Infunctional analysis, thedual norm is a measure of size for acontinuouslinear function defined on anormed vector space.

Let${\displaystyle X}$ be anormed vector space with norm${\displaystyle \Vert \cdot \Vert }$ and let${\displaystyle X^{\ast }}$ denote itscontinuous dual space. Thedual norm of a continuouslinear functional${\displaystyle f}$ belonging to${\displaystyle X^{\ast }}$ is the non-negative real number defined^[1] by any of the following equivalent formulas:where${\displaystyle sup}$ and${\displaystyle inf}$ denote thesupremum and infimum, respectively. The constant${\displaystyle 0}$ map is the origin of the vector space${\displaystyle X^{\ast }}$ and it always has norm${\displaystyle \Vert 0\Vert =0.}$ If${\displaystyle X=\{0\}}$ then the only linear functional on${\displaystyle X}$ is the constant${\displaystyle 0}$ map and moreover, the sets in the last two rows will both be empty and consequently, theirsupremums will equal${\displaystyle sup\varnothing =-\mathrm{\infty }}$ instead of the correct value of${\displaystyle 0.}$

Importantly, a linear function${\displaystyle f}$ is not, in general, guaranteed to achieve its norm${\displaystyle \Vert f\Vert =sup\{|f(x)|:\Vert x\Vert \le 1,x\in X\}}$ on the closed unit ball${\displaystyle \{x\in X:\Vert x\Vert \le 1\},}$ meaning that there might not exist any vector${\displaystyle u\in X}$ of norm${\displaystyle \Vert u\Vert \le 1}$ such that${\displaystyle \Vert f\Vert =|fu|}$ (if such a vector does exist and if${\displaystyle f\ne 0,}$ then${\displaystyle u}$ would necessarily have unit norm${\displaystyle \Vert u\Vert =1}$). R.C. James provedJames's theorem in 1964, which states that aBanach space${\displaystyle X}$ isreflexive if and only if every bounded linear function${\displaystyle f\in X^{\ast }}$ achieves its norm on the closed unit ball.^[2] 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, theBishop–Phelps theorem guarantees that the set of bounded linear functionals that achieve their norm on the unit sphere of aBanach space is a norm-dense subset of thecontinuous dual space.^[3][4]

The map${\displaystyle f\mapsto \Vert f\Vert }$ defines anorm on${\displaystyle X^{\ast }.}$ (See Theorems 1 and 2 below.) The dual norm is a special case of theoperator norm defined for each (bounded) linear map between normed vector spaces. Since theground field of${\displaystyle X}$ (${\displaystyle \mathbb{R}}$ or${\displaystyle \mathbb{C}}$) iscomplete,${\displaystyle X^{\ast }}$ is aBanach space. The topology on${\displaystyle X^{\ast }}$induced by${\displaystyle \Vert \cdot \Vert }$ turns out to be stronger than theweak-* topology on${\displaystyle X^{\ast }.}$

Thedouble dual (or second dual)${\displaystyle X^{\ast \ast }}$ of${\displaystyle X}$ is the dual of the normed vector space${\displaystyle X^{\ast }}$. There is a natural map${\displaystyle \phi :X\to X^{\ast \ast }}$. Indeed, for each${\displaystyle w^{\ast }}$ in${\displaystyle X^{\ast }}$ define$${\displaystyle \phi (v)(w^{\ast }):=w^{\ast }(v).}$$

The map${\displaystyle \phi }$ islinear,injective, anddistance preserving.^[5] In particular, if${\displaystyle X}$ is complete (i.e. a Banach space), then${\displaystyle \phi }$ is an isometry onto a closed subspace of${\displaystyle X^{\ast \ast }}$.^[6]

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

TheFrobenius norm defined by$${\displaystyle \Vert A{\Vert }_{\text{F}}=\sqrt{\sum _{i=1}^m\sum _{j=1}^n{\left|a_{ij}\right|}^2}=\sqrt{\mathrm{trace}(A^{\ast }A)}=\sqrt{\sum _{i=1}^{min\{m,n\}}{\sigma }_i^2}}$$is self-dual, i.e., its dual norm is${\displaystyle \Vert \cdot {\Vert }_{\text{F}}^{\prime }=\Vert \cdot {\Vert }_{\text{F}}.}$

