METRIC SPACE Let X is set of points and on X is defined real valued, nonnegative, two variable function, (x,y) x, y X. If this function is satisfying following conditions then tuple (X, ) is called metric space. 1. (x,y) = 0 x = y 2. (x,y) = δ(y,x) symmetric 3. (x,y) ≤ (x,z) + (z,y) , x, y, z X. Examples: A) The simplest metric space is a discrete metric space which expression following form 0, x y (x, y) 1, x y 1. and 2. conditions are easy to verifying. Let us check last condition 3. If If x y z (x, y) (x, z) (y, z) x y z (x, y) (x, z) ( y, z) B) Let us put metrics on real line as following form (R, ). (x,y) = | x –y |. This metric satisfies 1-3 conditions and it is called metric in real line. C) Let us identify metric on Rn. Rn has an element with n-real number coordinates. Rn = { x: x = (x1, x2, x3, . . . , xn), xi Є R} n (x If we take metric (x, y) i 1 i y i ) 2 then (Rn, ) is convert metric space. 1. If x y x i y i , i 1,2, . . ., n (x, y) 2. If (x, y) n (x i 1 i n (y yi ) 2 i 1 i n (x i 1 i yi )2 0 x i ) 2 (y, x) To verify last condition we need Cauchy-Schwarz Inequality in Rn. n n ( a k b k ) 2 a k k 1 k 1 2 n b k 1 2 k We have to prove ρ(x, y) ρ(x, z) ρ(z, y) - - - - - - - - - -(1) where x = (x1, x2, x3, . . . , xn) y = (y1, y2, y3, . . . , yn) z = (z1, z2, z3, . . . , zn) , xi, yi, zi Є R Let xi -zi = ai zi- yi = bi xi- yi = ai + bi. Then it is enough to prove. n (a i b i ) 2 i 1 n n i 1 i 1 n ai 2 i 1 n b i 1 (2) 2 i n n i 1 i 1 (a i b i ) 2 a i 2 a i b i b i 2 n ai 2 i 1 2 by Schwarz Inequality n ai 2 ai 2 i 1 2 root 2 b i i 1 n n n (a i b i ) 2 i 1 n 2 i 1 n b 2 i 1 i 1 2 i 1 n ai n bi bi 2 i 1 2 i Q.E.D D) 2 is collection of the point X which has infinite real coordinates. x = {x: x = (x1, x2, x3, . . . , xn, . . .), x x = (x1, x2, x3, . . . , xn, . . .) and 2 i 2 i } As an example (3) we can pıt metric as following n (x (x, y) i 1 i 2 y i ) 2 - - - - - - - - - - - - - (3) n 2 n 2 2 (x y ) 2 x i yi i i i 1 i 1 i1 Then by using last example we can say ( 2 , ) is metric space. First we can identify (3) takes finite value. We know E) Set of continuous functions in the closed interval. [a,b], C[a,b] is a metric space with the metrics ρ(f, g) max f(x) g(x) a x b 1. and 2. conditions are easy to prove. Let prove last condition. For every f(x), g(x), h(x) Є C[a,b]. We can write f(x) g(x) f(x) h(x) h(x) g(x) f(x) h(x) h(x) g(x) max f(x) h(x) max h(x) g(x) a x b a x b max f(x) g(x) max f(x) h(x) max h(x) g(x) a x b a x b (f, g) (f, h) (h, g) F) On the set of continuous functions C[a,b]. We can put metric following way. b ρ(f, g) (f(x) g(x)) 2 dx (Prove at home) a 2 b b b 2 Hint: f(x)g(x)dx f (x)dx g 2 (x)dx a a a Minkowsky Inequality 1/p n p a k bk k 1 1/p n p ak k 1 1/p n p bk k 1 a x b Hölder Inequality 1/p 1/q n n p q a b a . bk k k k k 1 k 1 k 1 n where the numbers p>1 and q>1 satisfy the condition 1 1 1 p q Open and Closed Set in Metric Space By the open sphere S(x0,r) is denoted collection of points X which satisfying ρ(x, x 0 ) r, S(x 0 , r) {x : ρ(x, x 0 ) r} By the closed sphere S(x0,r) is denoted collection of points X which satisfying ρ(x, x 0 ) r, S(x 0 , r) {x : ρ(x, x 0 ) r} Here x0 is called center and r is called radius of the sphere. Definition:(Closure Points) X is Closure point of the set M iff for any sphere with center X has at least one element from M. Definition:(Limit Point) X is limit point of the set M iff for any sphere with center X has at infinite number points from M. Definition: (Isolated Point) x Є M is called isolated point of the set M, iff for any sphere with center X has not point from M except itself. Closure point may be limit point or isolated point. Set of closure point of the M is called closure of the M and defined by M . Theorem: If A B then 1. A B 2. A B A B 3. Let us prove 2. 1. A A B A A B 2 AB AB B A B B A B is proved. Let us prove A B A B Assume converse: Let x Є A B and x A and x B . Then for any sphere with center X there aren’t point from A and B. It means there aren’t points from A B . X is not closure point A B It is inverse x Є A B . Definition:(Dence Subset) There are given two set A, B X . A is called Dence in B, if A B . Definition: (Everywhere Dence) Let A X . A is called everywhere dence if A X . Examples: 1) Set of rational numbers is dence in real line Q R and R Q Q' . 2) Set of polynomials with rational coefficient is dence. Set of continuous function. pn C . Remark: This example deduce from fames theorem that for any continuous function can be approximate by polynomials with rational coefficient. Seperable Spaces: The spaces is called seperable if it has countable everywhere dence subset. Examples: 1) R is seperable space. 2) C[a,b] is seperable. Example 1: En = {x: x = (x1, x2, x3, . . . , xn), xi Є R} n (x ρ(x, y) i 1 i y i ) 2 ; (R n , ρ) is metric space n ρ(x, y) x i y i ; (R 0n , ) is metric space i 1 ρ(x, y) max x i y i ; (R1n , ) is metric space 1i n Example 2: Set of numbers X = (x1, x2, . . . xn) with rational coordinates is dence in Rn, R 0n , R 1n . Example 3: Set of polynomials with rational coefficient is dence C[a,b]. Seperable Space 1) Set of rational number is countable dence in R. R is seperable space. R is seperable space. 2) Set of points x = (x1, x2, . . . xn), xi is rational number is countable dence subset for Rn , R 0n , R 1n . Then Rn , R 0n , R 1n are seperable metric space. Closed Set Subset of the metric space is called closed if it coinside with its closure. İ.e, M is metric space, A M, A A . Closed Ball in Metric Space S(a,r) = {x: ρ(x, a) r } [a,b] is closed interval. The set of all continuous function f Є C[a,b] with f(t) M is a closed set. Theorem: Intersection any number of closed set is closed and union of finite number set is closed. Proof: Let Fα is indexed as closed set by α. And we take X as X is limit point of Fα .Then α 1 for any ball contains X has element from Fα ; α = 1, 2, 3, . . . , ∞ ; open ball contains X has α 1 α 1 element from Fα and belongs Fα x Є Fα . α 1 n n i 1 i 1 Let us prove A i is closed. Assume contrary x is limit point of A i but n x A i x A i , i 1,2,. . . , n i 1 X is not limit point of Ai, i = 1,2, . . . , n There are δ1 0, δ 2 0, . . . , n 0 which ( x, 1 ) A1 ( x, 2 ) A2 . . . ( x, n ) An If we take δ min{δ1 , δ 2 ,..., δ n } S(x, δ) A i , for any i = 1, 2, . . . , n n n S(x, δ) ( A i ) X is not limit of A i i 1 i 1 n It is contrary of x is limit of A i . i1 Open Set in Metric Space Let M is metric space A is subset of M, A M. x A is called interior point of A iff, there is which S(x, δ) A . If any point of A is interior point then A is called open set in metric space. 