Dual Spaces. René van Hassel

Transcription

1 Dual Spaces René van Hassel October 1, 2006

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3 1 Spaces A little scheme of the relation between spaces in the Functional Analysis. FA spaces Vector space Topological Space Topological Metric Space Vector Space rrvhassel Normed Vector Space Complete Metric Space Inner Product Space Banach Space Hilbert Space rrvhassel 3

4 2 Dual spaces 2.1 Vector space X, dim X = n Let X be a finite dimensional vector space, dim X = n. There exists a basis {e 1,..., e n } of X. Every x X can be written in the form x = α 1 e α n e n. 2.2 Algebraïc dual space of X, denoted by X Take all the linear functionals f : X R and it is made a vectorspace by defining the following addition and scalar multiplication, f 1, f 2 are linear functionals on X and β is a scalar, (f 1 + f 2 )(x) = f 1 (x) + f 2 (x) (βf 1 )(x) = βf 1 (x) 2.3 Spaces X and X are essential identical First of all, you need a bijective mapping T between the spaces X and X. If T : X X, then T has to be onto and one to one, then T 1 exists. But in some cases, T has als to satisfy other conditions. T is called a isomorphism if it also preserves the structure on the space X, there are several possibilities MS: An isomorphism T between the metric space (X, d) and ( X, d). Besides a bijective mapping, also preserving the distance, d(t (x), T (y)) = d(x, y) for all x, y X. VS: An isomorphism T between vector spaces X and X. Besides a bijective mapping, also preserving the linearity, T (x + y) = T (x) + T (y) T (βx) = βt (x) for all x, y X, T : X X is a bijective linear operator. NVS: An isomorphism T between normed spaces X and X. Besides an isomorphism between vector spaces, also preserving the norm T (x) = x for all x X, also called an isometric ismorphism. ( ( MS: metric space, VS: vector space, NVS: normed vector space.) 4

5 2.4 Dual basis {f 1, f 2,..., f n } of {e 1,..., e n } Going back to the space X, dim X = n, with its base {e 1,..., e n } and the linear functionals f on X. For every linear functional f and for every x = n i=1 α ie i, you know that f(x) = f( α i e i ) = i=1 α i f(e i ) = i=1 α i γ i, i=1 with γ i = f(e i ), j = 1,..., n. The linear functional is uniquely determined by the values γ i at the basis vectors of X. Given n values of scalars γ 1,..., γ n, and a linear functional is determined on X. Take now the following n-tuples (1, 0,..., 0) (0, 1, 0,..., 0)... (0,...0, 1, 0,.., 0)... (0,..., 0, 1) By this are given n functionals, the functionals f 1,..., f n, with { 1 if j = k f k (e j ) = δ jk = 0 if j k. The set {f 1, f 2,..., f n } is called the dual basis of the basis {e 1, e 2,..., e n } for X. The functionals {f 1, f 2,..., f n } are linear independent, solving the following equation β k f k = 0, gives you β k = 0 for k = 1,..., n. Let the functional n β kf k work on e j and you get β j = 0, because f j (e j ) = 1 and f j (e k ) = 0 if j k. Every functional f X can be written as a linear combination of {f 1, f 2,..., f n }. Write the functional f = γ 1 f 1 + γ 2 f γ n f n and realize that when x = α 1 e 1 + α 2 e α n e n that f j (x) = f j (α 1 e 1 + α 2 e α n e n ) = α j, so f(x) = f(α 1 e 1 + α 2 e α n e n ) = α 1 γ α n γ n. It is interesting to note that dim X = dim X = n. 5

6 2.5 The dual space X of a normed space X X is a normed space, the norm of a linear functional f is defined by f = 8 < : sup x X x 0 9 = ; f(x) x = sup 8 < : x X x = 1 9 = ; f(x). The space of all bounded functionals, on the normed space X, is denoted by X. Interesting to note is that the dual space X, of a normed space X, is a Banach space. You can prove it on almost the same way, as you proved that C[0, 1], with the. -norm, is a Banach space. 2.6 Dual space of l 1 is l With the dual space of l 1 is meant (l 1 ), the space of bounded linear functionals of l 1. On both spaces you have a norm, and in this case there seems to be an isomorphism between two normed vector spaces. Let s realize that you have to do with two infinitely dimensional spaces. For l 1 there is a basis (e k ) and e k = δ kj, so every x l 1 can be written as x = α k e k. The norm of x l 1 is and the norm of x l is x 1 = α k (< ) x = sup α k (< ). Take a bounded linear functional f of l 1, (f : l 1 R) this functional can be written in the form f(x) = f( α k e k ) = α k γ k, remember f(e k ) = γ k. Take a look at the row (γ k ) and realize e k 1 = 1. Then γ k = f(e k ) f e k 1 6

