In functional analysis, the dual norm is a measure of size for a continuous linear function defined on a normed vector space.
Let be a normed vector space with norm and let denote its continuous dual space. The dual norm of a continuous linear functional belonging to is the non-negative real number defined [1] by any of the following equivalent formulas:
where and denote the supremum and infimum, respectively. The constant map is the origin of the vector space and it always has norm If then the only linear functional on is the constant map and moreover, the sets in the last two rows will both be empty and consequently, their supremums will equal instead of the correct value of
Importantly, a linear function is not, in general, guaranteed to achieve its norm on the closed unit ball meaning that there might not exist any vector of norm such that (if such a vector does exist and if then would necessarily have unit norm ). R.C. James proved James's theorem in 1964, which states that a Banach space is reflexive if and only if every bounded linear function achieves its norm on the closed unit ball. [2] It follows, in particular, that every non-reflexive Banach space has some bounded linear functional that does not achieve its norm on the closed unit ball. However, the Bishop–Phelps theorem guarantees that the set of bounded linear functionals that achieve their norm on the unit sphere of a Banach space is a norm-dense subset of the continuous dual space. [3] [4]
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. Since the ground field of ( or ) is complete, is a Banach space. The topology on induced by turns out to be stronger than the weak-* topology on
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. [5] In particular, if is complete (i.e. a Banach space), then is an isometry onto a closed subspace of . [6]
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.
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, that is,
has the nuclear norm as its dual norm, which is defined by
for any matrix where denote the singular values[ citation needed ].
If the Schatten -norm on matrices is dual to the Schatten -norm.
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 and [7] The dual of the dual norm is the original norm: we have for all (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 the value of 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 satisfies that is,
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. [8]
For p-norm (also called -norm) of vector is
If satisfy then the and norms are dual to each other and the same is true of the and norms, where is some measure space. In particular the Euclidean norm is self-dual since For , the dual norm is with positive definite.
For the -norm is even induced by a canonical inner product meaning that for all vectors This inner product can expressed in terms of the norm by using the polarization identity. On this is the Euclidean inner product defined by
while for the space associated with a measure space which consists of all square-integrable functions, this inner product is
The norms of the continuous dual spaces of and satisfy the polarization identity, and so these dual norms can be used to define inner products. With this inner product, this dual space is also a Hilbert spaces.
Given normed vector spaces and let [9] be the collection of all bounded linear mappings (or operators) of into Then can be given a canonical norm.
defines a norm on that makes into a normed space. Moreover, if is a Banach space then so is [10]
Proof |
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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 satisfying This fact together with the definition of implies the triangle inequality: Since is a non-empty set of non-negative real numbers, is a non-negative real number. If then for some which implies that and consequently This shows that is a normed space. [11] Assume now that is complete and we will show that is complete. Let be a Cauchy sequence in so by definition as This fact together with the relation implies that is a Cauchy sequence in for every It follows that for every the limit exists in and so we will denote this (necessarily unique) limit by that is: It can be shown that is linear. If , then for all sufficiently large integers n and m. It follows that for sufficiently all large Hence so that and This shows that in the norm topology of This establishes the completeness of [12] |
When is a scalar field (i.e. or ) so that is the dual space of
Theorem 2 — Let be a normed space and for every let
where by definition is a scalar. Then
Proof |
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Let denote the closed unit ball of a normed space When is the scalar field then so part (a) is a corollary of Theorem 1. Fix There exists [14] 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) [15] now follows directly. [16] |
As usual, let denote the canonical metric induced by the norm on and denote the distance from a point to the subset by
If is a bounded linear functional on a normed space then for every vector [17]
where denotes the kernel of
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This is a glossary for the terminology in a mathematical field of functional analysis.