1) Simplest example of open set is open interval in real line (a,b). Take any x Є (a,b), a < x < b denote δ min{x a, b x} (x - δ, x δ) C(a, b) 2) Open ball in metric space is open set. Let S(x, δ) {y; ρ(x, y) δ} Let take any y A and take r δ ρ(x, y) .Then S(x, r) S(x, δ) . Theorem: Intersection of finite number of open set is open and union of infinite number of open set is open. Why infinite intersection of open set is not open? 1 1 , {0} , {0} is unique element and closed. n 1 n n Complete Metric Space Completness real line is well-known. Let us give explanation Completness in metric space. Definition 1: The sequences Xn is satisfying Cauchy criterian if ε 0, n 0 which for each n,m > n0, we have |xn –xm| < . Definition 2: The sequence is called Cauchy Sequence (or fundamental sequence) if it is satisfying Cauchy Criterian. There is fames Cauchy Criterian for real line. Theorem (For the real line): In the real line any sequence is fundamental (Cauchy Sequence) if and only if it is convergent. Theorem: Any convergent sequence in metric space is fundamental. Proof: ε 0, n 0 , n n 0 , ρ(x, x n ) ε m, m n 0 , ρ(x, x m ) ε Let apply for Xn, Xm to the triangle inequality ρ(X m , X n ) ρ(X n , X) ρ(X m , X) ε ε 2ε n, m n 0 , ρ(X n , X m ) ε {X n } is fundamental sequence. Definition: If in metric space any fundamental sequence is convergent in this space then this metric space is called complete metric space. Example 1: n-dimensional Euclidean space with its metrics is complete metric space. Proof: Let {Xp} Є En, Xp = {X1p , X p2 ,..., X pn } and {Xp} is fundamental. ε 0, n 0 , p, ε n 0 ρ(X p , X q ) n (x i p i x iq ) 2 ε n (x ip x iq ) 2 ε 2 , p, q n 0 , x ip x iq ε, i 1 i the real line sequence X ip is fundamental {X p } is convergent for i 1,2,..., n lim Xip Xi (x1 , x 2 ,..., x n ) X p is convergent some x Є En which p X= (x1,x2, . . . , xn). Example 2: Sequence of continuous function in closed interval [a,b] is complete. Proof: Let {Xn(t)} is sequence of continuous function in [a,b] and {Xn(t)} is fundamental. Then by the definition we can write ε 0, n 0 which each n,m > n0, t Є[a,b] x n (t) x m (t) ε sequence of the function {Xn(t)} is uniform convergent if we take m→ ∞, x n (t) x m (t) ε Xm(t)→X(t). X(t) is continuous. C[a,b] is complete. Definition: Let M is metric space A M distance between the point x Є M and set A is ρ(X, A) inf ρ(x, y) distance between two sets. A and B is ρ(A, B) inf ρ(x, y) . yA xA yB Definition: The sequence of closed sphere S1[x1,r1], S2 [x2,r2], . . . , Sn[xn,rn], . . . is called nested (or decreasing) if S1[x1,r1] S2 [x2,r2] . . . Sn[xn,rn] . . . Using this concept, we can prove Completness theorem. Theorem (Nested Sphere Theorem): The metric space X is complete if and only if every sequence of closed nested sphere SA(xA,rn) which rn → 0, as n → ∞ has nonempty intersection. Proof: Assume X is complete metric space and Sn[xn,rn] are nested, closed sphere. Let take sequence of center of these sphere x1,x2, . . . , xn, . . . This sequence is fundamental; because n' n, ρ(x n' , x n ) rn and rn → 0 when n→ 0. Sn [xn,rn] = {x, ρ(x, x n ) rn } . Then {xn} is convergent lim xn x . n Assume x S n . Obviousely Sn has all point of sequence except x1,x2, . . . , xn-1 Sn is n 1 closed X is limit if Sn x belongs Sn. Assume any closed, nested sphere Sn[xn,rn] has nonempty intersection and take any fundamental sequence {xn}.We can choose sphere 1 1 S1 x n1 , ρ x, x n1 2 2 1 1 S2 x n 2 , ρ x, x n 2 2 2 2 . . . 1 1 Sk x n k , ρ x, x n k k 2 2 where n1 < n2 < n3 < . . . < nk < . . . 1 , n = 1, 2, 3 , . . . , n these sphere 2n 1 1 1 1 1 are nested.(Why?) Because 2 . S1 x n1 , S2 x n 2 , 2 ... Sk x n k , k 2 2 2 2 2 Then this sphere has nonempty intersection x Sn . Then x is limit of {x n k } because We can choose closed sphere center Xn with radius ρ(x, x n k ) 1 0 as k . Then x n k x . 2k Contraction Mapping Let A is a mapping defined on metric space (x, ρ ).The point x X is called fixed for A if X=A X mapping A is called contraction if there is 0 < α < 1 which we have ρ(Ax, Ay) ρ(x, y), x, y X. All this case we assume mapping A defined metric space X to itself A:X X. Theorem (Fixed Point): The contraction mapping A defined complete metric space X to itself has unique fixed point X; AX=X. Proof: Let take any x0 X and define X1=AX0, X2= AX1, ... , Xn= AXn-1 . . . Then we can build new sequence x0, x1, . . . , xn, . . . this sequence is fundamental. In fact, if we take n' n . ρ(x n , x n ' ) ρ(Ax n 1 , Ax n ' 1 ) α n ρ(x 0 , x n ' n ) α n (ρρ( 0 , x 1 ) ρ(x 1 , x 2 ) ... ρ(x n ' n 1 , x n ' n ) α n ρ(x 0 , x 1 )(1 α α 2 ... α n n 1 ) ' α n ρ(x 0 , x 1 ) 1 0 as n 1 α It mean that this sequence is fundamental. Then by conditional the theorem it is convergent (Because metric space is complete) Let us show this limit of the sequence is fixed point for the mapping A. lim x n = X n AX = A lim xn = lim Axn = lim x n+1= X n n n here we used continuous of the continuous mapping ρ(AX n , AX) ρ(X n , X) ; if xn→ x AXn → AX. We show that existence fixed point for the mapping A. Let us prove uniqueness.Assume there are two fixed points, Assume there are two fixed points, Ax = x Ay = y Then ρ(x, y) ρ(Ax, Ay) α ρ(x, y) ρ(x, y)(1- α) 0 ρ(x, y) 0 x y Application of the Contraction Mapping 1) Let f(x) is defined [a,b] to [a,b] and satisfying Lipschitz conditions with constant M |f(x)-f(y)| < M |x-y| If M < 1 then by theorem x = f(x) has unique solution. The limit of sequence x1=f(x0), x2=f(x1), . . . convergence to this solution. 2) Theorem (Piccard) Let f(x,y) is continuous some region G which contains point (x0,y0) and satisfying Lipschitz condition with constant M respect y. f(x,y1)- f(x,y2) < M |y1- y2 |.Then there exist δ , dy f(x, y) - - - - - 1 with initial which for the interval |x-x0| < δ , the differential equation dx condition y(x0) = y0 has unique solution y (x) . Proof (Piccard): We can reduce different equation (1) with its initial condition to the integral form as follow x y y 0 f(x, y)dy x0 It means that we can think some mapping A which described by x (x) y 0 f(x, (x)) dx f(x,y) is continuous in G, then there exist some G ' G and k>0 x0 which we have f(x, y) k (G ' continuous (x 0 , y0 )) Let us choose some which 1) for x x 0 δ, y - y 0 kδ (x, y) G ' 2) Mδ 1 Define G* the collect of continuous function (x) which for x x 0 δ and (x) y 0 kδ Take metrics on G* as ρ(1 2 ) max 1 (x) 2 (x) .Then metric space is complete and mapping x ψ(x) y 0 f(x, (x)) dx from G* to G * ' x0 In fact x ψ(x), y 0 y 0 f(x, (x)dx y 0 x0 x = f(x, (x))dx x0 x k dx k x x 0 kδ x0 We have to find which mapping to be contraction. Let 1 , 2 G * then x x x0 x0 ρ(A1 , A 2 ) max f(x, 1 (x) dx f(x, 2 (x) dx x max f(x, (x) f(x, 1 2 (x) dx x0 x max M 1 (x) 2 (x) M x x 0 max 1 (x) 2 (x) Mδδρ1 , 2 ) x0 means that if Mδ 1, δ 1 then differential equation (1) has unique solution. M Solution of Fredholm Integral Equation b The Fredholm Integral Equation is f(x) λ κ(x, y)f(y)dy (x) ______(1) a Here κ(x, y), (x) are given continuous function in some domain a x b, c y d, is parameter. f(x) is unknown function κ(x, y) is called kernel of this equation. We can also contradiction mapping for the integral equation (1) by using relations. b g(x) λ κ(x, y)f(y)dy (x) ______(2) a If f(y) is continuous [c,d] then we can build metrics by using ρ(f1 , f 2 ) max f1 (x) f 2 (x) xc, d We build mapping from C[a,b]. This metric space is complete. Then we can use contraction mapping theorem b b ρ(g 1 ,g 2 ) max λ k(x,y)f1(y)dy (x) λ k(x,y)f 2(y)dy (x) a a b b max λ k(x,y)f1(y)dy k(x,y)f 2(y)dy a a b λ k(x,y)max f 1(y) f 2(y)dy a λ M max f 1(y) f 2(y) (b a) λ M b a ρ(f 1 ,f 2 ) ρ(g 1 ,g 2 ) λ M(b a)ρ)ρ1 ,f 2 ) For last relation to be contract mapping it must be λ M(b a) 1 λ 1 ; here M(b a) k(x, y) M, k here k(x,y) is continuous. Problem 1) Let A be a mapping of metric space R into itself. Prove that the condition ρ(Ax, Ay) ρ(x, y), (x y) is insufficient for the existence of a fixed point of A. Solution: Equivalence it means that fixed point theorem is not hold if contraction mapping ρ(Ax, Ay) ρ(x, y) false when α 1 (means theorem of fixed point false ρ(Ax, Ay) ρ(x, y) ). If we use prove of the fixed point theorem we can write 1 ρ(Ax, Ay) ρ(x n , x 'n ) α n ρ(x 0 , x) ; 1 α if n ρ(x n , x 'n ) 0 x n is Cauchy Sequence but if α 1 . ρ(x n , x 'n ) 1n ρ(x 0 , x) 1 . 1 α Topological Space Definition: Let X is some set and collection of subset of X be following conditions. τ . If τ is satisfying 1) and X belongs to τ . 