FUNCTIONAL ANALYSIS DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR

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1 FUNCTIONAL ANALYSIS DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR Contents 1. About 1 2. Syllabus 1 References Orgn 2 3. Abstract space 2 4. Metrc Space 2 5. Sem-Metrc 6 6. Separable spaces 6 7. Completeness 8 1. About Functonal analyss s the abstract branch of mathematcs that has orgnated from classcal analyss. It s study of certan topologcal algebrac structures and of methods by whch knowledge of these structures can be appled to analytc problems. 2. Syllabus Normed Spaces, contnuty of a lnear mappng. Banach spaces, Lnear Transformatons and functonals and Normed bounded lnear transformaton, dual spaces, Hahn Banach theorem. Hlbert Spaces. Orthonormal sets, Bessels Inequalty, Parsevals relaton, Resz Representaton theorem, Relatonshp between Banach Spaces, Hlbert Spaces. Adjont operators n Hlbert Spaces, Self adjont operators, postve operators, Projecton Operators and orthogonal projectons n Banach & Hlbert spaces, Fxed pont theorems and ther applcatons, Best approxmatons n Hlbert Spaces. Gatebux and Frechat Dervatves. Soluton of boundary value problems. Optmzaton problems. Applcatons to Integral and dfferental equatons. References [1] Kreyszg, Introductory Functonal Analyss Wth Applcatons, Wley Inda Edton, [2] Walter Rudn, Functonal Analyss, Tata Mc-Graw Hll, New Delh. [3] Balmohan Vshnu Lmaye, Functonal analyss, New Age Internatonal, [4] Aldrc Loughman Brown anda. Page, Elements of functonal analyss, Van Nostrand-Renhold, 1970 [5] J.B. Conway, A Course n Functonal Analyss, 2nd Edton, Sprnger, Berln, Date: January 27, Emal:ragarwal.maths@mnt.ac.n 1 Web: drrtuagarwal.wordpress.com.

2 2 DR. RITU AGARWAL 2.1. Orgn. It orgnates from the calculus of varatons n the study of operators on functon spaces defned by dfferentaton and ntegraton. The name functonal analyss was coned by the French mathematcan Paul Levy ( ). Early poneers were: Italan Vto Volterra ( ), Swedsh Erk Ivar Fredholm ( ), German Davd Hlbert ( ) and Hungaran Frgyes Resz ( ) Influental contrbutons. The Polsh mathematcan Stefan Banach ( ) was nfluental n brngng the notons of topology nto functonal analyss, and he s known for the semnal book Theore des operatons lneares of Fundamental contrbutons to the study of operators on Hlbert spaces were made by the Hungaran mathematcan John von Neumann ( ) Inspraton. Inspraton to ths work came from many sdes, not least from the development of quantum mechancs n physcs n the 1920s by physcsts such as Nels Bohr ( ), Paul Drac ( ), Werner K. Hesenberg ( ) Erwn Schrodnger ( ) 3. Abstract space Defnton 3.1 (Mathematcal Analyss). Mathematcal analyss leads to exact results by approxmate computatons. It s based on the notons of approxmaton and lmt process. For nstance, the dervatve s the lmt of dfferental quotents, and the ntegral s the lmt of Remann sums. Defnton 3.2. Set of unspecfed elements satsfyng certan axoms Example: Groups Rngs Felds Vector space Metrc space and many more. 4. Metrc Space Defnton 4.1. Metrc Space A functon d : X X R s sad to be a metrc on non-empty set X f t satsfes the followng axoms: Non-negatve: d(x, y) 0. d(x, y) = 0 ff x = y

