Dual norm

Let ${\displaystyle X}$ be a normed vector space with norm ${\displaystyle |\cdot |}$ and let ${\displaystyle X^{*}}$ be the dual space. The dual norm of a continuous linear functional ${\displaystyle f}$ belonging to ${\displaystyle X^{*}}$ is defined to be the real number

${\displaystyle \|f\|:=\sup\{|f(x)|:x\in X,|x|\leq 1\}}$

where ${\displaystyle \sup }$ denotes the supremum.[1]

The map ${\displaystyle f\mapsto \|f\|}$ defines a norm on ${\displaystyle X^{*}}$. (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 ${\displaystyle X^{*}}$ induced by ${\displaystyle |\cdot |}$ turns out to be as strong as the weak-* topology on ${\displaystyle X^{*}}$.

If the ground field of ${\displaystyle X}$ is complete then ${\displaystyle X^{*}}$ is a Banach space.

The double dual (or second dual) ${\displaystyle X^{**}}$ of ${\displaystyle X}$ is the dual of the normed vector space ${\displaystyle X^{*}}$. There is a natural map ${\displaystyle \varphi :X\to X^{**}}$. Indeed, for each ${\displaystyle w^{*}}$ in ${\displaystyle X^{*}}$ define

${\displaystyle \varphi (v)(w^{*}):=w^{*}(v).}$

The map ${\displaystyle \varphi }$ is linear, injective, and distance preserving.[2] In particular, if ${\displaystyle X}$ is complete (i.e. a Banach space), then ${\displaystyle \varphi }$ is an isometry onto a closed subspace of ${\displaystyle X^{**}}$.[3]

In general, the map ${\displaystyle \varphi }$ is not surjective. For example, if ${\displaystyle X}$ is the Banach space ${\displaystyle L^{\infty }}$ consisting of bounded functions on the real line with the supremum norm, then the map ${\displaystyle \varphi }$ is not surjective. (See ${\displaystyle L^{p}}$ space). If ${\displaystyle \varphi }$ is surjective, then ${\displaystyle X}$ is said to be a reflexive Banach space. If ${\displaystyle 1<p<\infty ,}$ then the space ${\displaystyle L^{p}}$ is a reflexive Banach space.

Let ${\displaystyle \|\cdot \|}$ be a norm on ${\displaystyle \mathbb {R} ^{n}.}$ The associated dual norm, denoted ${\displaystyle \|\cdot \|_{*},}$ is defined as

${\displaystyle \|z\|_{*}=\sup\{z^{\intercal }x\;|\;\|x\|\leq 1\}.}$

(This can be shown to be a norm.) The dual norm can be interpreted as the operator norm of ${\displaystyle z^{\intercal }}$, interpreted as a ${\displaystyle 1\times n}$ matrix, with the norm ${\displaystyle \|\cdot \|}$ on ${\displaystyle \mathbb {R} ^{n}}$, and the absolute value on ${\displaystyle \mathbb {R} }$:

${\displaystyle \|z\|_{*}=\sup\{|z^{\intercal }x|\;|\;\|x\|\leq 1\}.}$

From the definition of dual norm we have the inequality

${\displaystyle z^{\intercal }x=\|x\|\left(z^{\intercal }{\frac {x}{\|x\|}}\right)\leq \|x\|\|z\|_{*}}$

which holds for all x and z.[4] The dual of the dual norm is the original norm: we have ${\displaystyle \|x\|_{**}=\|x\|}$ for all x. (This need not hold in infinite-dimensional vector spaces.)

The dual of the Euclidean norm is the Euclidean norm, since

${\displaystyle \sup\{z^{\intercal }x\;|\;\|x\|_{2}\leq 1\}=\|z\|_{2}.}$

(This follows from the Cauchy–Schwarz inequality; for nonzero z, the value of x that maximises ${\displaystyle z^{\intercal }x}$ over ${\displaystyle \|x\|_{2}\leq 1}$ is ${\displaystyle {\tfrac {z}{\|z\|_{2}}}}$.)

