x pn x qn 0 as n. Every convergent sequence is Cauchy. Not every Cauchy sequence in a normed space E converges to a vector in


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1 78 CHAPTER 3. BANACH SPACES 3.2 Banach Spaces Cauchy Sequence. A sequence of vectors (x n ) in a normed space is a Cauchy sequence if for everyε > 0 there exists M N such that for all n, m M, x m x n <ε. Theorem A sequence (x n ) is a Cauchy sequence if and only if for any two subsequences (x pn ) and (x qn ) x pn x qn 0 as n. Every convergent sequence is Cauchy. Not every Cauchy sequence in a normed space E converges to a vector in E. Example. Incompleteness. Lemma Let (x n ) be a Cauchy sequence of vectors in a normed space. Then the sequence ( x n ) or real numbers converges. 1. We have x m x n x m x n. Thus ( x n ) is Cauchy. Every Cauchy sequence is bounded. Exercise. Banach Space. A normed space E is complete (or Banach space) if every Cauchy sequence in E converges to an element in E. Examples.R n andc n (with any norm) are complete. Theorem The spaces L p (Ω) with p 1 and l p are complete. Theorem Completeness of l 2. The space of complex sequences with the norm 2 is complete. No proof. mathphyshass1.tex; September 25, 2013; 22:56; p. 76
2 3.2. BANACH SPACES 79 Theorem Completeness of C(Ω). The space of complexvalued continuous functions on a closed bounded setω R n with the norm is complete. No proof. f = sup f (x) x Ω Convergent and Absolutely Convergent Series. A series n=1 x n converges in a normed space E if the sequence of partial sums s n = n k=1 x k converges in E. If n=1 x n <, then the series converges absolutely. An absolutely convergent series does not need to converge. Theorem A normed space is complete if and only if every absolutely convergent series converges. in functional analysis books. Theorem A closed vector subspace of a Banach space is a Banach space. no proof. Completion of Normed Spaces Let (E, ) be a normed space. A normed space (Ẽ, 1 ) is a completion of (E, ) if there exists a linear injectionφ : E Ẽ, such that 1. for every x E 2.Φ(E) is dense in Ẽ, 3. Ẽ is complete. x = Φ(x) 1, The space Ẽ is defined as the set of equivalence classes of Cauchy sequences in E. Recall that if (x n ) is Cauchy then the sequence x n converges. mathphyshass1.tex; September 25, 2013; 22:56; p. 77
3 80 CHAPTER 3. BANACH SPACES We define an equivalence relation of Cauchy sequences as follows. Two Cauchy sequences (x n ) and (y n ) in E are equivalent, (x n ) (y n ), if lim x n y n = 0. The set of Cauchy sequences (y n ) equivalent to a Cauchy sequence (x n ) is the equivalence class of (x n ) [(x n )]={(y n ) E (y n ) (x n )} Obviously we can identify the space E with the set of constant Cauchy sequences. Then the set of all equivalent classes is Ẽ=E/ ={[(x n )] (x n ) is Cauchy sequence in E} The addition and multiplication by scalars in Ẽ are defined by [(x n )]+[(y n )]=[(x n + y n )], λ[(x n )]=[(λx n )] The norm in Ẽ is defined by the limit [(x n )] 1 = lim n x n, which exists for every Cauchy sequence. This definition is consistent since for any two equivalent Cauchy sequences (x n ) and (y n ) [(x n )] 1 = [(y n )] 1 Exercise. The linear injectionφ : E Ẽis defined by a constant sequence ThenΦis onetoone. Exercise. Φ(x)=[(x n )] such that x n = x, n N. mathphyshass1.tex; September 25, 2013; 22:56; p. 78
4 3.2. BANACH SPACES 81 Obviously, x E, x = Φ(x) 1. Claim:Φ(E) is dense in Ẽ. Since every element [(x n )] of Ẽ is the limit of a sequence (Φ(x n )). Claim: Ẽ is complete. Let ( x n ) be a Cauchy sequence in Ẽ. Then (x n ) in E such that Φ(x n ) x n 1 < 1 n. Claim: (x n ) is Cauchy sequence in E. x n x m = Φ(x n ) Φ(x m ) 1 Φ(x n ) x n 1 + x n x m 1 + x m Φ(x m ) 1 x n x m n + 1 m. Next, let x=[(x n )]. Claim: x n x 1 0. x n x 1 x n Φ(x n ) 1 + Φ(x n ) x 1 < 1 n + Φ(x n) x 1 0. Homeomorphism. Two topological spaces E 1 and E 2 are homeomorphic if there exists a bijectionψ : E 1 E 2 from E 1 onto E 2 such that bothψ andψ 1 are continuous. mathphyshass1.tex; September 25, 2013; 22:56; p. 79
