Dual norm

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In functional analysis, the dual norm is a measure of the "size" of continuous linear functionals. The formal proof for the dual norm depends upon the context in which it is introduced.



Let and be topological vector spaces, and [1] 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 , the norm on defines a topology which turns out to be stronger than its weak-*topology.

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 .[2]


  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.[3]
  2. Assume now that is complete, and that is a Cauchy sequence in .
    and it is assumed that as n and m tend to , is a Cauchy sequence in for every .
    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 [4]

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

for every

(a) This norm makes into a Banach space.[5]
(b) Let be the closed unit ball of . For every ,
Consequently, is a bounded linear functional on , of norm .
(c) is weak*-compact.
Since , when is the scalar field, (a) is a corollary of Theorem 1.
Fix . There exists[6] such that
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)[7] now follows directly.[8]

The second dual of a Banach space is an isometric isomorphism[edit]

The normed dual of a Banach space is also a Banach space, which means it has a normed dual, , of its own.

By part (b) of Theorem 2, every defines a unique by equation


It follows from the first and second equation that is linear and is an isometry. Given that is assumed to be complete, is closed in .

Thus, is an isometric isomorphism onto a closed subspace of .[9]

The members of are exactly the linear functionals on that are continuous with respect to its weak*-topology. Since the norm topology of is stronger, may happen that is a proper subspace of .

However, there are many important spaces, such as the Lp spaces with , where ; these are called reflexive.

It is stressed that, for to be reflexive, the existence of some isometric isomorphism of onto is not enough; it is crucial that satisfies first equation in this section.[10]

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.[11] 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.[12]


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 dual norm defined by for any matrix where denote the singular values.

See also[edit]


  1. ^ 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 .
  2. ^ Rudin 1991, p. 92
  3. ^ Rudin 1991, p. 93
  4. ^ Rudin 1991, p. 93
  5. ^ 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.
  6. ^ Rudin 1991, Theorem 3.3 Corollary, p. 59
  7. ^ Rudin 1991, Theorem 3.15 The Banach-Alaoglu theorem algorithm, p. 68
  8. ^ Rudin 1991, p. 94
  9. ^ Rudin 1991, Theorem 4.5 The second dual of a Banach space, p. 95
  10. ^ Rudin 1991, p. 95
  11. ^ 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.)
  12. ^ Boyd & Vandenberghe 2004, p. 637


External links[edit]