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


Definition


Let X be a normed vector space with norm \| \cdot \| and let X^* be the dual space. The dual norm of a continuous linear functional f belonging to X^* is the non-negative real number defined by any of the following equivalent formulas: \begin \| f \| &= \sup &&\ \\ &= \sup &&\ \\ &= \inf &&\ \\ &= \sup &&\ \\ &= \sup &&\ \;\;\;\text X \neq \ \\ &= \sup &&\bigg\ \;\;\;\text X \neq \ \\ \end where \sup and \inf denote the supremum and infimum, respectively. The constant map always has norm equal to and it is the origin of the vector space X^*. If X = \ then the only linear functional on X 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 . The map f \mapsto \| f \| defines a norm on X^*. (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 X^* induced by \| \cdot \| turns out to be as strong as the weak-* topology on X^*. If the ground field of X is complete then X^* is a Banach space.


The double dual of a normed linear space


The double dual (or second dual) X^ of X is the dual of the normed vector space X^*. There is a natural map \varphi: X \to X^. Indeed, for each w^* in X^* define : \varphi(v)(w^*): = w^*(v). The map \varphi is linear, injective, and distance preserving. In particular, if X is complete (i.e. a Banach space), then \varphi is an isometry onto a closed subspace of X^. In general, the map \varphi is not surjective. For example, if X is the Banach space L^ consisting of bounded functions on the real line with the supremum norm, then the map \varphi is not surjective. (See L^p space). If \varphi is surjective, then X is said to be a reflexive Banach space. If 1 < p < \infty, then the space L^p is a reflexive Banach space.


Mathematical optimization


Let \|\cdot\| be a norm on \R^. The associated ''dual norm'', denoted \| \cdot \|_*, is defined as :\|z\|_* = \sup\. (This can be shown to be a norm.) The dual norm can be interpreted as the operator norm of z^\intercal, interpreted as a 1 \times n matrix, with the norm \|\cdot\| on \R^n, and the absolute value on \R: :\|z\|_* = \sup\. From the definition of dual norm we have the inequality :z^\intercal x = \| x \| \left (z^\intercal \frac \right ) \leq \| x \| \| z \| _* which holds for all and . The dual of the dual norm is the original norm: we have \| x \|_ = \| x \| for all . (This need not hold in infinite-dimensional vector spaces.) The dual of the Euclidean norm is the Euclidean norm, since :\sup\ = \| z \| _2. (This follows from the Cauchy–Schwarz inequality; for nonzero , the value of that maximises z^\intercal x over \| x \| _2 \leq 1 is \tfrac.) The dual of the \ell_\infty -norm is the \ell_1-norm: :\sup\ = \sum_^n |z_i| = \| z \| _1, and the dual of the \ell_1-norm is the \ell_\infty-norm. More generally, Hölder's inequality shows that the dual of the \ell_p-norm is the \ell_q-norm, where, satisfies \tfrac + \tfrac = 1, i.e., q = \tfrac. As another example, consider the \ell_2- or spectral norm on \R^. The associated dual norm is :\| Z \| _ = \sup\, which turns out to be the sum of the singular values, :\| Z \| _ = \sigma_1(Z) + \cdots + \sigma_r(Z) = \mathbf (\sqrt), where r = \mathbf Z. This norm is sometimes called the ''nuclear'' norm.


Examples




Dual norm for matrices

The ''Frobenius norm'' defined by :\| A\|_ = \sqrt = \sqrt = \sqrt is self-dual, i.e., its dual norm is \| \cdot \|'_ = \| \cdot \|_. The ''spectral norm'', a special case of the ''induced norm'' when p=2, is defined by the maximum singular values of a matrix, i.e., :\| A \| _2 = \sigma_(A), has the nuclear norm as its dual norm, which is defined by :\|B\|'_2 = \sum_i \sigma_i(B), for any matrix B where \sigma_i(B) denote the singular values.


Some basic results about the operator norm


More generally, let X and Y be topological vector spaces and let L(X,Y) be the collection of all bounded linear mappings (or ''operators'') of X into Y. In the case where X and Y are normed vector spaces, L(X,Y) can be given a canonical norm. A subset of a normed space is bounded if and only if it lies in some multiple of the unit sphere; thus \| f \| < \infty for every f \in L(X,Y) if \alpha is a scalar, then (\alpha f)(x) = \alpha \cdot f x so that :\| \alpha f \| = | \alpha | \| f \| The triangle inequality in Y shows that :\begin \| (f_1 + f_2)x \| ~&=~ \| f_1 x + f_2 x \| \\ &\leq~ \| f_1 x \| + \| f_2 x \| \\ &\leq~ (\|f_1\| + \|f_2\|)\|x\| \\ &\leq~ \|f_1\| + \|f_2\| \end for every x \in X satisfying \| x \| \leq 1. This fact together with the definition of \| \cdot \| ~:~ L(X, Y) \to \mathbb implies the triangle inequality: :\| f_1 + f_2 \| \leq \|f_1\| + \|f_2\| Since \ is a non-empty set of non-negative real numbers, \| f \| = \sup \left\ is a non-negative real number. If f \neq 0 then f x_0 \neq 0 for some x_0 \in X, which implies that \| f x_0 \| > 0 and consequently \| f \| > 0. This shows that \left( L(X, Y), \| \cdot \|\right) is a normed space. Assume now that Y is complete and we will show that \left( L(X, Y), \| \cdot \|\right) is complete. Let f_ = \left( f_ \right)_^ be a Cauchy sequence in L(X, Y), so by definition \|f_n - f_m\| \to 0 as n, m \to \infty. This fact together with the relation :\| f_n x - f_m x \| = \| \left( f_n - f_m \right) x \| \leq \| f_n - f_m \| \| x \| implies that \left( f_ x \right)_^ is a Cauchy sequence in Y for every x \in X. It follows that for every x \in X, the limit \lim_ f_n x exists in Y and so we will denote this (necessarily unique) limit by f x, that is: :f x ~=~ \lim_ f_n x. It can be shown that f: X \to Y is linear. If \varepsilon > 0, then \| f_n - f_m \| \| x \| ~\leq~ \varepsilon \| x \| for all sufficiently large integers and . It follows that :\| fx - f_m x \| ~\leq~ \varepsilon \| x \| for sufficiently all large . Hence \| fx \| \leq \left( \| f_m \| + \varepsilon \right) \| x \|, so that f \in L(X, Y) and \| f - f_m \| \leq \varepsilon. This shows that f_m \to f in the norm topology of L(X, Y). This establishes the completeness of L(X, Y). When Y is a scalar field (i.e. Y = \Complex or Y = \R) so that L(X,Y) is the dual space X^* of X. Let B ~=~ \sup\denote the closed unit ball of a normed space X. When Y is the scalar field then L(X,Y) = X^* so part (a) is a corollary of Theorem 1. Fix x \in X. There exists y^* \in B^* such that :\langle\rangle = \|x\|. but, :|\langle\rangle| \leq \|x\|\|x^*\| \leq \|x\| for every x^* \in B^*. (b) follows from the above. Since the open unit ball U of X is dense in B, the definition of \|x^*\| shows that x^* \in B^* if and only if |\langle\rangle| \leq 1 for every x \in U. The proof for (c) now follows directly.


See also


* convex conjugate * operator norm * Lp spaces * Hölder's inequality


Notes





References


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External links


Notes on the proximal mapping by Lieven Vandenberge
{{DualityInLCTVSs Category:Linear algebra Category:Functional analysis Category:Mathematical optimization