Dual norm

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

Definition[edit]

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 is complete (i.e. a Banach space) then is also complete.[2]

The double dual of a normed linear space[edit]

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.[3] In particular, if is complete (i.e. a Banach space), then is an isometry onto a closed subspace of .[4]

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 Optimization[edit]

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.[5] 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.[6]

Examples[edit]

Dual norm for matrices[edit]

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 norm[edit]

More generally, let and be topological vector spaces, and [7] 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.

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

Theorem 1: Let and be normed spaces, and associate to each the number:

We first establish that is bounded (using the triangle inequality), and complete (using Cauchy sequences) using our definition of , thereby making a normed space. If is a Banach space, so is .[8]

Proof:

  1. 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.[9]
  2. Assume now that is complete, and that is a Cauchy sequence in .
    Since
    and it is assumed that as n and m tend to , 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 [10]

Theorem 2: Now suppose is the closed unit ball of normed space . Define

for every

(a) This norm makes into a Banach space.[11]
(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[12] 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)[13] now follows directly.[14]

See also[edit]

Notes[edit]

  1. ^ Rudin 1991, p. 87
  2. ^ Rudin 1991, p. 88
  3. ^ Rudin 1991, section 4.5, p. 95
  4. ^ Rudin 1991, p. 95
  5. ^ 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.)
  6. ^ Boyd & Vandenberghe 2004, p. 637
  7. ^ 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 .
  8. ^ Rudin 1991, p. 92
  9. ^ Rudin 1991, p. 93
  10. ^ Rudin 1991, p. 93
  11. ^ Aliprantis 2005, p. 230
    6.7 Definition The norm dual of a normed space is Banach space . The operator norm on is also called the dual norm, also denoted . That is,

    The dual space is indeed a Banach space by Theorem 6.6.
  12. ^ Rudin 1991, Theorem 3.3 Corollary, p. 59
  13. ^ Rudin 1991, Theorem 3.15 The Banach–Alaoglu theorem algorithm, p. 68
  14. ^ Rudin 1991, p. 94

References[edit]

  • 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 links[edit]