality twice in the second line and recognized that r S 57 gt gt r S minst lfpy 1 either xx2y2 1 or xy gt 0 gt 2y V17 2mg 3 1 Thus we get 17011702 S Va 03 Without loss of generality a1a2 31 00 If 1102 S 0 setting 7 01 7 02 7 21L 27 7 21L 27 V01 12 V01 12 NORMED LINEAR SPACES AND DUALS 9 we get xy 3 0 and May 1 and 1a1x a2y1 a a3 Therefore for 1102 S 0 we get pa17a2 a a3 lfalag gt07we havepz7y 1 gt 0 3x 1and0 y 1lfzy S 0buta1a2 gt07 it follows that am and agy do not have the same sign7 so that 1m a2y1 maxlt1a1z1i1a2y1gt max1a1171a21 If pg 2 07 we have am and agy have that May 1 gt 17 and by Ho39lder7s inequality7 we have 10196 02113 maxlt1al171az11961 1111 S maxlt1al171az1l Finally7 considering z 17y 0 and z 07y 17 we see that P01702 Z maxlt101171021l Cornbining all these argurnents7 we see that rnax1a1171a21 1102 gt 0 a a 1017 2 W otherwise 4 INFINITE DIMENSIONAL SPACES A key ingredient in all our computations for duals of nite dimensional spaces is the representation result that the set of all linear functions on V R is in a one to one correspondence with R where the correspondence is f V1 gtaER aifeZ7i17277n But not all linear functionals on in nite dimensional spaces can be represented this way In what follows el is the sequece that is 1 in the iith index7 but is zero otherwise Example 8 c is the space of all sequences that converge Show that Mp lirn xn Hoe de nes a bounded linear map on c7 Clearly7 c is a linear space and o is a linear rnapVERlFY THISl Also7 1 x Mi1 1 7 M1 sup 1 7 yn11196 7 1100 showing that o is indeed a bounded linear rnap Note also7 that Mei 0 for all i7 showing that there is no sequence a 6 RN such that 1595 Zi am So7 when can we assume that a linear function on a sequence space can be represented as Mp Salxi If this representation holds7 it is clear that al Mei Example 9 V lo the space of all compact sequences ie sequences that are eventually zero If t V a R is a linear map then Mp aim with al Mei 10 SHANKAR VENKATARAMANI If x E l07 then x mel is a nite linear combination Consequently7 N N 00 We Z5 1431 Za i Za i i1 i1 i1 where the last equality follows from the fact that the in nite series has only nitely many non zero terrns7 so it converges Note that7 there are no apriom39 restrictions on the coef cients 11 so the set of all linear functionals on l0 is in a one to one correspondence with RN and this is independent of any norrn that we choose to impose on l0 Example 10 Let V l07 Fmd V5 From the representation result in the previous exarnple7 we see that b E V implies that Z a i so that WM S llallqll llp Consequently7 if Hallq lt 007 the corresponding linear function b is a bounded linear rnap Also7 the standard argument taking signailailq 1 39 AN Ellilailqllp ZSN 0 otherwise along with 3 CHsz implies that N 2W 3 Cg i1 for all N Consequently7 a 6 lg and the induced norm is Ha llinduced Hallq We can use this result7 along with the fact that every bounded linear functional is continuous to compute the duals of the l sequence spaces For every y E l 7 it is clear that the sequence gm de ned by ygiv yi 239 S N l 0 otherwise Clearly7 gm 6 l0 and HyW 7 pr a 0 This implies that to de ne a bounded linear functional on l it is enough to de ne it on l0 and then extend the function to l by continuity ln particular7 every bounded linear function on l can be written as a i where a E lq7 guaranteeing that the sum converges7 and the same argument as in the previous example showing that the dual space is lq7 Hg with the correspondence b gt gt 0i Wei The same argument also works for the sequence space 00 to show that the dual is l1 VERIFY THIS NORMED LINEAR SPACES AND DUALS 11 This argument will not work for cl The sequence 60 17 17 17 17 E c For any y E l07 we see that 7 Bolloo 2 17 since for suf ciently large values of the index7 yn 0 So7 there is no sequence in lo that converges to 60 in the sup norm Example 11 Compute the dual ofV c7 The issue here is to nd a representation of all linear functions on c Observe7 if z 6 07 then x 7 limpH00 zn o E 007 where as above 60 17 17 17 17 As we argued earlier7 wow limpH00 xn is a bounded linear function on V Consequently7 if b is a bounded linear functional on V7 then M 050 W 0 o mlboW 45W IDOWWO where m 1560 Since zizbowmo E 007 it follows that there is a sequence xiiwt z such that N 7 N193 Hz 7 1009050 7 Emilee i 0 Consequentl7 if b is a bounded linear function N 7 l I Ms 7 m 0lt gt 1571 ami where 11 Mei If a E ll7 by Holders inequality7 we have WW S lmlll clloo 2llall1lllloo using the estimates S ll cllom Hill H96 7 lawman S H9ch l olt gtlll olloo 2lllloo showing that b is a bounded function if 2HaH1 lt oo Also7 considering the sequences 60 and lim xn Hoe N 7 signal 239 S n at 7 0 otherwise 7 it is easy to see that b is a bounded function only if lt 007 lt 00 We have thus identi es the space of bounded linear functionals as R gtlt l1 Computing the dual norm is tricky we dont have equality in the upper and lower bounds above7 but from the argument above7 it is clear that the dual norm of 77171 E R gtlt l1 is equivalent to llm7all E lmlllall1 E maXOmt llalll

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