2) infinite union and finite intersection in τ again belongs τ . Then it is said in X is given topology and it is denoted by ( X , ) .All the element of τ is called open set. Definition of closure point, limit point, isolated point and neighbourhood of the point X can be give in topology space. Let x X. Then neighbourhood of the point X is some G τ which x G . Let M X. X is called closure point of M if any open set G T contains X has nonempty intersect with M. (except x) Limit point, isolated point can be given as same way. Example 1) Metric space is special case of Topological space. Solution: First axiom is Obviousely. Second axiom can be obtain by using theorem in metric space about open sets. Example 2) Consider set with two elements {a,b}.Then we can give topology on this set as follows T ={{a,b},{a},{b}, }. Weak Topology Definition: Let T1, T2 are given two topologies on same set X. We will said T1 is weaker topology then T2 if T1 T2 . Theorem: Intersection of topology is also topology. If T1 T2 T3 . . . Tn are topologies then n T Ti ; T is also topology. İ1 n Proof: If and X belongs any Ti , X T Ti İ 1 If infinite union and finite intersection belongs any Ti infinite union and finite n intersection Ti İ 1 Base in Topological Space Let it is given Topological space T (X, τ) collection of elements from T, G α T is called base of the Topological space T if any subset M of X, can be shown as union of some elements of G . Remark: In metric space we first gave open ball and extended this to open set. Then we can say base in metric space is union of open ball. Because any element in metric space can be shown as open ball. Theorem : (Necessary and Sufficient Condition for Base) Let { G } are collection of element from T in Topological space, T (X, τ) . { G } is a base for Topological T if and only if for any G X, and x G there exist some G α' G α which x G α' G . Proof: ( ) Let { G } is a base. Then for any G X there exist some elements of { G } which G G α . Then if we take x G α ' which x G α' G . ( ) Let take G X, and take any x G .Then there exist some G , which x G α' G .Then we can fix all such element G , and union of then G , G α will base of T (X, τ) Topological space. In the topological space it is interesting to take no more countable base from the base of this topological space. Example: In metric space from the base we can choose countable base. In fact base in metric space one collection of open ball. If we choose balls with rational radius then collection of such balls is base in metric space. Theorem: (Countable every where base) Let T (X, τ) is topological space and G is countable base in T. Then there exist countable dense subset of T. Remark: If topological space T has countable base, then it is said T satisfies second axiom countability. Proof: Let O n is countability base for T and we take elements X n from O n , x n O n .Let us denote M {x1 , x 2 ,..., x n ,...} M is everywhere dense in T. If it is not then G T M 0. G is open set, then can be shown as union of some set of O n .(because O n is a base).It means G has some element from M. It is contradiction. In metric space, this theorem hold also inverse direction. Theorem: If metric space X has countable everywhere dense subset, then this metric space has countable base.(X satisfies second axiom of Countability) Proof: Let M {x1 , x 2 ,..., x n ,...} is countable everywhere dense subset of X. Then if we take 1 any open subset G of X, and x G , then we can choose open sphere S(x m, ) for some m n 1 and n which x S(x m, ) G .(Why? Because M X ) These open sphere can be take as n countable base. Corollary: (Necessary and Sufficient Condition) Confine let do theorems, we can say that metric space has countable base iff it is seperable. 2 Example 1) The sets real line, continuous function in [a,b] R, R1, Rn, 2 , C[a, b] which we showed that has countable dense subset all these sets has countable base. Example 2) The set of point with bounded coordinates are not seperable has not countable base. Remark: Set which elements on the sphere or inside sphere is uncountable. Proof: Let us take elements from M which has coordinate only zero and 1. We know that metrics in M is ρ(x, y) sup x i y i for such elements which coordinates only zero and one for to distinct coordinate metric will be 1. 1 .This sphere has not intersection. 2 This sphere is uncountable. If some set M is everywhere dense then for any sphere has element from M. This means M is not countable M is not seperable. Let set all sphere center at this coordinate and radius Countable neighbourhood base (First Axiom of Countability) In metric space X any point x X has countable neighbourhood base. For any x X , there exist countable open sets O which for any open G and x G , x C G , where C {O} . All the metric space has countable neighbourhood base. But in the topological space it may not hold. Definition: If topological space has countable neighbourhood base then it is said that this space has first axiom of countability. Definition: The collection of the set { M } is called cover of the topological space T, of T Mα . α If M is open then { M } is open cover. If M is closed then { M } is closed cover of T. Theorem: (To choose finite or countable subcover in topological space) If topological space T has countable base, then any open cover of T, can be choose finite are countable subcover. Proof: Let G {M α } is open cover of T and x G ( x M ).Then by using necessary and sufficient condition for the countable base in topological space T, we can choose countable base {K}. Then again by using this theorem for open M , and x M , there exist K β K , which x K β M α . Convergent Sequences in Topological Space In metric space (X, ρ) if M X is subset and X is closure point for M, then there exist convergent sequence x n M which x n x .But in topological space it may not satisfy.Of course, if T is topological space, M T is subset X is closure for M, x [M] but it may not exist convergent sequence, {x n } M. For this, we need other assumption on the topological space . Theorem: If topological space T satisfies first axiom of countability then for any closure point of the subset M τ has convergent sequence, i.e, {x n } M x n x . Proof: Let O is neighbourhood base at the point x. Then we can countable O n , and no loose n generality assume O n 1 O n . Then we can choose On be hold of O i . Take any point xn of i 1 M which contains On. We can always choose this because x is closure of M then x n x . Axiom of Seperation Now that most of properties of the metric space easily can be carry to the topological space. But there are some properties in topological space is not hold in metric space. For this topological space is the general concept of the analysis. Definition: If in topological space T for any two distinct point x y there exist neighbourhood Ox and Oy with x O y and y O x then it is said that T is satisfies first axiom of Seperation or it is T1 space. Example: Let T is space, take any subset T of T then (T, T) will be topological space. It is also T1 space. (It is easy to check this).But if T has two elements T = {a,b} and we topology as set T itself empty set and set {b}. T = {{a,b}, ,{b}} then (T, T) is topological space but is not T1 space. In fact if we take point a and b, then we have {a,b} which contains and {b}. Theorem: Every finite subset of T1 space is closed. Proof: Let {x} has singletion. Then for y x , y [x] . It means [x] = x, we can extend this for the set which has finite number of elements. Remark: Let us give strong from of the definition of T1. Definition: If in topological space T any two distinct points x and y, has two neighbourhood Ox and Oy which, O x O y , then T is called T2 space or Hausdorff Space. Example: Any T2 space is T1 space but is not hold conversely. Definition: T1 space is called normal if any closed set F1,F2 in T there are open set O1 containing F1 and open set O2 containing F2 which O1 O 2 . Example:1) Any normal space is T2 (Hausdorff) space. It is easy to check it. Example: 2) Hausdorff space (T2) may not to be normal space. Let take closed interval [0,1] and take any open neighbourhood of the point x 0 as usual, 1 1 1 and neighbourhood of x = 0 to be denote 1, , ,... ,... (1) deleted. 