7 for all k N, so you have this means (γ k ) l. sup k γ k f, is almost the same as speak- Speaking about the the functional f (l 1 ) ing over the row (γ k ) l. But now the other way around, can you construct linear functionals in (l 1 ) with the elements of l? If the element (γ k ) l is given, it is not difficult to construct the following linear functional on l 1 h(x) = α k γ k Linearity is no problem, the boundedness of the linear functional g g(x) α k γ k sup γ k α k sup γ k x 1. The functional g is linear and bounded on l 1, so g (l 1 ) 1. Looking at isomorphism between normed vector spaces, you have also to look if the norm is preserved. In this case, it is almost done, because f(x) = α k γ k sup γ k α k sup γ k x 1. Take now the supremum over all the x l 1 with x 1 = 1 and you find that f sup γ k, above you had the result sup γ k f, together you see that the norm is preserved, f = sup f(x) = sup γ k = (γ k ) x 1 =1 The isorphism between the two given normed spaces (l 1 ) and l is a fact. So taking a element out of (l 1 ) is in certain sense the same as speaking about an element out of l. Sorry for you, but (l ) and l 1 are not isomorph! But (c 0 ) and l 1 are isomorph. Here you have that just the linear functionals of (c 0 ), (c 0 ) is a particular subspace of l, are isomorph withe space l 1. 7

8 2.7 Dual space of c 0 is l 1 In certain sense, it is a little bit trikcy to prove that. Here also finite and infinite play an important rule. Just as above you take an x c 0 x = λ k e k, with (λ k ) c 0. Because of the fact that lim λ k = 0, k you can take the following approximation of x s n = λ k e k, because of the. -norm, you have lim s n x = 0. n Take a bounded functional f of c 0, this means that f is continuous, so f(s n ) f(x) as n. You already know that f(x) = λ k γ k, Take a special sequence x n 0 = (λ 0 k ) c 0, with λ 0 k = γ k γ k if γ k 0 and 1 k n and 0 otherwise. It is clear that x n 0 = 1 and f(x n 0) = f(e k ) f x n 0 f, but that is for every n N, so n f(e k) <, so (γ k ) l 1. Out of the last inequalities, you can derive that f(e k ) f. 8

9 That the norm is preserved is not so difficult f(x) λ k γ k x λ k x f(e k ) So from (c 0, x ) into (l 1, x 1 ) the norm is preserved. Is the mapping onto? Take any (α k ) l 1 and x = (λ k ) c 0, the series α lλ k is absolutely convergent. Define the linear functional f(x) = λ kα k. The defined linear functional f satifies f(x) α k when x 1, so the constructed functional is bounded ( and continuous). 2.8 Second algebraïc dual space of X, denoted by X Let X be a finite dimensional space again, so dim X = n. An element g X, which is a linear functional on X, can be obtained by g(f) = g x (f) = f(x), so x X is fixed and f X variable. You can see that g x (αf 1 + βf 2 ) = (αf 1 + βf 2 )(x) = αf 1 (x) + βf 2 (x) = αg x (f 1 ) + βg x (f 2 ) for all α, β R and f 1, f 2 X. Hence g x is an element of X. To each x X there corresponds a g x X. This defines the canonical mapping C of X into X, The mapping C is linear, because C : X X x g x (C(αx + βy))(f) = g (αx+βy) (f) = f(αx + βy) = αf(x) + βf(y) = αg x (f) + βg y (f) = α(c(x))(f) + β(c(y))(f) To prove that the linear operator C is injective (one to one) it is important to note that if x 0 X as the property that f(x 0 ) = 0 for all f X that x 0 = 0, if x 0 = ξ 0k e k then f(x 0 ) = ξ 0k γ k = 0, 9

10 for all choices of γ k, this means that ξ 0k = 0 for all k. If C(x) = C(y) then f(x) = f(y) for all f X. You have to do with a linear functional, so f(x y) = 0 for all f X and that gives x = y and you proved that C is injective. Result so far, that C is a (vector space) isomorphism of X onto its range R(C) X. The range R(C) is a linear vectorspace, because C is a linear operator on X, so R(C) is a linear subspace of X. Also is said that X is embeddable in X. The question becomes is C surjective, is C onto? (R(C) = X?) The domain of C is finite dimensional and the inverse mapping of C exists, so the dimension of the range of C and the dimension of the domain of C have to be equal, this gives that dim R(C) = dim X. Further you know that dim (X ) = dim X (= dim X) and you can conclude that dim R(C) = dim X, the mapping C is onto the space X. C is vector isomorphism, it preserves the linearity but about preservation of other structures is not spoken. We have only looked to the algebraic operations, but not to the preservation of a norm for instance. The result is that X and X look algebraic identical. So speaking about X or X, it doesn t matter, but be careful, dim X = n <. They mention X algebraic reflexive, if R(C) = X. ( If X and X are identical, then we have also to look to the preservation of the norm, if X is a normed space.) Important to note is that the canonical mapping C defined at the beginning of this section, is also called a natural embedding of X into X. There are examples of Banach spaces (X,, ), which are isometric isomorph with (X,, ), but not reflexive. For reflexivity, you need the natural embedding. 10