Dual norm

 ( Functional Analysis ) 

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

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Definition Let $X$ be a normed vector space with norm $\backslash |\backslash cdot\backslash |$ 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:

       &= \sup &&\{\,|f(x)| &&~:~ \|x\| < 1 ~&&~\text{ and } ~&&x \in X\} \\
       &= \inf &&\{\,c \in [0, \infty) &&~:~ |f(x)| \leq c \|x\| ~&&~\text{ for all } ~&&x \in X\} \\
       &= \sup &&\{\,|f(x)| &&~:~ \|x\| = 1 \text{ or } 0 ~&&~\text{ and } ~&&x \in X\} \\
       &= \sup &&\{\,|f(x)| &&~:~ \|x\| = 1 ~&&~\text{ and } ~&&x \in X\} \;\;\;\text{ this equality holds if and only if } X \neq \{0\} \\
       &= \sup &&\bigg\{\,\frac{\|x\|} ~&&~:~ x \neq 0 &&~\text{ and } ~&&x \in X\bigg\} \;\;\;\text{ this equality holds if and only if } X \neq \{0\} \\
     

\end{alignat} where $\backslash sup$ and $\backslash inf$ denote the supremum and infimum, respectively. The constant $0$ map is the origin of the vector space $X^*$ and it always has norm $\backslash |0\backslash |\; =\; 0.$ If $X\; =\; \backslash \{0\backslash \}$ 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 will equal $\backslash sup\; \backslash varnothing\; =\; -\; \backslash infty$ instead of the correct value of $0.$

Importantly, a linear function $f$ is not, in general, guaranteed to achieve its norm $\backslash |f\backslash |\; =\; \backslash sup\; \backslash$

on the closed unit ball $\backslash \{x\; \backslash in\; X\; :\; \backslash |x\backslash |\; \backslash leq\; 1\backslash \},$ meaning that there might not exist any vector $u\; \backslash in\; X$ of norm $\backslash |u\backslash |\; \backslash leq\; 1$ such that $\backslash |f\backslash |\; =\; |f\; u|$ (if such a vector does exist and if $f\; \backslash neq\; 0,$ then $u$ would necessarily have unit norm $\backslash |u\backslash |\; =\; 1$).>

R.C. James proved James's theorem in 1964, which states that a Banach space $X$ is reflexive space if and only if every bounded linear function $f\; \backslash in\; X^*$ achieves its norm on the closed unit ball. It follows, in particular, that every non-reflexive Banach space has some bounded linear functional that does not achieve its norm on the closed unit ball. However, the Bishop–Phelps theorem guarantees that the set of bounded linear functionals that achieve their norm on the unit sphere of a Banach space is a norm-Dense set of the continuous dual space. The map $f\; \backslash mapsto\; \backslash$

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. Since the ground field of $X$ ($\backslash Reals$ or $\backslash Complex$) is complete, $X^*$ is a Banach space. The topology on $X^*$ induced by $\backslash$

turns out to be stronger than the weak-* topology on $X^*.$ 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 $\backslash varphi:\; X\; \backslash to\; X^\{**\}$. Indeed, for each $w^*$ in $X^*$ define $$\backslash varphi(v)(w^*):\; =\; w^*(v).$$ The map $\backslash varphi$ is linear map, injective, and Isometry. In particular, if $X$ is complete (i.e. a Banach space), then $\backslash varphi$ is an isometry onto a closed subspace of $X^\{**\}$. In general, the map $\backslash varphi$ is not surjective. For example, if $X$ is the Banach space $L^\{\backslash infty\}$ consisting of bounded functions on the real line with the supremum norm, then the map $\backslash varphi$ is not surjective. (See Lp spaces). If $\backslash varphi$ is surjective, then $X$ is said to be a reflexive Banach space. If then the Lp spaces is a reflexive Banach space. Examples Dual norm for matrices The defined by $$\backslash$$

^2} = \sqrt{\operatorname{trace}(A^*A)} = \sqrt{\sum_{i=1}^{\min\{m,n\}} \sigma_{i}^2} is self-dual, i.e., its dual norm is $\backslash$

_{\text{F}}. The , a special case of the induced norm when $p=2$, is defined by the maximum singular values of a matrix, that is, $$\backslash$$

_2 = \sigma_{\max}(A), has the nuclear norm as its dual norm, which is defined by $$\backslash$$

'_2 = \sum_i \sigma_i(B), for any matrix $B$ where $\backslash sigma\_i(B)$ denote the singular values. If $p,\; q\; \backslash in\; 1,$ the Schatten $\backslash ell^p$-norm on matrices is dual to the Schatten $\backslash ell^q$-norm. Finite-dimensional spaces Let $\backslash$

_*, is defined as $$\backslash$$

\leq 1 \}. (This can be shown to be a norm.) The dual norm can be interpreted as the operator norm of $z^\backslash intercal,$ interpreted as a $1\; \backslash times\; n$ matrix, with the norm $\backslash$

on $\backslash R^n$, and the absolute value on $\backslash R$: $$\backslash$$

\leq 1 \}. From the definition of dual norm we have the inequality $$z^\backslash intercal\; x\; =\; \backslash$$

\right) \leq \|x\| \|z\|_* which holds for all $x$ and $z.$This inequality is tight, in the following sense: for any $x$ there is a $z$ for which the inequality holds with equality. (Similarly, for any $z$ there is an $x$ that gives equality.) The dual of the dual norm is the original norm: we have $\backslash |x\backslash |\_\{**\}\; =\; \backslash |x\backslash |$ for all $x.$ (This need not hold in infinite-dimensional vector spaces.)

