Dual norm

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


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 space

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 Optimization

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]


Dual norm for matrices

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.

Some basic results about the operator norm

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:
(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


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 also


  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


  • 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.
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