The dual of the ${\displaystyle \ell _{\infty }}$-norm is the ${\displaystyle \ell _{1}}$-norm:

${\displaystyle \sup\{z^{\intercal }x\;|\;\|x\|_{\infty }\leq 1\}=\sum _{i=1}^{n}|z_{i}|=\|z\|_{1},}$

and the dual of the ${\displaystyle \ell _{1}}$-norm is the ${\displaystyle \ell _{\infty }}$-norm.

More generally, Hölder's inequality shows that the dual of the ${\displaystyle \ell _{p}}$-norm is the ${\displaystyle \ell _{q}}$-norm, where, q satisfies ${\displaystyle {\tfrac {1}{p}}+{\tfrac {1}{q}}=1}$, i.e., ${\displaystyle q={\tfrac {p}{p-1}}.}$

As another example, consider the ${\displaystyle \ell _{2}}$- or spectral norm on ${\displaystyle \mathbb {R} ^{m\times n}}$. The associated dual norm is

${\displaystyle \|Z\|_{2*}=\sup\{\mathrm {\bf {tr}} (Z^{\intercal }X)|\|X\|_{2}\leq 1\},}$

which turns out to be the sum of the singular values,

${\displaystyle \|Z\|_{2*}=\sigma _{1}(Z)+\cdots +\sigma _{r}(Z)=\mathrm {\bf {tr}} ({\sqrt {Z^{\intercal }Z}}),}$

where ${\displaystyle r=\mathrm {\bf {rank}} Z.}$ This norm is sometimes called the nuclear norm.[5]

More generally, let ${\displaystyle X}$ and ${\displaystyle Y}$ be topological vector spaces, and ${\displaystyle L(X,Y)}$[6] be the collection of all bounded linear mappings (or operators) of ${\displaystyle X}$ into ${\displaystyle Y}$. In the case where ${\displaystyle X}$ and ${\displaystyle Y}$ are normed vector spaces, ${\displaystyle L(X,Y)}$ can be normed in a natural way.

Theorem 1. Let ${\displaystyle X}$ and ${\displaystyle Y}$ be normed spaces, and associate to each ${\displaystyle f\in L(X,Y)}$ the number:

${\displaystyle \|f\|=\sup\{|f(x)|:x\in X,\|x\|\leq 1\}.}$

This turns ${\displaystyle L(X,Y)}$ into a normed space. Moreover if ${\displaystyle Y}$ is a Banach space, so is ${\displaystyle L(X,Y)}$.[7]

Proof. A subset of a normed space is bounded if and only if it lies in some multiple of the unit sphere; thus ${\displaystyle \|f\|<\infty }$ for every ${\displaystyle f\in L(X,Y)}$ if ${\displaystyle \alpha }$ is a scalar, then ${\displaystyle (\alpha f)(x)=\alpha \cdot fx}$ so that

${\displaystyle \|\alpha f\|=|\alpha |\|f\|}$

The triangle inequality in ${\displaystyle Y}$ shows that

${\displaystyle {\begin{aligned}\|(f_{1}+f_{2})x\|&=\|f_{1}x+f_{2}x\|\\&\leq \|f_{1}x\|+\|f_{2}x\|\\&\leq (\|f_{1}\|+\|f_{2}\|)\|x\|\\&\leq \|f_{1}\|+\|f_{2}\|\end{aligned}}}$

for every ${\displaystyle x\in X}$ with ${\displaystyle \|x\|\leq 1}$. Thus

${\displaystyle \|f_{1}+f_{2}\|\leq \|f_{1}\|+\|f_{2}\|}$

If ${\displaystyle f\neq 0}$, then ${\displaystyle fx\neq 0}$ for some ${\displaystyle x\in X}$; hence ${\displaystyle \|f\|>0}$. Thus, ${\displaystyle L(X,Y)}$ is a normed space.[8]

Assume now that ${\displaystyle Y}$ is complete, and that ${\displaystyle \{f_{n}\}}$ is a Cauchy sequence in ${\displaystyle L(X,Y)}$. Since