The dual of the Euclidean norm is the Euclidean norm, since $$\backslash sup\backslash \{z^\backslash intercal\; x\; :\; \backslash |x\backslash |\_2\; \backslash leq\; 1\; \backslash \}\; =\; \backslash |z\backslash |\_2.$$

(This follows from the Cauchy–Schwarz inequality; for nonzero $z,$ the value of $x$ that maximises $z^\backslash intercal\; x$ over $\backslash |x\backslash |\_2\; \backslash leq\; 1$ is $\backslash tfrac\{z\}\{\backslash |z\backslash |\_2\}.$)

The dual of the $\backslash ell^\backslash infty$-norm is the $\backslash ell^1$-norm: $$\backslash sup\backslash \{z^\backslash intercal\; x\; :\; \backslash |x\backslash |\; \_\backslash infty\; \backslash leq\; 1\backslash \}\; =\; \backslash sum\_\{i=1\}^n\; |z\_i|\; =\; \backslash |z\backslash |\; \_1,$$ and the dual of the $\backslash ell^1$-norm is the $\backslash ell^\backslash infty$-norm.

More generally, Hölder's inequality shows that the dual of the Lp spaces is the $\backslash ell^q$-norm, where $q$ satisfies $\backslash tfrac\{1\}\{p\}\; +\; \backslash tfrac\{1\}\{q\}\; =\; 1,$ that is, $q\; =\; \backslash tfrac\{p\}\{p-1\}.$

As another example, consider the $\backslash ell^2$- or spectral norm on $\backslash R^\{m\backslash times\; n\}$. The associated dual norm is $$\backslash |Z\backslash |\; \_\{2*\}\; =\; \backslash sup\backslash \{\backslash mathbf\{tr\}(Z^\backslash intercal\; X)\; :\; \backslash |X\backslash |\_2\; \backslash leq\; 1\backslash \},$$ which turns out to be the sum of the singular values, $$\backslash |Z\backslash |\; \_\{2*\}\; =\; \backslash sigma\_1(Z)\; +\; \backslash cdots\; +\; \backslash sigma\_r(Z)\; =\; \backslash mathbf\{tr\}\; (\backslash sqrt\{Z^\backslash intercal\; Z\}),$$ where $r\; =\; \backslash mathbf\{rank\}\; Z.$ This norm is sometimes called the nuclear operator.

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L^p and ℓ^p spaces For $p\; \backslash in\; 1,,$ -norm (also called $\backslash ell\_p$-norm) of vector $\backslash mathbf\{x\}\; =\; (x\_n)\_n$ is $$\backslash |\backslash mathbf\{x\}\backslash |\_p\; ~:=~\; \backslash left(\backslash sum\_\{i=1\}^n\; \backslash left|x\_i\backslash right|^p\backslash right)^\{1/p\}.$$

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

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

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Properties Given normed vector spaces $X$ and $Y,$ let $L(X,Y)$Each $L(X,Y)$ is a vector space, with the usual definitions of addition and scalar multiplication of functions; this only depends on the vector space structure of $Y$, not $X$. be the collection of all Bounded operator (or ) of $X$ into $Y.$ Then $L(X,Y)$ 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 for every $f\; \backslash in\; L(X,Y)$ if $\backslash alpha$ is a scalar, then $(\backslash alpha\; f)(x)\; =\; \backslash alpha\; \backslash cdot\; f\; x$ so that $$\backslash |\backslash alpha\; f\backslash |\; =\; |\backslash alpha|\; \backslash |f\backslash |.$$

The triangle inequality in $Y$ shows that

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

Since $\backslash \{\; |f(x)|\; :\; x\; \backslash in\; X,\; \backslash |x\backslash |\; \backslash leq\; 1\; \backslash \}$ is a non-empty set of non-negative real numbers, $\backslash |f\backslash |\; =\; \backslash sup\; \backslash left\backslash \{\; |f(x)|\; :\; x\; \backslash in\; X,\; \backslash |\; x\; \backslash |\; \backslash leq\; 1\; \backslash right\backslash \}$ is a non-negative real number. If $f\; \backslash neq\; 0$ then $f\; x\_0\; \backslash neq\; 0$ for some $x\_0\; \backslash in\; X,$ which implies that $\backslash left\backslash |f\; x\_0\backslash right\backslash |\; >\; 0$ and consequently $\backslash |f\backslash |\; >\; 0.$ This shows that $\backslash left(\; L(X,\; Y),\; \backslash |\; \backslash cdot\; \backslash |\backslash right)$ is a normed space.