Thespectral norm, a special case of theinduced norm when${\displaystyle p=2}$, is defined by the maximumsingular values of a matrix, that is,$${\displaystyle \Vert A{\Vert }_2={\sigma }_{max}(A),}$$has the nuclear norm as its dual norm, which is defined by$${\displaystyle \Vert B{\Vert }_2^{\prime }=\sum _i{\sigma }_i(B),}$$ for any matrix${\displaystyle B}$ where${\displaystyle {\sigma }_i(B)}$ denote the singular values^{[citation needed]}.

If${\displaystyle p,q\in [1,\mathrm{\infty }]}$ theSchatten${\displaystyle {\ell }^p}$-norm on matrices is dual to the Schatten${\displaystyle {\ell }^q}$-norm.

Let${\displaystyle \Vert \cdot \Vert }$ be a norm on${\displaystyle {\mathbb{R}}^n.}$ The associateddual norm, denoted${\displaystyle \Vert \cdot {\Vert }_{\ast },}$ is defined as$${\displaystyle \Vert z{\Vert }_{\ast }=sup\{z^{⊺}x:\Vert x\Vert \le 1\}.}$$

(This can be shown to be a norm.) The dual norm can be interpreted as theoperator norm of${\displaystyle z^{⊺},}$ interpreted as a${\displaystyle 1\times n}$ matrix, with the norm${\displaystyle \Vert \cdot \Vert }$ on${\displaystyle {\mathbb{R}}^n}$, and the absolute value on${\displaystyle \mathbb{R}}$:$${\displaystyle \Vert z{\Vert }_{\ast }=sup\{|z^{⊺}x|:\Vert x\Vert \le 1\}.}$$

From the definition of dual norm we have the inequality$${\displaystyle z^{⊺}x=\Vert x\Vert \left(z^{⊺}\frac{x}{\Vert x\Vert }\right)\le \Vert x\Vert \Vert z{\Vert }_{\ast }}$$which holds for all${\displaystyle x}$ and${\displaystyle z.}$^[7][8] The dual of the dual norm is the original norm: we have${\displaystyle \Vert x{\Vert }_{\ast \ast }=\Vert x\Vert }$ for all${\displaystyle x.}$ (This need not hold in infinite-dimensional vector spaces.)

The dual of theEuclidean norm is the Euclidean norm, since$${\displaystyle sup\{z^{⊺}x:\Vert x{\Vert }_2\le 1\}=\Vert z{\Vert }_2.}$$

(This follows from theCauchy–Schwarz inequality; for nonzero${\displaystyle z,}$ the value of${\displaystyle x}$ that maximises${\displaystyle z^{⊺}x}$ over${\displaystyle \Vert x{\Vert }_2\le 1}$ is${\displaystyle \frac{z}{\Vert z{\Vert }_2}.}$)

The dual of the${\displaystyle {\ell }^{\mathrm{\infty }}}$-norm is the${\displaystyle {\ell }^1}$-norm:$${\displaystyle sup\{z^{⊺}x:\Vert x{\Vert }_{\mathrm{\infty }}\le 1\}=\sum _{i=1}^n|z_i|=\Vert z{\Vert }_1,}$$and the dual of the${\displaystyle {\ell }^1}$-norm is the${\displaystyle {\ell }^{\mathrm{\infty }}}$-norm.