3 Symmetry: d(x, y) = d(y, x) for all x, y FUNCTIONAL ANALYSIS 3 Trangle nequalty: d(x, y) d(x, z) + d(z, y) for all x, y, z X. A set X wth metrc d defned on t s called Metrc Space and s denoted by (X, d). Example 1. Examples of metrc spaces: a. Real lne R s a metrc space wth metrc d(x, y) = x y, x, y R. b. Eucldean space R 2 s a metrc space wth metrc d 1 (x, y) = (ξ 1 η 1 ) 2 + (ξ 2 η 2 ) 2 and d 2 (x, y) = ξ 1 η 1 + ξ 2 η 2 x = (ξ 1, ξ 2 ), y = (η 1, η 2 ) R 2. c. Eucldean space R n s a metrc space wth metrc d(x, y) = (ξ 1 η 1 ) (ξ n η n ) 2, x = (ξ 1,..., ξ n ), y = (η 1,..., η n ) R n. d. Untary Space C n s a metrc space wth metrc d(x, y) = ξ 1 η ξ n η n 2, x = (ξ 1,..., ξ n ), y = (η 1,..., η n ) C n. e. Functon space C[a, b]: Ths s the set of all real valued functons x, y,... whch are functons of ndependent real varable t and are defned and contnuous on a gven closed nterval [a, b]. The metrc on ths set can be defned as: d 1 (x, y) = max t [a,b] x(t) y(t) d 2 (x, y) = b x(t) y(t) dt represents the area between two curves bounded by the ordnates a x = a and x = b. f. Sequence space S of all bounded and unbounded sequences of complex numbers wth metrc d defned as 1 ξ η d(x, y) = ξ η where x = (ξ ), y = (η ) S, s a metrc space. [Hnt: To prove trangle nequalty n ths case, use the result a + b a + b a 1+ a + b 1+ b as f(t) = t 1+t =1 s monotoncally ncreasng functon of t.] a+b 1+ a+b a + b 1+ a + b g. B[a, b] set of all bounded functons defned on [a, b] wth metrc d defned as d(x, y) = sup t [a,b] x(t) y(t), x, y B[a, b] s a metrc space. h. Sequence space l { } l = x = (ξ ) =1 s.t. sup ξ < wth metrc d defned as d(x, y) = sup ξ η, x, y l.. Sequence space l p l p = { x = (ξ ) =1 s.t. } ξ p < =1

4 4 DR. RITU AGARWAL ( ) 1/p, wth metrc d defned as d(x, y) = ξ η p x, y l p, p 1. Remark 4.2. Dfferent metrc can be defned on a gven set X. =1 In order to prove trangle nequalty for Example 1(), frst we are requred to prove followng results. Defnton 4.3 (Conjugate Exponents). Let p > 1 and q R be such that 1 p + 1 = 1. (4.1) q Then p and q are called conjugate exponents. Also, (4.1) can be wrtten as (p 1)(q 1) = 1 Theorem 4.4. Let α and β be any non-negatve real numbers and f p and q are conjugate exponents then αβ αp p + βq (4.2) q Proof: If α = 0, β = 0, the result s obvous. Let α 0, β 0, consder the functon u = t p 1.e. t = u q 1. Then, αβ = Total area of the regon 1 and 2 α 0 tp 1 dt + β 0 uq 1 du = αp + βq. p q Theorem 4.5 (Hölder s nequalty). Let x = (ξ j ) l p, p > 1 and y = (η j ) l q, p and q are conjugate exponents. Then (ξ j η j ) l 1 and ( ) 1/p ( ) 1/q ξ j η j ξ k p η m q (4.3) k=1 m=1 Proof: Let x = ( ξ j ) l p, and y = ( η j ) l q be two sequences such that ξ k p = 1 and η m q = 1 k=1 Settng α = ( ξ j ), and β = ( η j ) n (4.2), we get ξ j η j ξ j p p + η j q q m=1 Now, let x = (ξ j ) l p, p > 1 and y = (η j ) l q. Then substtutng ξ j η j 1. (4.4) ξ j = ( k=1 ξ and k p ) 1/p n (4.4), we get the desred result (4.3). ξ j η j ( m=1 η m q ) 1/q Theorem 4.6 (Mnkowsk Inequalty). Let x = (ξ j ), y = (η j ) l p p 1. Then ( ) 1/p ( ) 1/p ( ) 1/p ξ j + η j p ξ k p + η m p k=1 m=1 (4.5)

