The double 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. In particular, if is complete, then is an isometry onto a closed subspace of. 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.. 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 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. The dual of the dual norm is the original norm: we have for all. The dual of the Euclidean norm is the Euclidean norm, since 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, 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.
Examples
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 be the collection of all bounded linear mappings of into. In the case where and are normed vector spaces, can be normed in a natural way. 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. 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 and. It follows for sufficiently large. Hence, so that and. Thus in the norm of. This establishes the completeness of When is a scalar field so that is the dual space of. Proof. Since, when is the scalar field, is a corollary of Theorem 1. Fix. There exists such that but, for every. 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 now follows directly.
