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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 : $\varphi\left(v\right)\left(w^*\right): = w^*\left(v\right).$ 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 :$z^\intercal x = \| x \| \left \left(z^\intercal \frac \right \right) \leq \| x \| \| z \| _*$ 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, :$\| Z \| _ = \sigma_1\left(Z\right) + \cdots + \sigma_r\left(Z\right) = \mathbf \left(\sqrt\right),$ 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 $\| \cdot \|\text{'}_ = \| \cdot \|_.$ 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., :$\| A \| _2 = \sigma_\left(A\right),$ has the nuclear norm as its dual norm, which is defined by :$\|B\|\text{'}_2 = \sum_i \sigma_i\left(B\right),$ for any matrix $B$ where $\sigma_i\left(B\right)$ denote the singular values.

Some basic results about the operator norm

More generally, let $X$ and $Y$ be topological vector spaces and let $L\left(X,Y\right)$ 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, $L\left(X,Y\right)$ 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 $f \in L\left(X,Y\right)$ if $\alpha$ is a scalar, then $\left(\alpha f\right)\left(x\right) = \alpha \cdot f x$ so that :$\| \alpha f \| = | \alpha | \| f \|$ The triangle inequality in $Y$ shows that :$\begin \| \left(f_1 + f_2\right)x \| ~&=~ \| f_1 x + f_2 x \| \\ &\leq~ \| f_1 x \| + \| f_2 x \| \\ &\leq~ \left(\|f_1\| + \|f_2\|\right)\|x\| \\ &\leq~ \|f_1\| + \|f_2\| \end$ for every $x \in X$ satisfying $\| x \| \leq 1.$ This fact together with the definition of $\| \cdot \| ~:~ L\left(X, Y\right) \to \mathbb$ 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 $\left\left( L\left(X, Y\right), \| \cdot \|\right\right)$ is a normed space. Assume now that $Y$ is complete and we will show that $\left\left( L\left(X, Y\right), \| \cdot \|\right\right)$ is complete. Let $f_ = \left\left( f_ \right\right)_^$ be a Cauchy sequence in $L\left(X, Y\right),$ so by definition $\|f_n - f_m\| \to 0$ as $n, m \to \infty.$ This fact together with the relation :$\| f_n x - f_m x \| = \| \left\left( f_n - f_m \right\right) x \| \leq \| f_n - f_m \| \| x \|$ implies that $\left\left( f_ x \right\right)_^$ 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 $\| fx \| \leq \left\left( \| f_m \| + \varepsilon \right\right) \| x \|,$ so that $f \in L\left(X, Y\right)$ and $\| f - f_m \| \leq \varepsilon.$ This shows that $f_m \to f$ in the norm topology of $L\left(X, Y\right).$ This establishes the completeness of $L\left(X, Y\right).$ When $Y$ is a scalar field (i.e. $Y = \Complex$ or $Y = \R$) so that $L\left(X,Y\right)$ 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 $L\left(X,Y\right) = X^*$ 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.

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

Notes

References

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