5 Proof: Result s obvous for p = 1. For p > 1, let ξ j + η j = ω j. FUNCTIONAL ANALYSIS 5 ω j p = ω j ω j p 1 = ξ j + η j ω j p 1 ξ j ω j p 1 + η j ω j p 1 Now, snce (p 1)(q 1) = 1, ( ( ω j p 1 ) q = ω j p ωj (p 1)q) = ω j p < ω j p 1 l q Applyng Hölder s nequalty to x = (ξ j ) l p and ω j p 1 l q, we get ( ) 1/p ( ) 1/q ( ) 1/p ( ) 1/q ξ j ω j p 1 ξ k p ( ω m p 1 ) q = ξ k p ω m p k=1 m=1 k=1 m=1 Smlarly, ( ) 1/p ( ) 1/q η j ω j p 1 η k p ω m p k=1 m=1 Addng (4.6) and (4.7) and smplfyng, we get the desred result. (4.6) (4.7) Remark 4.7. In order to prove the trangle nequalty for l p space, replace ξ j by ξ j ζ j and η j by ζ j η j n (4.5). Defnton 4.8 (Metrc Subspace). Let (X, d) be a metrc space and Y X s non-empty proper subset of X. Suppose d s the restrcton of d on Y Y such that d(y 1, y 2 ) = d(y 1, y 2 ) y 1, y 2 Y Example 2. If A s a subspace of l consstng of all sequences of 0 and 1. The nduced metrc on A s a dscrete metrc.e. { 0, x = y d(x, y) = 1, x y Defnton 4.9 (Contnuous mappng). Let (X, d) and (Y, d) be two metrc spaces. A mappng T : X Y s sad to be contnuous at a pont x 0 X, f for every ɛ > 0, δ > 0 such that d(t x, T x 0 ) < ɛ whenever d(x, x 0 ) < δ Theorem A mappng T of a metrc (x, d) nto a metrc space (Y, d) s contnuous f and only f nverse amge of an open subset of y s an open subset of X. Proof: Let f s contnuous Y s open set n T. Also, let p f 1 (Y ). Then y = f(p) for some y Y. Snce Y s open and y Y, for some ɛ > 0, B T (y; ɛ) Y. Snce f s contnuous, δ > 0 such that f(b S (p; δ)) B T (y; ɛ). B S (p; δ) f 1 (f(b S (p; δ))) f 1 (B T (y; ɛ)) f 1 (Y ) p s an nteror pont of f 1 (Y ). Conversely let f 1 (Y ) s open n S for every open subset Y n T. Choose p S and y = f(p). For every ɛ > 0, B T (y; ɛ) s open n T. f 1 (B T (y; ɛ)) s open n S. Now p f 1 (B T (y; ɛ)) δ > 0 such that B S (p; δ) f 1 (B T (y; ɛ)) f(b S (p; δ)) B T (y; ɛ) f s contnuous at p. Remark The mage of an open set under contnuous mappng s not necessarly open. E.g. Constant functon

6 6 DR. RITU AGARWAL 5. Sem-Metrc Defnton 5.1 (Sem-Metrc). A functon d : X X R s sad to be sem-metrc f d(x, y), x, y X s a real valued, non-negatve fnte number such that t s symmetrc and satsfes trangle nequalty,.e. d(x, x) = 0, d(x, y) = d(y, x) and d(y, z) d(x, y) + d(y, z). A space X equpped wth sem-metrc d s sad to be sem-metrc space (X, d). Example 3. X = L[0, 1] s the set of all Lebesgue ntegrable functons f on [0, 1].e. f L[0, 1] = 1 f(x)dx s fnte. 0 The metrc on ths space can be defned as : d(f, g) = 0 = f = g almost everywhere. (L[0, 1], d) s sem-metrc. d(f, g) = 1 0 (f g)(x) dx (5.1) Remark 5.2. For the same defnton gven by (5.1), the space (C[0,1],d) s a metrc space. Theorem 5.3 (Equvalence relaton usng sem-metrc). Let (X, d) be a sem-metrc space. Defne a relaton R as xry or x y f d(x, y) = 0. Then R s an equvalence relaton and decomposes the set X nto dsjont equvalent classes E x = {x : d(x, x ) = 0}. Take E = {E x, E y,...}. Defne dstance functon ρ on E as ρ(e x, E y ) = d(x, y). The ρ s well-defned and (E, ρ) s a metrc space. Proof: Frst prove that R s an equvalence relaton where xry f d(x, y) = 0. Next, to show that ρ s well-defned, consder x Rx and y Ry. Then d(x, y ) = d(x, y). Further, (E, ρ) s a metrc space. Ths can be proved usng the concept that two equvalence classes are ether dentcal or dsjont. 