Dual norm

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 $X^*$ denote its continuous 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 $0$ map is the origin of the vector space $X^*$ and it always has norm $\, 0\, = 0.$
If $X = \$ then the only linear functional on $X$ is the constant $0$ map and moreover, the sets in the last two rows will both be empty and consequently, their supremums will equal $\sup \varnothing = - \infty$ instead of the correct value of $0.$

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.

Examples

Dual norm for matrices

The ' defined by
$\, A\, _ = \sqrt = \sqrt = \sqrt$
is self-dual, i.e., its dual norm is $\, \cdot \, '_ = \, \cdot \, _.$

The ', a special case of the ''induced norm'' when $p=2$, is defined by the maximum singular values of a matrix, that is,
$\, A \, _2 = \sigma_(A),$
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.

If p, q \in , \infty/math> the Schatten $\ell^p$-norm on matrices is dual to the Schatten $\ell^q$-norm.

Finite-dimensional spaces

Let $\, \cdot\,$ be a norm on $\R^n.$ 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 $x$ and $z.$ The dual of the dual norm is the original norm: we have $\, x\, _ = \, x\,$ for all $x.$ (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 $z,$ the value of $x$ 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 $q$ satisfies $\tfrac + \tfrac = 1,$ that is, $q = \tfrac.$

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 '.

''L^p'' and ℓ^''p'' spaces

For p \in , \infty -norm (also called $\ell_p$-norm) of vector $\mathbf = (x_n)_n$ is
$\, \mathbf\, _p ~:=~ \left(\sum_^n \left, x_i\^p\right)^.$

If p, q \in , \infty/math> satisfy $1/p+1/q=1$ then the $\ell^q$ and $\ell^q$ norms are dual to each other and the same is true of the $L^q$ and $L^q$ norms, where $(X, \Sigma, \mu),$ is some measure space.
In particular the Euclidean norm is self-dual since $p = q = 2.$
For $\sqrt$, the dual norm is $\sqrt$ with $Q$ positive definite.

For $p = 2,$ the $\, \,\cdot\,\, _2$-norm is even induced by a canonical inner product $\langle \,\cdot,\,\cdot\rangle,$ meaning that $\, \mathbf\, _2 = \sqrt$ for all vectors $\mathbf.$ This inner product can expressed in terms of the norm by using the polarization identity.
On $\ell^2,$ this is the ' defined by
$\langle \left(x_n\right)_, \left(y_n\right)_ \rangle_ ~=~ \sum_n x_n \overline$
while for the space $L^2(X, \mu)$ associated with a measure space $(X, \Sigma, \mu),$ which consists of all square-integrable functions, this inner product is
$\langle f, g \rangle_ = \int_X f(x) \overline \, \mathrm dx.$
The norms of the continuous dual spaces of $\ell^2$ and $\ell^2$ 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.

Properties

More generally, let $X$ and $Y$ be topological vector spaces and let $L(X,Y)$ be the collection of all bounded linear mappings (or ) 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
$\begin \, \left(f_1 + f_2\right) x \, ~&=~ \, f_1 x + f_2 x\, \\ &\leq~ \, f_1 x\, + \, f_2 x\, \\ &\leq~ \left(\, f_1\, + \, f_2\, \right) \, x\, \\ &\leq~ \, f_1\, + \, f_2\, \end$

for every $x \in X$ satisfying $\, x\, \leq 1.$ This fact together with the definition of $\, \cdot \, ~:~ L(X, Y) \to \mathbb$ implies the triangle inequality:
$\, f + g\, \leq \, f\, + \, g\, .$

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 $\left\, f x_0\right\, > 0$ 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 $( L(X, Y), \, \cdot \, )$ is complete. Let $f_ = \left(f_n\right)_^$ be a Cauchy sequence in $L(X, Y),$ so by definition $\left\, f_n - f_m\right\, \to 0$ as $n, m \to \infty.$ This fact together with the relation
$\left\, f_n x - f_m x\right\, = \left\, \left( f_n - f_m \right) x \right\, \leq \left\, f_n - f_m\right\, \, x\,$

implies that $\left(f_nx \right)_^$ 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 $\left\, f_n - f_m\right\, \, x \, ~\leq~ \varepsilon \, x\,$ for all sufficiently large integers and . It follows that
$\left\, fx - f_m x\right\, ~\leq~ \varepsilon \, x\,$
for sufficiently all large $m.$ Hence $\, fx\, \leq \left( \left\, f_m\right\, + \varepsilon \right) \, x\, ,$ so that $f \in L(X, Y)$ and $\left\, f - f_m\right\, \leq \varepsilon.$ 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

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Notes

References

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External links

Notes on the proximal mapping by Lieven Vandenberge

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