${\displaystyle \|f_{n}x-f_{m}x\|\leq \|f_{n}-f_{m}\|\|x\|}$

and it is assumed that ${\displaystyle \|f_{n}-f_{m}\|\to 0}$ as ${\displaystyle n,m\to \infty }$, ${\displaystyle \{f_{n}x\}}$ is a Cauchy sequence in ${\displaystyle Y}$ for every ${\displaystyle x\in X}$. Hence

${\displaystyle fx=\lim _{n\to \infty }f_{n}x}$

exists. It is clear that ${\displaystyle f:X\to Y}$ is linear. If ${\displaystyle \varepsilon >0}$, ${\displaystyle \|f_{n}-f_{m}\|\|x\|\leq \varepsilon \|x\|}$ for sufficiently large n and m. It follows

${\displaystyle \|fx-f_{m}x\|\leq \varepsilon \|x\|}$

for sufficiently large m. Hence ${\displaystyle \|fx\|\leq (\|f_{m}\|+\varepsilon )\|x\|}$, so that ${\displaystyle f\in L(X,Y)}$ and ${\displaystyle \|f-f_{m}\|\leq \varepsilon }$. Thus ${\displaystyle f_{m}\to f}$ in the norm of ${\displaystyle L(X,Y)}$. This establishes the completeness of ${\displaystyle L(X,Y).}$[9]

When ${\displaystyle Y}$ is a scalar field (i.e. ${\displaystyle Y=\mathbb {C} }$ or ${\displaystyle Y=\mathbb {R} }$) so that ${\displaystyle L(X,Y)}$ is the dual space ${\displaystyle X^{*}}$ of ${\displaystyle X}$.

Theorem 2. Suppose ${\displaystyle B}$ is the closed unit ball of normed space ${\displaystyle X}$. For every ${\displaystyle x^{*}\in X^{*}}$ define:

${\displaystyle \|x^{*}\|=\sup\{|\langle {x,x^{*}}\rangle |:x\in B\}}$

Then

(a) This norm makes ${\displaystyle X^{*}}$ into a Banach space.[10]

(b) Let ${\displaystyle B^{*}}$ be the closed unit ball of ${\displaystyle X^{*}}$. For every ${\displaystyle x\in X}$,

${\displaystyle \|x\|=\sup\{|\langle {x,x^{*}}\rangle |:x^{*}\in B^{*}\}.}$

Consequently, ${\displaystyle x^{*}\to \langle {x,x^{*}}\rangle }$ is a bounded linear functional on ${\displaystyle X^{*}}$ of norm ${\displaystyle \|x\|}$.

(c) ${\displaystyle B^{*}}$ is weak*-compact.

Proof. Since ${\displaystyle L(X,Y)=X^{*}}$, when ${\displaystyle Y}$ is the scalar field, (a) is a corollary of Theorem 1. Fix ${\displaystyle x\in X}$. There exists[11] ${\displaystyle y^{*}\in B^{*}}$ such that

${\displaystyle \langle {x,y^{*}}\rangle =\|x\|.}$

but,

${\displaystyle |\langle {x,x^{*}}\rangle |\leq \|x\|\|x^{*}\|\leq \|x\|}$

for every ${\displaystyle x^{*}\in B^{*}}$. (b) follows from the above. Since the open unit ball ${\displaystyle U}$ of ${\displaystyle X}$ is dense in ${\displaystyle B}$, the definition of ${\displaystyle \|x^{*}\|}$ shows that ${\displaystyle x^{*}\in B^{*}}$ if and only if ${\displaystyle |\langle {x,x^{*}}\rangle |\leq 1}$ for every ${\displaystyle x\in U}$. The proof for (c)[12] now follows directly.[13]

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

• Aliprantis, Charalambos D.; Border, Kim C. (2007). Infinite Dimensional Analysis: A Hitchhiker's Guide (3rd ed.). Springer. ISBN 9783540326960.
• Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization. Cambridge University Press. ISBN 9780521833783.
• Kolmogorov, A.N.; Fomin, S.V. (1957). Elements of the Theory of Functions and Functional Analysis, Volume 1: Metric and Normed Spaces. Rochester: Graylock Press.
• Rudin, Walter (1991), Functional analysis, McGraw-Hill Science, ISBN 978-0-07-054236-5.

• Notes on the proximal mapping by Lieven Vandenberge