5 82 CHAPTER 3. BANACH SPACES Isomorphism of Normed Spaces. Two normed spaces (E 1, 1 ) and (E 2, 2 ) are isomorphic if there exists a linear homeomorphismψ : E 1 E 2 from E 1 onto E 2. Theorem Any two completions of a normed space are isomorphic. Read elsewhere. mathphyshass1.tex; September 25, 2013; 22:56; p. 80
6 3.3. LINEAR MAPPINGS Linear Mappings Let L : E 1 E 2 be a mapping from a vector space E 1 into a vector space E 2. If x E 1, then L(x) is the image of the vector x. If A E 1 is a subset of E 1, then the set is the image of the set A. L(A)={y E 2 y=l(x) for some x A} If B E 2 is a subset of E 2, then the set is the inverse image of the set B. L 1 (B)={x E 1 L(x) B} A mapping L : D(L) E 2 may be defined on a proper subset (called the domain) D(L) E 1 of the vector space E 1. The image of the domain, L(D(L)), of a mapping L is the range of L. That is the range of L is R(L)={y E 2 y=l(x) for some x D(L)}. The null space N(L) (or the kernel Ker(L)) of a mapping L is the set of all vectors in the domain D(L) which are mapped to zero, that is N(L)={x D(L) L(x)=0}. The graphγ(l) of a mapping L is the set of ordered pairs (x, L(x)), that is Γ(L)={(x, y) E 1 E 2 x D(L) and y=l(x)}. Continuous Mappings. A mapping f : E 1 E 2 from a normed space E 1 into a normed space E 2 is continuous at x 0 E 1 if any sequence (x n ) in E 1 converging to x 0 is mapped to a sequence f (x n ) in E 2 that converges to f (x 0 ). mathphyshass1.tex; September 25, 2013; 22:56; p. 81
7 84 CHAPTER 3. BANACH SPACES That is f is continuous at x 0 if x n x 0 0 implies f (x n ) f (x 0 ) 0. A mapping f : E 1 E 2 is continuous if it is continuous at every x E 1. Theorem The norm : E R in a normed space E is a continuous mapping from E intor. If x n x 0, then x n x x n x 0 Theorem Let f : E 1 E 2 be a mapping from a normed space E 1 into a normed space E 2. The following conditions are equivalent: 1. f is continuous. 2. The inverse image of any open set of E 2 is open in E The inverse image of any closed set of E 2 is closed in E 1. Exercise. Linear Mappings. Let S E 1 be a subset of a vector space E 1. A mapping L : S E 2 is linear if x, y S and α,β Fsuch thatαx+βy S, L(αx+βy)=αL(x)+βL(y). Theorem If S is not a vector subspace of E 1, then there is a unique extension of L : S E 2 to a linear mapping L : span S E 2 from the vector subspace span S to E 2. The extension L is defined by linearity. Thus, one can always assume that the domain of a linear mapping is a vector space. Theorem The range, the null space and the graph of a linear mapping are vector spaces. Exercise. mathphyshass1.tex; September 25, 2013; 22:56; p. 82
8 3.3. LINEAR MAPPINGS 85 For any linear mapping L, L(0)=0. Thus, 0 N(L) and the null space N(L) is always nonempty. Theorem A linear mapping L : E 1 E 2 from a normed space E 1 into a normed space E 2 is continuous if and only if it is continuous at a point. 1. Assume L is continuous at x 0 E Let x E 1 and (x n ) x. 3. Then (x n x+ x 0 ) x Thus L(x n ) L(x) = L(x n x+ x 0 ) L(x 0 ) 0 Bounded Linear Mappings. A linear mapping L : E 1 E 2 from a normed space E 1 into a normed space E 2 is bounded if there is a real number K R such that for all x E 1, L(x) K x. Theorem A linear mapping L : E 1 E 2 from a normed space E 1 into a normed space E 2 is continuous if and only if it is bounded. 1. (I). Assume that L is bounded. 2. Claim: L is continuous at Indeed, x n 0 implies 4. Hence, L is continuous. 5. (II). Assume that L is continuous. L(x n ) K x n 0 6. By contradiction, assume that L is unbounded. 7. Then, there is a sequence (x n ) in E 1 such that L(x n ) > n x n mathphyshass1.tex; September 25, 2013; 22:56; p. 83
