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Definition of the dual space of a normed space containing all linear operators.

Definition

The set of linear functionals f on X is itself a vector space:

  • ( f + g ) ( x ) = f ( x ) + g ( x ) (closed under addition),
  • ( a f ) ( x ) = a f ( x ) (closed under scalar multiplication),
  • z ( x ) = 0 is linear and f + z = f (features an addition zero).

Definition 1 The dual space (or algebraic dual) X * of X is the vector space of all linear function on X .

Example 1 If X = R N then f ( x ) = i = 1 N a i x i X * and any linear functional on X can be written in this form.

Definition 2 Let X be a normed space. The normed dual X * of X is the normed space of all bounded linear functionals with norm f = sup x 1 | f ( x ) | .

Since in the sequel we refer almost exclusively to the normed dual, we will abbreviate to “dual” (and ignore the algebraic dual in the process).

Example 2 Let us find the dual of X = R N . In this space,

x = x 1 x 2 x n and x = i = 1 n | x i | 2 , where x i R .

From Example [link] , a linear functional f can be written as:

f ( x ) = i = 1 n a i x i = x , a , a R n

There is a one-to-one mapping between the space X and the dual of X :

f ( R n ) * a R n

For this reason, R n is called self-dual (as there is a one-to-one mapping ( R n ) * R n ). Using the Cauchy-Schwartz Inequality, we can show that

| f ( x ) | = | x , a | x a ,

with equality if a is a scalar multiple of x . Therefore, we have that f = a , i.e., the norms of X and X * match through the mapping.

Theorem 1 All normed dual spaces are Banach.

Since we have shown that all dual spaces X * have a valid norm, it remains to be shown that all Cauchy sequences in in X * are convergent.

Let { f n } n = 1 X * be a Cauchy sequence in X * . Thus, we have that for each ϵ > 0 there exists n 0 Z + such that if n , m > n 0 then f n ( x ) - f m ( x ) 0 as n , m . We will first show that for an arbitrary input x X , the sequence { f n ( x ) } n = 1 R is Cauchy. Pick an arbitrary ϵ > 0 , and obtain the n 0 Z + from the definition of Cauchy sequence for { f n * } n = 1 . Then for all n , m > n 0 we have that | f n ( x ) - f m ( x ) | = | ( f n - f m ) ( x ) | f n ( x ) - f m ( x ) · x ϵ x .

Now, since R = R and R = C are complete, we have that the sequence { f n ( x ) } n = 1 converges to some scalar f * ( x ) . In this way, we can define a new function f * that collects the limits of all such sequences over all inputs x X . We conjecture that the sequence f n f * : we begin by picking some ϵ > 0 . Since { f n ( x ) } n = 1 is convergent to f * ( x ) for each x X , we have that for n > n 0 , x , | f n ( x ) - f n * ( x ) | ϵ . Now, let n 0 * = sup x X , x = 1 n 0 , x . Then, for n > n 0 * ,

f n - f * = sup x X , x = 1 | f n ( x ) - f * ( x ) | < ϵ .

Therefore, we have found an n 0 * for each ϵ that fits the definition of convergence, and so f n f * .

Next, we will show that f * is linear and bounded. To show that f * is linear, we check that

f * ( a x + b y ) = lim n f n ( a x + b y ) = a lim n f n ( x ) + b lim n f n ( x ) = a f * ( x ) + b f * ( y ) .

To show that f * is bounded, we have that for each ϵ > 0 there exists n 0 * Z + (shown above) such that if n > n 0 * then

| f * ( x ) | | f n ( x ) - f * ( x ) | + | f n ( x ) | ϵ | | x | | + | | f n | | · | | x | | ( ϵ + | | f n | | ) | | x | | .

This shows that f * is linear and bounded and so f * X * . Therefore, { f n * } n = 1 converges in X * and, since the Cauchy sequence was arbitrary, we have that X * is complete and therefore Banach.