6. Separable spaces Defnton 6.1 (Dense Set). A subset M of a metrc space X s sad to be dense n X f M = X Example: Q R s dense n R. Defnton 6.2 (Separable Space). A metrc space X s sad to be separable f t has a countable subset M whch s dense n X. Example 4. The real lne R s separable. The set Q of ratonal numbers s countable and dense n R. Example 5. The complex plane C s separable. A countable dense subset of C s the set of all complex numbers whose real and magnary parts are both ratonal. Example 6. A dscrete metrc space X s separable f and only f X s countable. The type of metrc mples that no proper subset of X can be dense n X. Hence the only dense set n X s X tself. Example 7. The space l s not separable. Proof: Let A l be the set contanng sequences of zeros and ones as y = (η 1, η 2,...). Then y l. Wth each y, we assocate a pont y [0, 1] whose bnary representaton s η η η

7 FUNCTIONAL ANALYSIS 7 Snce the nterval [0, 1] s uncountable, there are uncountably many sequences of zeros and ones n A and metrc wll be dscrete metrc (see example 2). Any two unequal elements n A are unt dstance apart. Let each of these sequences be the centre of a small ball of radus 1/3, then these balls do not ntersect and we have uncountably many of them. If M s dense n l, each of these non-ntersectng balls must contan an element of M. Hence M cannot be countable. Snce M was arbtrary dense set, l cannot have dense subsets whch are countable. Consequently, l s not separable. Example 8. The space l p, p 1 s separable. Proof: Let M be the set of all sequences y of the form y = (η 1, η 2,..., η n, 0, 0,...) where n s any postve nteger and η s are ratonal. Then M s countable. We show that M s dense n l p. Let x = (ξ ) l p be arbtrary.. Then for every ɛ > 0 there s an n (dependng on ɛ) such that ξ j p < ɛp 2 j=n+1 where the L.H.S. s the remander of a convergng seres. Snce the ratonals are dense n R, for each ξ j, there s a ratonal η j close to t. Hence, we can fnd y M satsfyng n ξ j η j p < ɛp 2 It follows that {d(x, y)} p = n ξ j η j p < ɛp 2 + We thus have d(x, y) < ɛ and see that M s dense n l p. j=n+1 ξ j p < ɛ p Exercse 6.3. Show that the space (B[a, b], d) of bounded functons defned over closed nterval [a, b] and metrc d(f, g) = sup t [a,b] f(t) g(t) s not separable. Theorem 6.4. If (X, d) s a separable metrc space and Y X, then (Y, d) s separable n the nduced metrc d. Proof: Gven (X, d) s a separable metrc space. So t has a countable subset E = {x 1, x 2,...} whch s dense n X. Case I: If E s contaned n Y, then there s nothng to prove. Case II: If E s not contaned n Y. We construct a countable subset y n,m of Y whose ponts are arbtrary close to the set E. To show: There exsts a countable subset of Y whch s dense n Y. For postve ntegers n and m, let B n,m = B(x n, 1/m) be open ball centered at x n E wth radus 1/m. Choose y n,m B n,m Y whenever t s non-empty. A sequence wll be obtaned as y 1,m, y 2,m, y 3,m,... B n,m Y. The balls B(x n, 1/m) are countable n number as x 1, x 2,... are countable and some of these have empty ntersecton wth Y, leave those balls. {y n,m } s a countable subset of Y. It remans to show that y n,m s dense n Y. Let y Y and r > 0. Let m be so large that 1 m < r 2 and fnd x n B(y, 1/m) then y B(x n, 1/m) = B n,m and hence y Y B n,m. Now d(y, y n,m ) d(y, x n ) + d(x n, y n,m ) < 1 m + 1 m < r Therefore, y n,m B(y, r). Snce y Y and r > 0 are arbtrary, we see that {y n,m } s dense n Y.

