In functional analysis, the dual norm is a measure of the "size" of each continuous linear functional defined on a normed vector space.

Contents

DefinitionEdit

Let   be a normed vector space with norm   and let   be the dual space. The dual norm of a continuous linear functional   belonging to   is defined to be the real number

 

where   denotes the supremum.[1]

The map   defines a norm on  . (See Theorems 1 and 2 below.)

The dual norm is a special case of the operator norm defined for each (bounded) linear map between normed vector spaces.

The topology on   induced by   turns out to be as strong as the weak-* topology on  .

If the ground field of   is complete then   is a Banach space.

The double dual of a normed linear spaceEdit

The double dual (or second dual)   of   is the dual of the normed vector space  . There is a natural map  . Indeed, for each   in   define

 

The map   is linear, injective, and distance preserving.[2] In particular, if   is complete (i.e. a Banach space), then   is an isometry onto a closed subspace of  .[3]

In general, the map   is not surjective. For example, if   is the Banach space   consisting of bounded functions on the real line with the supremum norm, then the map   is not surjective. (See   space). If   is surjective, then   is said to be a reflexive Banach space. If   then the space   is a reflexive Banach space.

Mathematical OptimizationEdit

Let   be a norm on   The associated dual norm, denoted   is defined as

 

(This can be shown to be a norm.) The dual norm can be interpreted as the operator norm of  , interpreted as a   matrix, with the norm   on  , and the absolute value on  :

 

From the definition of dual norm we have the inequality

 

which holds for all x and z.[4] The dual of the dual norm is the original norm: we have   for all x. (This need not hold in infinite-dimensional vector spaces.)

The dual of the Euclidean norm is the Euclidean norm, since

 

(This follows from the Cauchy–Schwarz inequality; for nonzero z, the value of x that maximises   over   is  .)

The dual of the  -norm is the  -norm:

 

and the dual of the  -norm is the  -norm.

More generally, Hölder's inequality shows that the dual of the  -norm is the  -norm, where, q satisfies  , i.e.,  

As another example, consider the  - or spectral norm on  . The associated dual norm is

 

which turns out to be the sum of the singular values,

 

where   This norm is sometimes called the nuclear norm.[5]

ExamplesEdit

Dual norm for matricesEdit

The Frobenius norm defined by

 

is self-dual, i.e., its dual norm is  

The spectral norm, a special case of the induced norm when  , is defined by the maximum singular values of a matrix, i.e.,

 

has the nuclear norm as its dual norm, which is defined by

 

for any matrix   where   denote the singular values[citation needed].

Some basic results about the operator normEdit

More generally, let   and   be topological vector spaces, and  [6] be the collection of all bounded linear mappings (or operators) of   into  . In the case where   and   are normed vector spaces,   can be normed in a natural way.

Theorem 1. Let   and   be normed spaces, and associate to each   the number:
 
This turns   into a normed space. Moreover if   is a Banach space, so is  .[7]

Proof. A subset of a normed space is bounded if and only if it lies in some multiple of the unit sphere; thus   for every   if   is a scalar, then   so that

 

The triangle inequality in   shows that

 

for every   with  . Thus

 

If  , then   for some  ; hence  . Thus,   is a normed space.[8]

Assume now that   is complete, and that   is a Cauchy sequence in  . Since

 

and it is assumed that   as  ,   is a Cauchy sequence in   for every  . Hence

 

exists. It is clear that   is linear. If  ,   for sufficiently large n and m. It follows

 

for sufficiently large m. Hence  , so that   and  . Thus   in the norm of  . This establishes the completeness of  [9]

When   is a scalar field (i.e.   or  ) so that   is the dual space   of  .

Theorem 2. Suppose   is the closed unit ball of normed space  . For every   define:
 
Then
(a) This norm makes   into a Banach space.[10]
(b) Let   be the closed unit ball of  . For every  ,
 
Consequently,   is a bounded linear functional on   of norm  .
(c)   is weak*-compact.

Proof. Since  , when   is the scalar field, (a) is a corollary of Theorem 1. Fix  . There exists[11]   such that

 

but,

 

for every  . (b) follows from the above. Since the open unit ball   of   is dense in  , the definition of   shows that   if and only if   for every  . The proof for (c)[12] now follows directly.[13]

See alsoEdit

NotesEdit

  1. ^ Rudin 1991, p. 87
  2. ^ Rudin 1991, section 4.5, p. 95
  3. ^ Rudin 1991, p. 95
  4. ^ This inequality is tight, in the following sense: for any x there is a z for which the inequality holds with equality. (Similarly, for any z there is an x that gives equality.)
  5. ^ Boyd & Vandenberghe 2004, p. 637
  6. ^ Each   is a vector space, with the usual definitions of addition and scalar multiplication of functions; this only depends on the vector space structure of  , not  .
  7. ^ Rudin 1991, p. 92
  8. ^ Rudin 1991, p. 93
  9. ^ Rudin 1991, p. 93
  10. ^ Aliprantis 2005, p. 230
  11. ^ Rudin 1991, Theorem 3.3 Corollary, p. 59
  12. ^ Rudin 1991, Theorem 3.15 The Banach–Alaoglu theorem algorithm, p. 68
  13. ^ Rudin 1991, p. 94

ReferencesEdit

  • Aliprantis, Charalambos D.; Border, Kim C. (2007). Infinite Dimensional Analysis: A Hitchhiker's Guide (3rd ed.). Springer. ISBN 9783540326960.
  • Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization. Cambridge University Press. ISBN 9780521833783.
  • Kolmogorov, A.N.; Fomin, S.V. (1957). Elements of the Theory of Functions and Functional Analysis, Volume 1: Metric and Normed Spaces. Rochester: Graylock Press.
  • Rudin, Walter (1991), Functional analysis, McGraw-Hill Science, ISBN 978-0-07-054236-5.

External linksEdit