2 3 n 1 1 1 If we take set {0} and set which has element {1, , ,... ,...} these sets are closed. But has 2 3 n without disjoint neighbourhoods. It means that this space is not normal. Theorem: (Normality of the metric space) Metric space is normal. Homomorph Topological Space Let there are given two topological space X and Y and mapping between then f : X Y . If this mapping f is one-to-one and f, f-1 are continuous, then X and Y called Homomorph space. Homomorph space has same characters. Continuity Topological Space Let X and Y two topological space and f is mapping between them. It is said f is continuous at the point x 0 X. If for any open set G Y containing y0=f(x0), there exist open set E containing x0, which f(E) G. If f is continuous any point of X, then it is said that f is continuous on X. Theorem: (Necessary and Sufficient Condition for Continuity) The mapping f : X Y is continuous in X iff the pre image f-1(G) for any open set G Y is open in X. Proof: Let f be continuous mapping from X to Y take any open set G Y and consider f-1(G) f is continuous means for any y 0 G , the open set containing y 0 f(x 0 ) (we can take open set containing y0 as G) there exist open set x 0 X and f (x 0 ) G . We took any x0 and found open x 0 f 1 (G) . It means f-1(G) is open. Compactness in Topological Space In real line we know Heine-Borel theorem as follow any open cover of the closed interval [a,b] has a finite subcover. Let us give definition of the compactness in topological space. Definition: Topological space T is called compact if any cover of T has finite subcover. Hausdorff (T2) space is called compactum. Definition: The sequence of subset {A α } of the topological space T is called centred, if any n finite intersection A k is nonempty. k 1 Theorem:(Necessary and Sufficient Condition of the Compactness of the topogical space) The topological space T is compact iff any centered closed sequence subset {A α } of T has nonempty intersection. Theorem: If f is continuous mapping between two topological spaces X and Y, and X is compact then y so is. Proof: We have to prove that any cover of Y has finite subcover. Let { G } is open cover of Y. Then by Continuity of f pre image {f (G α )} is also open and cover of X. X is compact then by definition of compactness we can take finite subcover n n i1 i 1 i.e, f 1 (G1 ) f -1 (G2 )... f -1 (Gn ) which x {f(G i )}. Then G i is also cover of Y and finite. It means that Y is compact. Theorem 7: If T is compact topological space then any infinite subset of T has at least ne limit point. Proof: Let X is infinite subset of T and has not limit point. Then there is countable subset of T. M {X1 , X 2 ,..., X n ,...} which has not limit. Take set of from X n {X n , X n 1 ,...} .These sets are centered, closed and empty intersection X n .The T is not compact. It is α contradiction of the theorem condition. Theorem: Closed subset of the compact topological space is compact. Definition (Countable Compact): Topological space T is said countable compact if any infinite subset of T has at least one limit point. It is clear that if topological space is compact then it is countable compact but inverse is not hold. But we can write a condition which countable compactness reduce to compactness. Theorem: (Necessary and Sufficient for Countable Compactness) Each following two conditional one necessary and sufficient topological space T to be countable compact. 1) Every countable open cover of T has finite subcover. 2) Every countable centred system of closed subset of T has nonempty intersection. Theorem: The concepts compactness and countable compactness coincide for a topological space T with a countable base. Relatively Compact Topological Space In topological space it is interesting to think about subset is closed. Definition: The subset M of the topological space T is called relatively compact if M is compact in T. If we use theorem.(closed subset of the compact topological space is compact) We can say any closed subset of the topological space is relatively compact. Compactness in Metric Space Since metric space are topological space with special kind, definitions and results in preceding section apply to metric as well. But compactness concepts in metric space related with total bounded. Definition (_net) : Let R be is metric space and is positive number, subset A R is _net for the subset M R , if for x M there is at least one point a A , which ρ(a, x) ε . Definition (Total Boundness): The subset M R is total bound if for ε 0 it has finite _net . From this definition, it is clear that total bound set M is bound. Because M is union of finite bounded set. Example: In Euclidean Rn space Boundness is equivalent to total boundness. Indeed, let M is bounded is positive number then it is setting in some cube, Q. Let us divided cube by small ε cubes with side . The all vertexes of these cubes is finite set and make n . -net set for 2 ε Q, then it also make n -net set for M. Then M is total bound. 2 Remark: If subset M is bound then it may not total bound. Let us take x 2 , x = (x1, x2, . . . , xn, . . . ) with x k 1 2 k 1 . It is sphere in 2 of course bounded. If we take base in 2 such e1 = (1,0,0, ...), e2 = (0,1,0, ...) . . . en = (0,0, ... , 1,0,0). These points setting on M then is not total boundness because distance any ei, ej, ρ(e i , e j ) 2 . Indeed if we take ε 2 then definition of total boundness is not hold! 2 Example: Let us consider Hilbert cube in 2 .i.e, x 2 , x = (x1, x2, . . . , xn, . . . ) and 1 1 1 x 1 1, x 2 , x 3 2 , ..., x k k 1 , ... and denote this subset by . is denote this 2 2 2 1 ε subset by . is total bound for this let us take ε 0 , and such n that n 1 and for 2 2 * x = (x1, x2, . . . , xn, . . . ) consider such point x = (x1, x2, . . . , xn,0,0, ...) ρ(x, x * ) x k n 1 2 k k n 1 1 4 k 1 1 2 n 1 ε . 2 Theorem: Every countable compact metric space is total bound. Proof: Assume conversion i.e R is . Then it mean that there exist which R has not _net .Choose a 1 R .Then there exist a 2 R which ρ(a1 , a 2 ) ε .Otherwise a1 is _net for R. Let fix a1;a2 there exist a3 R which ρ(a 1 , a 3 ) ε ρ(a 2 , a 3 ) ε .Otherwise a1 and a2 is _net of R. Of course, we choosed a1, a2,..., ak and there exist a n 1 R which ρ(a1 , a k 1 ) ε , ρ(a 2 , a k 1 ) ε , ..., ρ(a k , a k 1 ) ε . Then we build sequence a1, a2,..., an, ... which has not limit point then infinite subset {a1, a2,..., ak, ...} of R is not countably,compact. Theorem: Every countable compact metric space has countable everywhere dence subset and countable base. From this theorem we can say following corollary. Corollary: Every countable metric space is compact 1) Countably compact metric space has countable base 2) Compact and countably compact equivalent in T space has countable base. We shown that boundness is necessary condition for compact metric space. But inverse is not true. For example, let take set of rational numbers in closed interval [0,1] with distance metris. This set is total bound. Let take sequence of rational number 0, 0.4, 0.41, 0.414, 0.4142, ... this sequence is convergent to irrational number 2 1 .But this has not limit point in rational number.