The dual of p

Recall the space p = { ( x 1 , x 2 , ... ) : i = 1 | x i | p < } , with the p -norm | | x | | p = ( i = 1 | x i | p ) 1 / p . The dual of p is ( p ) * = q , with q = p p - 1 . That is, every linear bounded functional f ( p ) * can be represented in terms of f ( x ) = n = 1 a n x n , where a = ( a 1 , a 2 , ... ) q . Note that if p = 2 then q = 2 , and so 2 is self-dual.

Bases for self-dual spaces

Consider the example self-dual space R n and pick a basis { e 1 , e 2 , . . . , e n } for it. Recall that for each a R n there exists a bounded linear functional f a ( R n ) * given by f a ( x ) = x , a . Thus, one can build a basis for the dual space ( R n ) * as { f e 1 , f e 2 , . . . , f e n } .

Riesz representation theorem

Since it is easier to conceive a dual space by linking it to elements of a known space (as seen above for self duals), we may ask if there are large classes of spaces who are self-dual.

Theorem 2 (Riesz Representation Theorem) If f H * is a bounded linear functional on a Hilbert space H , there exists a unique vector y H such that for all x H we have f ( x ) = x , y . Furthermore, we have | | f | | = | | y | | , and every y H defines a unique bounded linear functional f y ( x ) = x , y .

Thus, since there is a one-to-one mapping between the dual of the Hilbert space and the Hilbert space ( H * H ), we say that all Hilbert spaces are self-dual. Pick a linear bounded functional f H * and let N H be the set of all vectors x H for which f ( x ) = 0 . Note that N is a closed subspace of H , for if { x n } N is sequence that converges to x H then, due to the continuity of f , f ( x n ) f ( x ) = 0 and so x N . We consider two possibilities for the subspace N :

  • If N = H then y = 0 and f ( x ) = x , 0 = 0 .
  • If N H then we can partition H = N + N and so there must exist some z 0 N , z 0 0 . Then, define z = z o | f ( z o ) | to get z N such that f ( z ) = 1 . Now, pick an arbitrary x H and see that
    f ( x - f ( x ) z ) = f ( x ) - f ( x ) f ( z ) = f ( x ) - f ( x ) = 0 .
    Therefore, we have that x - f ( x ) z N . Since z N , we have that
    x - f ( x ) z , z = 0 , x , z - f ( x ) z , z = 0 ,
    and so
    f ( x ) = x , z z , z = x , z | | z | | 2 = x , z | | z | | 2 .
    Thus, we have found a vector y = z z 2 such that f ( x ) = x , y for all x H . Now, to compute the norm of the functional, consider
    | f ( x ) | = | x , y | | | x | | · | | y | | ,
    with equality if x = α y for α R . Thus, we have that f = y .

We now show uniqueness: let us assume that there exists a second y ^ y , y ^ H , for which f ( x ) = x , y ^ for all x H . Then we have that for all x H ,

x , y = x , y ^ , x , y - x , y ^ = 0 , x , y - y ^ = 0 , y - y ^ = 0 , y = y ^ .

This contradicts the original assumption that y y ^ . Therefore we have that y is unique, completing the proof.

Linear functional extensions

When we consider spaces nested inside one another, we can define pairs of functions that match each other at the overlap.

Definition 3 Let f be a linear functional on a subspace M X of a vector space X . A linear functional F is said to be an extension of f to N (where N is another subspace of X that satisfies M N X ) if F is defined on N and F is identical to f on M .

Here is another reason why we are so interested in bounded linear functionals.

Theorem 3 (Hahn-Banach Theorem) A bounded linear functional f on a subspace M X can be extended to a bounded linear functional F on the entire space X with

F = sup x X | F ( x ) | x = f = sup x M | f ( x ) | x = sup x M | F ( x ) | x .

The proof is in page 111 of Luenberger.

Questions & Answers

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Source:  OpenStax, Signal theory. OpenStax CNX. Oct 18, 2013 Download for free at http://legacy.cnx.org/content/col11542/1.3
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