8 8 DR. RITU AGARWAL 7. Completeness Defnton 7.1 (Convergent sequence). A sequence (x n ) n a metrc space (X, d) s sad to be convergent f x X such that d(x n, x) 0 as n Such an x s called lmt of (x n ) and we wrte lm n x n = x. Defnton 7.2 (Cauchy sequence). A sequence (x n ) n a metrc space (X, d) s sad to be a Cauchy sequence f d(x n, x m ) < ɛ for n, m n 0 ( dependng on ɛ) Remark 7.3. Every convergent sequence s Cauchy sequence but every Cauchy sequence s not necessarly convergent. (Why??) A non-empty subset M X s sad to be bounded f ts dameter δm = sup x,y M d(x, y) s fnte. Defnton 7.4 (Bounded sequence). A sequence (x n ) n a metrc space (X, d) s sad to be bounded f correspondng pont set {x 1, x 2,...} s a bounded subset of X. Remark 7.5. If M s bounded, M B(x 0 ; r) where x 0 s arbtrary and r s suffcently large postve real number. Theorem 7.6. Let (X,d) be a metrc space. Then (.) A convergent sequence n X s bounded and ts lmt s unque. (.) If x n x and y n y n X then d(x n, y n ) d(x, y). Proof: () Suppose that x n x. Then takng ɛ = 1, we can fnd an n 0 such that d(x n, x) < 1 for all n n 0. Hence by trangle nequalty d(x n, x) < 1 + a where a = max{d(x 1, x), d(x 2, x),..., d(x n0, x)} Ths shows that (x n ) s bounded. Further, assume that x and z be two lmts of (x n ). Then 0 d(x, z) d(x, x n ) + d(x n, z) Hence, follows the unqueness of the lmt of (x n ). () Snce, x n x and y n y, by trangle nequalty, we get d(x n, y n ) d(x n, x) + d(x, y) + d(y, y n ) Smlarly, d(x n, y n ) d(x, y) d(x n, x) + d(y, y n ) d(x, y) d(x n, y n ) d(x, x n ) + d(y n, y) d(x n, y n ) d(x, y) d(x n, x) + d(y, y n ) 0 as n. Defnton 7.7 (Completeness). A metrc space (X,d) s sad to be complete f every Cauchy sequence n X s convergent (.e. the lmt s an element of X). Theorem 7.8. Let M be a nonempty subset of a metrc space (X,d) and M s closure of M. Then (a) x M ff there s a sequence {x n } n M such that x n x. (b) M s closed ff the stuaton x n M, x n x mples that x M.

9 FUNCTIONAL ANALYSIS 9 Proof: (a) Let x M. If x M, a sequence of type s (x, x,...). If x / M, t s an accumulaton pont of M. Hence for each n = 1, 2,... the ball B(x; 1/n) contans x n M and x n x because 1/n 0 as n. Conversely, f (x n ) s n M and x n x then x M or every neghbourhood of x contans ponts x n x so that x s a pont of accumulaton of M. Hence, x M, by the defnton of closure. (b) M s closed f and only f M = M, so that (b) follows readly from (a). Theorem 7.9. A subspace M of a complete metrc space (X,d) s complete ff M s closed n X. Proof: Frst let M be complete. Let x M then there s a sequence (x n ) convergng to x. Snce (x n ) s a Cauchy sequence and M s complete, t converges n M. Ths mples, x M.e. M M M s closed. Conversely, let M be closed and (x n ) s a Cauchy sequence n M. Snce X s complete, (x n ) x X whch mples x M = M as M s closed. Hence the arbtrary Cauchy sequence (x n ) converges n M whch proves completeness of M. Example 