(decimal approximation) It means that this set is not countable compact. Let us give necessary & sufficient condition for compactness, of the metric space. Theorem: (Necessary & Sufficient Condition for Compactness) A metric space R is compact iff it is total bound and complete. Its interested to learn compactness on set of continuous function which defined on closed interval [a,b]. For this we have to give some definition. Definition (Uniformly Bounded of the set of function): The set of of the functions which defined on closed interval [a,b] is said to be uniformly bounded if (x) M, for all Φ and all x [a, b] , some constant M. Definition: (Equicontinuous of set of functions) The set of the functions defined on [a,b] is said to be equicontinuous if any given ε 0 there exist δ 0 , which for any x1, x2 [a,b] x 1 x 2 δ implies (x1 ) (x 2 ) ε for all ψ Φ . Theorem: (Arzela) Set of continuous functions defined on [a,b] is compact iff uniformly bound and equicontinuous. Theorem: (Peano) Let function f(t,y) is continuous on plane domain G in differential equation dy f(x, y) (1) dx Then there is at least one integral curve of the differential equation (1) which passes point (x 0 , y 0 ) G . Real Valued Function on Metric and Topological Space Let T to be topological space and f is function defined on T.We can put special case of topological space, metric space R and collection of the function defined on R. Simple definition of the functional is mapping defined on such collection of functions. Example: Let X(t) be continuous function defined on [0,1]. (S0 , S1 ,...S n ) is a n+1 real variable function, ψ(t, u) is defined on two real variable then following mapping are functional. F1 (x) sup X(t) 0 t 1 F2 (x) inf X(t) 0 t 1 F3 (x) x(t0 ) where t0 [0,1] F4 (x) [x(t0 ), x(t1 ),..., x(t n ) 1 F5 (x) ψ[t,x(t)]dt 0 F6 (x) x(t0 ) where t0 [0,1] 1 F7 1 x' (t)dt 2 0 1 F8 x' (t) dt 0 All these relations are functionals. F1, F2, F3, F4, F5 are continuous functionals. F1 is continuous, because ρ(x, y) sup x y and supx supy sup x y F1 (x) sup x(t) 0 t 1 ε 0 δ x(t) y(t) δ ρ F1 (x), F1 (y) ε sup x sup y sup x y F6 is not continuous function. Let us prove this. Let take any differentiable X(t) with x(t) ε and x ' (t 0 ) 1 For example we can take x(t) x ' (t) 1 2 t 2t 0 t t0 x ' (t 0 ) 1 denote y(t) x 0 (t) x(t) for any differentiable x0(t). It is clear y(t) x 0 (t) x(t) ε . We know that y ' (t 0 ) x '0 (t 0 ) x ' (t 0 ) x '0 (t 0 ) 1. y ' (t 0 ) x '0 (t 0 ) 1 . Then F6 is not continuous. Show that F6 is continuous function on the C.But we can show F6 is continuous C’ (set of function which its and first derivative are continuous) which the metrics ( x, y ) sup x(t ) y (t ) x ' (t ) y ' (t ) [ 0 , 1] Example: F7 is discontinuous functional in fact let take X0(t) = 0, Xn(t) = 0 ρ(x 0 , x n ) 1 sin2π n t 2 1 1 1 sin2π nt sin2π nt 0 n 2 n But we can shown that. 1 F7 (x 0 ) dt 1 0 1 F7 (x n ) 1 4π 2 sin 2 2π nt dt 4 0 not F7 (x n ) F7 (x 0 ) F7 is discontinuous Definition: Let real valued function defined on metric space R. It is said f(x) to be uniform continuous on R. If ε 0, δ () , the ρ(x1 , x 2 ) δ implies f(x1 ) - f(x 2 ) ε for all x1,x2 R uniform continuous continuous We know that function defined on real line and continuous on closed and bounded set then this function is uniform continuous. We can generalized this also in metric space. Theorem: The continuous function defined on compact metric space R is uniform continuous. Proof: Let f(x) is continuous but is not uniform continuous. Then ε 0, {xn}; x n k ρ(x n , x n k ) 1 there exist n 1 (1) implies f(xn ) f(xn k ) ε n R is compact then {x n' }{x n } convergent some x0. By (1) x 'n k {x n k } is also convergent to x0. Then ε f(x 0 ) f(x n ' ) (2) ; 2 We know that f(x) is continuous then one of (2) is not hold it means that f(x) is uniform continuous. There are other properties of the continuous function which defined on compact set. Then let introduce: Theorem: The continuous function defined on compact metric space (topological space) R is bounded. Moreover it takes upper bowd and lower bounded value. Proof: Let f defined a compact topalagical space T and takes real valued. By using theorem we can say mapping T by f is compact R1. Then image of f in R1 is closed and bounded then f(x) takes supremum and infimum. Definition:(Cluster Point) The number is called cluster point of the sequence {xn} if >0 and n | n- |< . The sequence {xn} maybe has cluster points bu not limit point. Example: xn = (-1)n lim x2n = 1 lim x2n+1 = -1 xn has two cluster point 1, -1 but has no limit point if there are cluster point if there are cluster points and equal each other then it is limit point. Definition : (Upper Semi Continuity) The function f(x) is said to be upper semi continuous at x0 if given any , there exist neighbourhood of x0, which f(x) f(x 0 ) ε for every x in this neighbourhood. X0 Definition : (Lower Semi Continuity) The function f(x) is said to be lower semi continuous at x0 if given any , there exist neighbourhood of x0, which f(x) f(x 0 ) ε for every x in this neighbourhood. X0 If function lower and upper semi-continuous at a point x0 then this function is continuous at x0. If function continuous then lower and upper semi-continuous but inverse is not correct. It means that semi-continuous function covers larger class of the function that the continuous function. Example 1) Integral part of x, i.e [x] f(x) = [x]. f(x) = [x] is upper semi-continuous but is not continuous. 1, if x 1 1 Example 2) f(x) , if x 1 2 1 2 , if x 1 is lower semi-continuous. 1 sin ; if x 0 Example 3) f(x) x 0; if x 0 is upper semi-continuous. Theorem: Lower semi-continuous function defined on compact metric space bounded from below takes it is upper bound value. Proof: Let assume contrary, i.e, inf f(x) = . Then there exist xn, which f(x n ) n . Let take E = {x1, x2, ..., xn, .. } space is compact, then there exist x0, which x n x 0 . f(x) is lower semi-continuous (x 0 ) which f(x) f(x 0 ) ε has finite number of element from E. Because f(x) = . Then x0 is not limit point. It is contradiction, i.e, f(x) is bounded below. Definition: (Limit Superior) The number is said to be limit Superior of f(x) at x0. Definition: (Limit Interior) The number is said to be limit interior of f(x) at x0. Difference is called osculation f(x) at the point x0. For the sequence {xn} the limit superior is defined by lim x n inf supx n and k n k interpreted as 1) ε 0, n 0 , n n 0 , x n ε 2) ε 0, n 0 , n, xn ε CHAPTER 4 LINEAR SPACE Definition: The set L of the elements x,y,z,w,... are called linear space, if it satisfies following three identity: 1) For any two elements x,y of the set L corresponding element x + y L which satisfies x+y=y+x (x+y)+z = x + (y + z) (asso.) There exist element 0 L is called zero element with the property that x + 0 = x For any element x of L, there exist – x is called negative of x, with the property x + (-x) = 0 2) For any element of L and real numbers , the following properties satisfies α(βx) ()x 1x x 3) Distributive Law ( ) x x x ( x y) x y Note 1: Linear space obeys two operation, addition of elements and scalar multiples. Note 2: If scalar multiples is real numbers then L is called real linear space it complex numbers then L is complex linear space. Example 1: The real line is a simple example with usual operation of sum of real numbers and scalar multiplication. Example 2: Set of n-tuple ordered numbers x = ( x1, x2, ... , xn) and y = ( y1, y2, ... , yn) are linear space with operation of sum of elements x + y = ( x1+y1, x2+y2, ... , xn+yn) and multiplication of numbers αx (α x1 , α x 2 ,...