Assume now that $Y$ is complete and we will show that $(\; L(X,\; Y),\; \backslash |\; \backslash cdot\; \backslash |)$ is complete. Let $f\_\{\backslash bull\}\; =\; \backslash left(f\_n\backslash right)\_\{n=1\}^\{\backslash infty\}$ be a Cauchy sequence in $L(X,\; Y),$ so by definition $\backslash left\backslash |f\_n\; -\; f\_m\backslash right\backslash |\; \backslash to\; 0$ as $n,\; m\; \backslash to\; \backslash infty.$ This fact together with the relation $$\backslash left\backslash |f\_n\; x\; -\; f\_m\; x\backslash right\backslash |\; =\; \backslash left\backslash |\; \backslash left(\; f\_n\; -\; f\_m\; \backslash right)\; x\; \backslash right\backslash |\; \backslash leq\; \backslash left\backslash |f\_n\; -\; f\_m\backslash right\backslash |\; \backslash |x\backslash |$$

implies that $\backslash left(f\_nx\; \backslash right)\_\{n=1\}^\{\backslash infty\}$ is a Cauchy sequence in $Y$ for every $x\; \backslash in\; X.$ It follows that for every $x\; \backslash in\; X,$ the limit $\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; f\_n\; x$ exists in $Y$ and so we will denote this (necessarily unique) limit by $f\; x,$ that is: $$f\; x\; ~=~\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\; f\_n\; x.$$

It can be shown that $f:\; X\; \backslash to\; Y$ is linear. If $\backslash varepsilon\; >\; 0$, then $\backslash left\backslash |f\_n\; -\; f\_m\backslash right\backslash |\; \backslash |\; x\; \backslash |\; ~\backslash leq~\; \backslash varepsilon\; \backslash |x\backslash |$ for all sufficiently large integers and . It follows that $$\backslash left\backslash |fx\; -\; f\_m\; x\backslash right\backslash |\; ~\backslash leq~\; \backslash varepsilon\; \backslash |x\backslash |$$ for sufficiently all large $m.$ Hence $\backslash |fx\backslash |\; \backslash leq\; \backslash left(\; \backslash left\backslash |f\_m\backslash right\backslash |\; +\; \backslash varepsilon\; \backslash right)\; \backslash |x\backslash |,$ so that $f\; \backslash in\; L(X,\; Y)$ and $\backslash left\backslash |f\; -\; f\_m\backslash right\backslash |\; \backslash leq\; \backslash varepsilon.$ This shows that $f\_m\; \backslash to\; f$ in the norm topology of $L(X,\; Y).$ This establishes the completeness of $L(X,\; Y).$

When $Y$ is a scalar field (i.e. $Y\; =\; \backslash Complex$ or $Y\; =\; \backslash R$) so that $L(X,Y)$ is the dual space $X^*$ of $X.$

Let $B\; ~=~\; \backslash sup\backslash \{\; x\; \backslash in\; X\; ~:~\; \backslash |\; x\; \backslash |\; \backslash le\; 1\; \backslash \}$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 $x\; \backslash in\; X.$ There exists $y^*\; \backslash in\; B^*$ such that $$\backslash langle\{x,y^*\}\backslash rangle\; =\; \backslash |x\backslash |.$$ but, $$|\backslash langle\{x,x^*\}\backslash rangle|\; \backslash leq\; \backslash |x\backslash |\backslash |x^*\backslash |\; \backslash leq\; \backslash |x\backslash |$$ for every $x^*\; \backslash in\; B^*$. (b) follows from the above. Since the open unit ball $U$ of $X$ is dense in $B$, the definition of $\backslash |x^*\backslash |$ shows that $x^*\; \backslash in\; B^*$ if and only if $|\backslash langle\{x,x^*\}\backslash rangle|\; \backslash leq\; 1$ for every $x\; \backslash in\; U$. The proof for (c) now follows directly.

As usual, let $d(x,\; y)\; :=\; \backslash |x\; -\; y\backslash |$ denote the canonical metric induced by the norm on $X,$ and denote the distance from a point $x$ to the subset $S\; \backslash subseteq\; X$ by $$d(x,\; S)\; ~:=~\; \backslash inf\_\{s\; \backslash in\; S\}\; d(x,\; s)\; ~=~\; \backslash inf\_\{s\; \backslash in\; S\}\; \backslash |x\; -\; s\backslash |.$$ If $f$ is a bounded linear functional on a normed space $X,$ then for every vector $x\; \backslash in\; X,$ $$|f(x)|\; =\; \backslash |f\backslash |\; \backslash ,\; d(x,\; \backslash ker\; f),$$ where $\backslash ker\; f\; =\; \backslash \{k\; \backslash in\; X\; :\; f(k)\; =\; 0\backslash \}$ denotes the kernel of $f.$

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See also

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Notes

• (2025). 9783540326960, Springer. ISBN 9783540326960
• (2025). 9780521833783, Cambridge University Press. ISBN 9780521833783
• Joe Diestel (1984). 9780387908595, Springer-Verlag. ISBN 9780387908595

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External links

• Notes on the proximal mapping by Lieven Vandenberge

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