9. Completeness for R s already proved. Set R s complete ordered feld. Eucldean space (R n, d) s complete for the usual metrc d(x, y) = n ξ η 2, where x = (ξ 1,..., ξ n ), y = (η 1,..., η n ) R n. Proof: Consder a Cauchy sequence (x m ), x m = (ξ (m) 1,..., ξ n (m) ) n R n. For gven ɛ > 0, N(ɛ) such that for m, r > N n d(x m, x r ) < ɛ ξ (r) 2 < ɛ For each fx ξ (m) =1 ξ (m) ξ (r) < ɛ.e. the sequence ξ (1), ξ (2),... behaves as a Cauchy sequence of real numbers whch s a complete space. Hence the sequence (ξ (m) ) s convergent n R and converges to ξ R(say).e. ξ (m) ξ < ɛ/n for all m > N Clearly x = (ξ 1,..., ξ n ) R n. To show x m x as m. From above x m x n =1 Untary space C n s complete. (Prove) ξ (m) ξ 2 < ɛ for m > N. Space of bounded sequences l s complete Proof: Consder a Cauchy sequence (x m ) l, x m = (ξ (m) 1, ξ (m) 2,...), ξ (m) 1 R or C. For gven ɛ > 0, N(ɛ) such that for m, r > N For each fx, for all m, r > N d(x m, x r ) < ɛ ξ (m) sup ξ (m) =1 ξ (r) < ɛ ξ (r) < ɛ.e. the sequence ξ (1), ξ (2),... behaves as a Cauchy sequence of real numbers whch s a complete space. Hence the sequence (ξ (m) ) s convergent n R (or C) and converges to ξ R(say).e. =1

10 10 DR. RITU AGARWAL ξ < ɛ for all m > N Let x = (ξ 1, ξ 2,...). To show x m x as m and x l. From above ξ (m) Further, snce ξ = ξ ξ (m) sup ξ (m) ξ < ɛ for m > N d(x m, x) < ɛ x m x. + ξ (m), ξ = ξ ξ (m) + ξ (m) < ɛ + K m where K m = sup ξ (m) <. sup ξ < x l. Space of bounded sequences l p, 1 p < s complete Space of convergent sequences C, wth the metrc d nduced from l, s complete. Some spaces may be complete wth respect to one metrc but not wth another metrc defned on t. Example 10. Functon space C[a, b]: Ths s the set of all real valued functons x, y,... whch are functons of ndependent real varable t and are defned and contnuous on a gven closed nterval [a, b]. The metrc on ths set can be defned as: d 1 (x, y) = max t [a,b] x(t) y(t) d 2 (x, y) = b x(t) y(t) dt represents the area between two curves bounded by the ordnates a x = a and x = b. (C[a, b], d 1 ) s complete metrc space but (C[a, b], d 2 ) s not. Proof: (a) Let (x n ) be a Cauchy sequence n C[a, b]. So gven ɛ > 0, n 0 N such that d 1 (x m, x n ) < ɛ, m, n > n 0. That s max a t b x m(t) x n (t) < ɛ Ths mples x m (t) x n (t) < ɛ t [a, b]... () Fx t = t 0, x m (t 0 ) R, therefore (x m (t 0 )) s a Cauchy sequence of real numbers and hence t converges to some x(t 0 ) R (say). (x s well defned): In ths way, we can assocate wth each t [a, b], an unque real number x(t). Ths defnes a pontwse functon x on [a, b]. Lettng n n (), t follows that, x m (t) x(t) < ɛ t [a, b]... () (x m ) converges to x unformly on [a, b] and (x m ) s a sequence of contnuous functons. Hence the lmt functon x s also contnuous on [a, b]. That s x C[a, b]. Further, () = max a t b x m(t) x n (t) < ɛ d = x 1 m x. Remark Ths metrc on C[a, b] s sometmes called unform metrc also snce convergence of x m x s unform.