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

As another example, consider the${\displaystyle {\ell }^2}$- or spectral norm on${\displaystyle {\mathbb{R}}^{m\times n}}$. The associated dual norm is$${\displaystyle \Vert Z{\Vert }_{2\ast }=sup\{\mathbf{t}\mathbf{r}(Z^{⊺}X):\Vert X{\Vert }_2\le 1\},}$$which turns out to be the sum of the singular values,$${\displaystyle \Vert Z{\Vert }_{2\ast }={\sigma }_1(Z)+\cdots +{\sigma }_r(Z)=\mathbf{t}\mathbf{r}(\sqrt{Z^{⊺}Z}),}$$where${\displaystyle r=\mathbf{r}\mathbf{a}\mathbf{n}\mathbf{k}Z.}$ This norm is sometimes called thenuclear norm.^[9]

For${\displaystyle p\in [1,\mathrm{\infty }],}$p-norm (also called${\displaystyle {\ell }_p}$-norm) of vector${\displaystyle \mathbf{x}=(x_n{)}_n}$ is$${\displaystyle \Vert \mathbf{x}{\Vert }_p:={\left(\sum _{i=1}^n{\left|x_i\right|}^p\right)}^{1/p}.}$$

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

For${\displaystyle p=2,}$ the${\displaystyle \Vert \cdot {\Vert }_2}$-norm is even induced by a canonicalinner product${\displaystyle ⟨\cdot ,\cdot ⟩,}$ meaning that${\displaystyle \Vert \mathbf{x}{\Vert }_2=\sqrt{⟨\mathbf{x},\mathbf{x}⟩}}$ for all vectors${\displaystyle \mathbf{x}.}$ This inner product can expressed in terms of the norm by using thepolarization identity. On${\displaystyle {\ell }^2,}$ this is theEuclidean inner product defined by$${\displaystyle ⟨{\left(x_n\right)}_n,{\left(y_n\right)}_n{⟩}_{{\ell }^2}=\sum _n{x}_n\overline{y_n}}$$while for the space${\displaystyle L^2(X,\mu )}$ associated with ameasure space${\displaystyle (X,\mathrm{\Sigma },\mu ),}$ which consists of allsquare-integrable functions, this inner product is$${\displaystyle ⟨f,g{⟩}_{L^2}={\int }_{X}f(x)\overline{g(x)}\mathrm{d}x.}$$The norms of the continuous dual spaces of${\displaystyle {\ell }^2}$ and${\displaystyle {\ell }^2}$ satisfy thepolarization identity, and so these dual norms can be used to define inner products. With this inner product, this dual space is also aHilbert space.

Given normed vector spaces${\displaystyle X}$ and${\displaystyle Y,}$ let${\displaystyle L(X,Y)}$^[10] be the collection of allbounded linear mappings (oroperators) of${\displaystyle X}$ into${\displaystyle Y.}$ Then${\displaystyle L(X,Y)}$ can be given a canonical norm.

When${\displaystyle Y}$ is ascalar field (i.e.${\displaystyle Y=\mathbb{C}}$ or${\displaystyle Y=\mathbb{R}}$) so that${\displaystyle L(X,Y)}$ is thedual space${\displaystyle X^{\ast }}$ of${\displaystyle X.}$

As usual, let${\displaystyle d(x,y):=\Vert x-y\Vert }$ denote the canonicalmetric induced by the norm on${\displaystyle X,}$ and denote the distance from a point${\displaystyle x}$ to the subset${\displaystyle S\subseteq X}$ by$${\displaystyle d(x,S):=\underset{s\in S}{inf}d(x,s)=\underset{s\in S}{inf}\Vert x-s\Vert .}$$If${\displaystyle f}$ is a bounded linear functional on a normed space${\displaystyle X,}$ then for every vector${\displaystyle x\in X,}$^[18]$${\displaystyle |f(x)|=\Vert f\Vert d(x,\ker f),}$$where${\displaystyle \ker f=\{k\in X:f(k)=0\}}$ denotes thekernel of${\displaystyle f.}$

• Convex conjugate – Generalization of the Legendre transformation
• Hölder's inequality – Inequality between integrals in Lp spaces
• Lp space – Function spaces generalizing finite-dimensional p norm spaces
• Operator norm – Measure of the "size" of linear operators
• Polarization identity – Formula relating the norm and the inner product in a inner product space

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• Notes on the proximal mapping by Lieven Vandenberge

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