9 86 CHAPTER 3. BANACH SPACES 8. Let 9. Then y n = x n n x n, n N y n = 1 n, and L(y n) > Then y n 0 but L(y n ) Thus, L is not continuous at zero. Remark. For linear mappings, continuity and uniform continuity are equivalent. The set L(E 1, E 2 ) of all linear mappings from a vector space E 1 into a vector space E 2 is a vector space with the addition and multiplication by scalars defined by (L 1 + L 2 )(x)=l 1 (x)+ L 2 (x), and (αl)(x)=αl(x). The set B(E 1, E 2 ) of all bounded linear mappings from a normed space E 1 into a normed space E 2 is a vector subspace of the space L(E 1, E 2 ). Theorem The space B(E 1, E 2 ) of all bounded linear mappings L : E 1 E 2 from a normed space E 1 into a normed space E 2 is a normed space with norm defined by 1. Obviously, L 0. L(x) L = sup x E 1,x 0 x 2. L =0 if and only if L=0. 3. Claim: L satisfies triangle inequality. 4. Let L 1, L 2 B(E 1, E 2 ). = sup L(x). x E 1, x =1 mathphyshass1.tex; September 25, 2013; 22:56; p. 84
10 3.3. LINEAR MAPPINGS Then L 1 + L 2 = sup L 1 (x)+ L 2 (x) x =1 sup L 1 (x) +sup L 2 (x) x =1 x =1 = L 1 + L 2 For any bounded linear mapping L : E 1 E 2 L(x) L x, x E 1. L is the least real number K such that L(x) K x for all x E 1. The norm defined by L = sup x E1, x =1 norm. L(x) is called the operator Convergence with respect to the operator norm is called the uniform convergence of operators. Strong convergence. A sequence of bounded linear mappings L n B(E 1, E 2 ) converges strongly to L B(E 1, E 2 ) if for every x E 1 we have L n (x) L(x) 0 as n. Theorem Uniform convergence implies strong convergence. Follows from Converse is not true. L n (x) L(x) L n L x Theorem Let E 1 be a normed space and E 2 be a Banach space. Then B(E 1, E 2 ) is a Banach space. see functional analysis books. mathphyshass1.tex; September 25, 2013; 22:56; p. 85
11 88 CHAPTER 3. BANACH SPACES Theorem Let E 1 be a normed space and E 2 be a Banach space. Let S E 1 be a subspace of E 1 and L : S E 2 be a continuous linear mapping from S into E 2. Then: 1. L has a unique extension to a continuous linear mapping L : S E 2 defined on the closure of the domain of the mapping L. 2. If S is dense in E 1, then L has a unique extension to a continuous linear mapping L : E 1 E 2. see functional analysis. Theorem Let E 1 and E 2 be normed spaces, S E 1 be a subspace of E 1 and L : S E 2 be a continuous linear mapping from S into E 2. Then: 1. the null space N(L) is a closed subspace of E If the domain S is a closed subspace of E 1, then the graphγ(l) of L is a closed subspace of E 1 E 2. Exercise. A bounded linear mapping L : E F from a normed space E into the scalar fieldfis called a functional. The space B(E,F) of functionals is called the dual space and denoted by E or E. The dual space is always a Banach space. sincefis complete. mathphyshass1.tex; September 25, 2013; 22:56; p. 86
12 3.4. BANACH FIXED POINT THEOREM Banach Fixed Point Theorem Contraction Mapping. Let E be a normed space and A Ebe a subset of E. A mapping f : A E from A into E is a contraction mapping if there exists a real numberα, such that 0<α<1and x, y A f (x) f (y) α x y. A contraction mapping is continuous. Exercise. If x, y A f (x) f (y) x y, then it is not necessarily a contraction since the contraction constantαmay not exist. Theorem Banach Fixed Point Theorem. Let E be a Banach space and A E be a closed subset of E. Let f : A A be a contraction mapping from A into A. Then there exists a unique z A such that f (z) = z. Examples. Homework Exercises: DM[9,10,11,12,14,21,23,26,31,34,36,37,38,39,42,44] mathphyshass1.tex; September 25, 2013; 22:56; p. 87
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