11 FUNCTIONAL ANALYSIS 11 (b) To show that (C[a, b], d 2 ) s not complete, we here gve a counter example. Let (x m ) be a sequence n C[0, 1] defned as follows: 0, t [0, 1/2] x m (t) = (t 1/2)/m, t (1/2, a m ] 1, t (a m, 1] where a m = m. (x m) s a Cauchy sequence under the metrc d 2 n C[0, 1]. Let m > n then 1 m < 1 n. d 2 (x m, x n ) = 1 Let m n (7.1) then 0 x m (t) x n (t) dt = 1/m 1/2 x m (t) x n (t) dt + 1/n 1/m { 0, t [0, 1/2] x(t) = lm x m(t) = m 1, t (1/2, 1] (7.1) x m (t) x n (t) dt 0 as m, n Clearly x(t) s not a contnuous functon whle x m x not n C[0, 1]. Hence, (C[a, b], d 2 ) s not complete. Other examples of ncomplete metrc spaces: 1. X = Q, the set of ratonal numbers s ncomplete. But Q can be enlarged to R a complete metrc space. So, ths gves a way to convert t to complete metrc space. 2. Set of polynomals s ncomplete. Ths can be proved by gvng counter example as follows: Consder the sequence of polynomals defned by (P n (z)), P n (z) = 1+z + z2 + z lm 2! 3! n P n (z) = e z whch s clearly not a polynomal but a transcendental functon. Defnton 7.11 (Isometrc mappng). Let (X, d) and (Y, d) be two metrc spaces and T be a mappng from X to Y. T s sad to be sometrc or an sometry f t preserves the dstance,.e. d(t x, T y) = d(x, y) Isomtery s one-one for f T x = T y = d(t x, T y) = 0 = d(x, y) = x = y. If the sometry T s onto, the spaces x and y are sad to be somterc spaces. Remark Two sometrc spaces may be regarded as two copes of same abstract space n the study where nature of ponts does not matter. Example 11. Show that C[0, 1] and C[a, b] are sometrc spaces. Proof: Defne T (t) = t a, t [a, b] and T (t) [0, 1]. So that whenever x(t) C[a, b], T x = x(t t) b a C[0, 1]. Then d(t t a t a x, T y) = d(x( ), y( )) = d(x, y). b a b a Defnton 7.13 (Homeomorphsm). A homeomorphsm s a contnuous bjectve mappng T : X Y whose nverse s also contnuous. Remark If two spaces are sometrc then they have same nature regardng completeness but ths s not necessary n case of homeomorphsm. Example 12. Illustrate wth one example that a complete and ncomplete metrc spaces may be homeomorphc. Soluton: Consder the metrc spaces X = R and Y = ( 1, 1). Clearly, X s complete whle Y not. Defne a mappng T : X Y as T (x) = y = 2 π tan 1 x. Then T s a contnuous mappng and bjecton also. Further T 1 y = x = tan(πy/2) s also contnuous. T s a homeomorphsm. Defnton 7.15 (Equvalent metrc). A metrc d s stronger than other metrc d defned on a non-empty set X f for every ɛ > 0, δ > 0 such that B d (x, δ) B d (x, ɛ) that s every open subset of X w.r.to d s also open w.r.to d. Two metrcs defned on X are sad to be equvalent f each s stronger than the other. (7.2)

12 12 DR. RITU AGARWAL Mathematcally, two metrcs d and d are equvalent to each other f there exsts constants α and β such that αd(x, y) d (x, y) βd(x, y) (7.3) Remark If a sequence s convergent n the strong metrc, t has to converge n a weak metrc..e. whenever x d d n x = x n x. Example 13. Show that the completeness of a metrc may not be shared by an equvalent metrc. Soluton: Consder X = (0, 1]. Let d (x, y) = x y and d(x, y) = 1 1 x y for x, y (0, 1]. Clearly, x n d x x n d x. Hence d and d are equvalent metrcs. (X, d ) s complete but (X, d) not. Exercse Show that the Cauchy sequences n (X, d) and (X, d ) are same where d and d are equvalent metrcs. Exercse Show that Cauchy Schwarz nequalty mples ( ξ 1 + ξ ξ n ) 2 n( ξ ξ ξ n 2 ). Show that the metrc d(x, y) = (ξ 1 η 1 ) 2 + (ξ 2 η 2 ) 2 and d (x, y) = ξ 1 η 1 + ξ 2 η 2, x = (ξ 1, ξ 2 ), y = (η 1, η 2 ), x, y R 2 are equvalent.