α x n ) .This space is called n-space and denoted by Rn. If numbers are complex then it is called complex linear space and denoted by Cn. Example 3: Set of ordered infinite sequence of numbers x= ( x1, x2, ... , xn) with the property x k 1 2 k is a linear space and denoted by 2 space. The property x, y 2 x y 2 can be shown by using following simple inequality x y 2(x2 y 2 ) .In fact, if x y 2 ; X= ( x1, x2, ... , xn), y = ( y1, y2, ... , yn).Then x + y = ( x1+y1, x2+y2, ... , xn+yn) and 2 2 2 2 2 (x y ) 2(x y ) 2 x k yk 2 . k k k k k 1 k 1 k k 1 2 Let again see the set of ordered sequence of infinite numbers x= ( x1, x2, ... , xn, ...) (1) If this sequence convergent lim x k , it defined the Complex space, if sequence of (1) is k convergent to zero lim x k 0. Then it defined C0 space, if (1) is bounded x k M , the it k defined by M, if it divergence then it defined by Rn. All these space are linear space. If we talk about the linear space then first the think about operation of sum of elements and multiplication of elements by numbers. Then we can define isomorphism between linear space. Definition: Two linear space L and L* is called isomorphic if it preserves operation of addition of elements and multiplication by scalar: x x* y y* x y x * y* α x α x * for any numbers where x, y L, x * , y * L* . Definition: Elements x1, x2, . . . , xn of the linear space L are called dependent if there exist numbers 1 , 2 ,... n not all are zero which satisfies following relation. α x1 , α x 2 ,...α x n 0 Converse case is called independent of course if last relation hold only if α1 α 2 ... α n 0 then elements x1, x2, . . . , xn are called independent. Dimensional of the Linear Space The linear space L is called n-dimensional then always it can be found n independent elements in L. But if we add one more element then it converge a linear dependent. Linear space is called infinite dimensional if any n elements of this space are independent for every n. Definition: (Subspace) Let L is linear space and L’ is subset of L. L' L . If L' also satisfies operations of L then L' is called subspace of L. Exactly L' is subspace of linear space L, if x y L' , when x, y L' . Example: Let L is linear space fix any element x L and take all elements from of x , where changes all numbers on real line. Then L' {x, R} is subspace. Definition: (Linear Functionals) The real valued function f which defined on linear space L is called linear functional if it satisfy following condition: f ( x y ) x y Example 1: Let us define linear functional on n-space; Rn. Fix any element a R n , a = (a1, a2, . . . , an ) and define functional as following n f(x) a k x k , x = (x1, x2, . . . , xn) k 1 x, y R n n f (x y) ak (xk yk ) k 1 n n α a k x k β a k y k k 1 k 1 αf(xk ) βf(yk ) Example 2: Let us define on C[a,b] linear functional as follow b I(X(t)) x(t)dt a Or more generally, fix any a(x) C[a, b] b I(X) x(t)a(x)dt a b I(x(t) y (t ))) (x(t) y(t))dt a b b a a x(t)dt y(t)dt α I(x(t)) β I(y(t)) Definition:(Null Space) Let L is linear space and f is linear functional defined on it. Then it defines space as follows L f {x L; f(x) 0}. Lf is subspace of the space L and is called null space. There is following interesting theorem between linear space and null space. L – Lf is linear space. Theorem: Let x0 is fixed as x 0 L L f then for any x L has unique representation X αx 0 y , where y L f Proof: Clearly f(x0 ) 0 . No less generality, we assume that f(x0) = 1. Another case we need x x0 1 f(x 0 ) 1 . and notion f 0 f(x 0 ) f(x 0 ) f(x 0 ) Then take any element x L , denote α f(x) and consider new element y x αx 0 . Then y L f since only replace x0, by f(y) f(x αf(x 0 )) f(x) f(α( 0 ) f(x) - αf(x 0 ) α - α.1 0 y Lf We proved existence representation, i.e, existence y and such that satisfying X α x 0 y (1), where y L f , x 0 L L f is fixed. Let us prove uniqueness of this representation. Assume that there exist another choice of y’ and ' , i.e, x α' x 0 y ' (2) Substracting from (2) to (1), we can get y' y y' y ( ' α)x 0 x 0 then α'α y' y f(y') f(y) 0 0 f(x0 ) f 0 α'α α'α α'α x 0 L f . It is contrary of choice of x0. Definition: (Convex Set) The set M is called Convex if for any x, y M the expressions αx βy M , where α, β 0 and α β 1 . The set of all points which αx βy, α, β 0 , α β 1 is called segment connecting between point x and y. Definition: (Interior of the Convex Set) Let E is convex set M is called interior of the E, if we take x M , and y M then there exist ; which x ty M where t ε . Convex set is called a convex body if it has nonempty interior. Example: Let C[a,b] and take of the subset M of C[a,b] which f M ; f(x) 1 . M is convex f, g M f 1 , g 1 . Let take αf βg for any α β 1 , α, β 0 . Then αf βg αf βg α f β g α.1 β.1 α β 1 αf βg 1 αf βg M Example: Let take 2 space and take subset M of 2 with the property x M, x = (x1, x2, . . . , xn) M is convex set. x k 1 2 k 1 for Definition: (Convex Functionals): Let L to be linear space and f(x) is functional defined on it. The functional f(x) is convex functional, if f(α1x α 2 y) α1f(x) α 2 f(y) for any x, y L α1 , α 2 0 , α 1 α 2 1 . Note: Unlike in the case of linear function, we don’t assume convex functional take finite value, the case of f(x) is allowed for some Example: The length of the vector Euclidean n-space is convex functional. Let us prove this in Euclidean n-space length of the vector x = (x1, x2, . . . , xn) is d(x) x12 x 22 ... x 2n . Let us prove d(x) is convex functional. Let x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) then d(λ1 x λ 2 y) (λ1x1 λ 2 y1 ) 2 (λ1x 2 λ 2 y 2 ) 2 ... (λ1 x n λ 2 y n ) 2 d 2 (λ1 x λ 2 y) (λ1 x 1 λ 2 y1 ) 2 (λ1 x 2 λ 2 y 2 ) 2 ... (λ1 x n λ 2 y n ) 2 n n n n k 1 k 1 k 1 (λ1 x k λ 2 y k ) 2 λ12 x 2k λ 22 y 2k 2λ1λ 2 x k y k k 1 Cauchy Schwartz n n λ12 x 2k λ 22 y 2k 2λ1λ 2 k 1 k 1 n n x 2k k 1 n y k 1 2 k n λ12 x 2k λ 22 y 2k k 1 k 1 (λ1d(x) λ 2 d(y)) d(λ1 x λ 2 y) λ1d(x) λ 2 d(y) 2 Example: We can define convex functional on the m-space, i.e,the set of all points which xi ; i = 1, 2, . . . is convergent. Let define functional on M as follow f(x) sup x k . k This functional is convex. Minkowski Function In this last chapter we will give connection between convex set and convex function. Theorem: Let p is a convex functional defined on linear space and k is a positive number. Then the set E {x L; p(x) k} is convex set. If p is also finite then E is body with interior I(E) {x L; p(x) k} there 0 I(E) , i.e, I(E) contain zero element. Note: The meaning of this theorem such that any convex functional define convex set. Inverse of this theorem is also hold. Let E is convex body which contains zero element then following functional is called Minkowski functional. Theorem(Minkowski):The Minkowski functional is finite and convex. About the Hahn-Banach Theorem: Let the space L to be linear space and L 0 L is subspace f0(x) is linear functional defined on subspace L0.The linear functional f(x) which define whole space L and satisfying equality f0(x) = f(x) on L0 is called extention of the linear f0(x) whole space L. Note: In many problems of the convex analysis in countered is that extension of the function, origional defined on subspace, to the more general space. Theorem: (Hahn-Banach) Let p is infinite convex functional defined on linear space L and f0(x) is a linear functional defined on subspace L 0 L where f 0 (x) p(x) (4), x L 0 .Then there extension of the functional f(x) which defined whole space L satisfying inequality (4) on it. Euclidean Space (Inner Product Space) By a scalar product in the real linear space R we meant the function of the two pairs x, y R with the following properties: 1) (X, X) 0 for all x, y R and (X, X) 0 iff x 0 2) (x, y) (y, x) 3) (λλxy) λ(x, y) for all x, y R and for any real λ 4) (x y, z) (x, z) (y, z) Definition: (Euclidean Space) The real linear space equipped with the scalar product is called Euclidean or iner product space. Theorem: (Schwartz- Inequality) For any two elements from Euclidean space the following Schwartz- Inequality is hold. x, y x y where x (x, x) and y (y, y) . Proof: The case, when x and y are zero element is obviously. Then, we assume that x 0, y 0 . Let us define new function as follow: (λ x y, λx y) 0 If we take some calculation on (x ) , we can get λ 2 x, x λx, y y, λx y, y 0 x, x λ 2 2x, y λ y, y 0 We know that x, x λ 2 2x, y y, y 0 x, y2 x, x . y, y 0 x, y2 x, x y, y x 2 is equal to the (x ) is quadratic polynomials. y x, y x y 2 Theorem: The Euclidean space R becomes normed space if we define norm, x for any x X . x, x , Proof: First & second properties of the Euclidean space. Let consequence of the definition of Euclidean space. Let us prove triangle by using Schwartz-Inequality xy x y If we take two elements from the Euclidean space R, it follows that x y x y, x y x, x 2x, y y, y 2 x, x 2 x, y y, y x 2 x y y 2 2 x y 2 If we consider first and last part of the expression, then we get xy x y For any two vectors of the Euclidean space, x, y R it can be define angle between them. x, y ; 0 θ π (1) .This angle is unique, because the SchwartzLet x, y R , then cosθ x y Inequality right side of the last relation x, y x y x, y i.e, x y is not exceed 1. x, y x y If (x,y) = 0, then from (1) 1 . In this case the vectors x and y is called ortogonal. 2 Definition: The set of vectors x α of the Euclidean space R is called ortogonal if 0, α β , x β 0, when .Ortonormal if x α , x β 1, α β The Ortonormal system x α is called ortonormal base if it is complete set. x α Example: Let take real n-space Rn. If we take any two element x, y R . x = (x1, x2, . . . , xn) , y = (y1, y2, . . . , yn) and define scalar product as follows. n x, y x i y i , then Rn becomes Euclidean space. It is easy to check all the properties of i 1 the Euclidean space. One of the ortonormal systems in Euclidean space Rn is e1 = (1,0,0,0,...0) e2 = (0,1,0,0,...0) . . . en = (0,0,0,0,...,1) Example: The set of continuous function in [a,b], i.e, C[a,b] with the scalar product b f, g f.g.dx is a Euclidean space. The main ortonormal base in this space is a 1, cos 2π nx 2π nx , sin ,..., ba ba n 1,2,3,... (2) b cos a 2π nx (b a).sin2π nx dx b-a 2π nx (b - a) b a b 1 sin2nx 0 2ππ a If we take a π, b π then (2) becomes simplify 1, cosnx, sinnx,..., n 1,2,3,... 2π nx 2π nx , sin ,..., n 1,2,3,... is complete. ba ba The most important notion is seperable Euclidean space.i.e, Euclidean space on seperable space Rn, 2 ,C[a,b]. Theorem: The system 1, cos Theorem: Seperable Euclidean space has no more then countable elements. Proof: Let x α be system of countable elements Euclidean space R and convert this ortogonal system to the ortonormal system as follows: y α x α . We know y α 1 . x α Let take spheres with the center at y α and radius 2 .These spheres, are not intersect 2 y1, y2, . . . , yn, . . . y1 1 , y 2 1 , . . . y1 y 2 y1 y 2 , y1 y 2 y1 , y1 2y1 , y 2 y 2 , y 2 2 If we take any x R by density there exist element from y α which neighbourhood of the yα . Theorem of Orthogonalization In preceding section, we show that in the Euclidean Rn space. There exist ortogonal vectors. In this section we will generalize this opinion to the infinite dimensional Euclidean Space. Theorem: Let f1,f2, . . . , fn, . . . are linear independent elements in Euclidean space R. Then there exist vectors 1 , 2,...,n ,... in Rn such that 1) 1 , 2,...,n ,... are ortonormal. 2) Any vectors n can be describe linear combination of the vectors f1,f2, . . . , fn, . . . n an 1f1 an 2 f 2 ... an n f n (a n n 0) 3) Every vectors fn is a linear combination. f n bn 11 bn 2 2 ... bn n n (bn n 0) These orthogonalization process is unique within factor of 1 . This process can be describe as analytical following form 1 f1 n f , 2 f 2 2 21 1, i ,2,..., n 1 i 1 . . . n n 1 f n 1 f n 1 , i , i i 1 i 2 i ,2,..., n Theorem: Every seperable Euclidean space has countable orthonormal base. Proof: We can choose countable elements f1,f2, . . . , fn, . . . due to Euclidean space is seperable. Then we will try to find linear independent vectors between elements f1,f2, . . . , fn, . . . as follows: If k is a linear combination of the elements f1,f2, . . . , fk, . . . then we eleminate vectors f k . After his process remain elements will be independent. Then we can use orthoganalization process to choose orthonormal vectors. Bessel Inequality Closed Orthogonal Systems: We show that in Rn there exist orthonormal base e1,e2, . . . , en, . . . for which any f R n , n f c k e k , where ck = (f, ek), k = 1,2, . . . , n k 1 Let us generalize this opinion to the infinite dimensional Euclidean space R, and f R . Then we can define following quantities: a) a k (f,k ) , k = 1,2, . . . which is called Fourier coefficients of the vector f R respect to the elements 1 , 2 ,... n ,... b) The series a k k 1 k k (2) which is called Fourier Series of the element f R respect to the It is natural to ask question that the series (2) is convergent?_ if it is convergence, coinside with the vector f.Then we can think about the following theorem. Theorem: Let 1 , 2 ,... n ,... is orthonormal system in the Euclidean space R and f any n element of the space R. Then the quantity f ak k takes its minimum value for k 1 ck = ak = (f k , k ) this means value is f 2 n c 2k , and k 1 Proof: Let us take S n ak k and calculate k 1 f Sn 2 n n (f a k k , f a k k ) k 1 k 1 n n k 1 n k 1 n (f,f) 2(f, a k k ) a k (k , k ) f 2 n n k 1 k 1 2(f, a k k ) ak2 ck2 ck2 k 1 k 1 n c k 1 2 k f (Bessel Inequality). 2 f f 2 n n n n k 1 k 1 k 1 k 1 n n k 1 k 1 c 2k c 2k 2 a k (f, k ) c 2k 2 c 2k (a k c k ) 2 (4) by the orthogonality of the system 1 , 2 ,... n ,... last relation takes its minimum when n ck = ak = (f k , k ) (5). In fact minimum value is min f a k k f k 1 2 n c 2k .Let us prove k 1 Bessel’s Inequality from (4) we get n 0 min f a k k f 2 k 1 n c 2k f 2 k 1 If we pass limit as n then we get n n k 1 k 1 c 2k 0 c 2k f n c k 1 2 k f 2 2 (6).The case of Bessel’s Inequality becomes equality is particulary interesting. n Definition: If inequality (6) becomes equality c k 1 2 k f 2 for any f R , then it is said the orthonormal system 1 , 2 ,... n ,... is closed. The equality (6) is called Parsevalle Theorem. Note: It is another meaning of the word, “closed” is not confused with the word “closed” in preceding section.(Orthonormal system is closed.) Theorem :(Necessary & Sufficient Condition for Closedness): The orthonormal system 1 , 2 ,... n ,... in Euclidean space iff f R is its Fourier series. Proof: ( ) Let 1 , 2 ,... n ,... is closed then by the Parseval Theorem we get, f 2 n c 2k 2k (8) where c k (f,k ) , k . It is known in last theorem that k 1 n n c 2k (9). If we pass limit in (9) and consider (8) get f c 2k 2k =f f c 0f = k 1 n k 1 2 k 2 k 2 k 1 n c k 1 k k . Riesz-Fischer Theorem In previous section we learn that for given orthonormal system 1 , 2 ,... n ,... and any element R the necessary condition the sequence of number c1 , c2 ,...cn ,... to be Fourier n coefficient is the series c k 1 condition is also sufficient. 2 k convergent. It turns out that in the complete Euclidean space this Theorem: (Riesz-Fischer) Let 1 , 2 ,... n ,... is orthonormal space in Euclidean space R and n the sequence of numbers such that c k 1 2 k is convergent. Then there exist an element f R with c1 , c2 ,...cn ,... are its Fourier coefficients, i.e, f 2 n c 2k ; k 1 f c k k . k 1 Proof: Let us take ortonormal system 1 , 2 ,... n ,... and sequence of number c1 , c2 ,...cn ,... n such that c 2k ; in complete Euclidean space. Let denote f n c k k .Then k 1 k 1 f n p f n n p c c k 1 n p n k k k k k = c k n 1 k k n p c k n 1 2 k p 0 f n p f n 0 , by fundamental of fn and Completness of R fn is convergent sum f R f n f .Let us prove c1 , c2 ,...cn ,... are Fourier coefficient of f. (f k ) (f,f n , k ) (f k , k ) (10) limit then we get (f f n ), k f - f n k 0 and (f, k )= ck for n k, f, k c k . Hilbert Space Definition: The set H of elements x, y, z, . . . is called Hilbert space if 1) 2) 3) 4) H is Euclidean space, i.e, H is linear space with scalar product. H is complete with the metrics ρ(x, y) x y H is seperable space, i.e H contains countable dence subset. H is infinite dimensional, i.e for any n, H has n linear independent elements. Theorem: 2 is Hilbert space. Proof: In fact, 2 is infinite dimensional space. Let us prove 2 is seperable space take all the X which has finite numbers nonzero rational coordinates and denote D. Set of this vector is countable because coordinates. 2 is dense because let take any x = ( x1 , x 2 ,...x n ,... ) 2 and sequence xk = ( x1 , x 2 ,...x k ,... ).We know xk D, k . We know that lim x k x .Then D is k seperable. Isomorphic Space Definition: Two Euclidean space R and R* are called isomorphic, if there is one-to-one corresponding between theirs elements x x * , y y * x, y R, x * , y * R * with the preserving linear operation and scalar product as x y x * y * , x x * , x, y x * , y * Note: Any two finite dimensional Euclidean space is isomorphic and any finite dimensional Euclidean space is isomorphic to n-tuple space. But maybe infinite dimensional Euclidean spaces may not isomorphic. For example; let us take 2 and C[a,b] this spaces are not isomorphic due to 2 is complete C[a,b] is not. Theorem: Any two Hilbert spaces are isomorphic. Characterization of the Euclidean Space Theorem: (Necessary and sufficient condition Normed Space) E to be Euclidean is for any two elements f, g E preserves the parallegrom rules, i.e; f g f g 2( f 2 2 2 g ) 2 An example; Let’s take R np normed space with the norm, x = ( x1 , x 2 ,...x n ,... ) R np , x p n x k 1 k 1 p .In preceding sections, we know that R np is normed space when p 1 . For the p = 1, it is easy to check that R np is normed space. 1 1 p p n n n p p p For the p > 1, by using Minkowski Inequality x i y i x i y i i 1 i 1 i 1 n n We can prove R p is normed space. R p is not Euclidean space, in the case p 2 .Let us prove this. Let take f = (1, 0, 0, . . . , 0) R n f 2 1 1 p p g = (1, -1, 0, . . . , 0) R n g 2 1 p f + g = (2, 0, 0, . . . ) f 2 f – g = (0, 2, 0, . . . ) f 2 Let us prove parallelogram rule. f g f g 4 2 f g 2.2.2 2 2 2 p 2 p 2 2 2 p 8 if p 2 n p R is Euclidean space but others p is not. π Example: Let us take C[a,b], i.e, set of all continuous function in 0, . Our aim to check that 2 this space is Euclidean or not. C π is normed space with the norm f(x) max f(x) .Let us 0, 2 take two functions f(x) = sinx, g(x) = cosx. 0 x π 2 f(x) = sinx f(x) max sinx 1 0 x π 2 g(x) = cosx g(x) max cosx 1 0 x π 2 max sin x cos x max 2 sin x 2 4 π max sinx cosx max 2 sin x 1 4 Definition: The topological linear space is meant the set E with following properties. 1) E is linear space. 2) E is topological space. 3) The operation of addition of elements and multiply by scalar in E are continuous with topological in this space. It means that the following conditions hold; a) If, z0 = x0 + y0, then for any neighbourhood. V of the z0, there exist neighbourhood u and w of x0 and y0, which x + y v, where x U and y W b) If y 0 αx 0 , then for any neighbourhood of the y0, there exist neighbourhood u of the x0 and ε 0 , with αx V for α α 0 ε and x U. Theorem-1: Let U is neighbourhood of the zero in linear topological space. Then the set U + x0 = {z, z = x + x0, x U} is the neighbourhood of x0. Example: Let K[a,b] be at a set of infinitely times differentiable functions. Then let us define by N0 set of following space. U r,ε { : K[a, b] , 0 (x) ε, 1 (x) ε,..., r (x) ε; x [a, b]} Definition: (Bounded in linear space) The subset M of the linear space E is called bounded if for any neighbourhood U of the zero there exist constant α 0 with M αU {z : z αx, x U} . Definition: (Total Bounded) The subset M of the linear space E is called total bounded if it contains nonempty, open and bounded set. For example; Any normed space is total bounded. In fact, if we take ε 0 , then set N which elements satisfying x ε is open and bounded. CHAPTER 5 Continuous Linear Functional Definition: Functional F defined on linear topological space E is called a linear if f αx βy αf(x) βf(y) for any constant , and x, y E. The linear functional f(x) is called continuous at the point x 0 E. If ε 0 neighbourhood U of the x0, with f(x) f(x0 ) ε , x U . Theorem: If linear functional defined on linear topological space E is continuous some point x 0 E, then this functional is continuous on E.i.e, all points of E. Proof: Let f(x) is continuous at x 0 E, take any ε 0 then there exist neighbourhood U of x0, with property f(x) f(x0 ) ε , x U .Let take any points y E and consider such set V u x 0 y z : z x x 0 y, x V V is neighbourhood of y.Let take any z V then z x x 0 y ,x U then f(z) f(y) f(x x 0 y f(y) f(x) f(x0 ) f(y) f(y) f(x) f(x0 ) . Corollary: To check of Continuity of the linear functional in linear topological space is enough to check at only one point. i.e;for example at the zero. Theorem: Necessary & sufficient for the linear functional in the topological space be continuous is bounded on some neighbourhood of zero. Proof: ( ) Let f(x) is continuous on E. Then ε 0 , neighbourhood U of the zero which f(x) f(0) ε we know for the linear functional f(0) = 0.If we take M = then last relation means f(x) M neighbourhood U of zero. ( ) Let f(x) is bounded in some neighbourhood U of zero. We will proof f(x) is continuous. F is bounded some neighbourhood of zero, i.e, c f(x) c, x U. Let take ε 0 and built ε ε neighbourhood set v u {z : z x; x u} and consider c c εf(x) ε ε e f(z) f(0) f x 0 f(x) .c ε v neighbourhood zero f(z) ε , z V c c c c Theorem: Necessary & sufficient condition in the topological space to be continuous is f is bounded for any bounded set. This condition is sufficient if E satisfy first axiom countability. Note: Linear functional is called bound if it bounded on any bounded subset. Continuous Linear functional on Normed linear space We know that normed space first axiom countability then by using last theorem linear functional in normed space is continuous iff it is bounded. Then we can radius to following opinion f is bounded functional radius to f bounded on any bounded set means f(x) M , for any x c .By linearity we can say f bounded on unit sphere. Then we can think about f sup f(x) (1). If (1) is finite then f is called norm of the functional. x 1 Theorem: Norm of the linear functional is normed space satisfies following to properties: f(x) f sup , f(x) x f . x x 0 Proof: We know that sup f(x) sup f(x) right side of the equality can be substitute by x 1 x 1 x x f(x) sup f then f sup f(x) sup f sup x 0 x 1 x 0 x x x 0 x In last inequality f sup x 0