The Continuous Dual of a Normed Space is Complete

Measurement on a normed vector space

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

Definition [edit]

Let ${\displaystyle X}$ be a normed vector space with norm ${\displaystyle \|\cdot \|}$ and let ${\displaystyle X^{*}}$ denote its continuous dual space. The dual norm of a continuous linear functional ${\displaystyle f}$ belonging to ${\displaystyle X^{*}}$ is the non-negative real number defined^[1] by any of the following equivalent formulas:

$${\displaystyle {\begin{alignedat}{5}\|f\|&=\sup &&\{\,|f(x)|&&~:~\|x\|\leq 1~&&~{\text{ and }}~&&x\in X\}\\&=\sup &&\{\,|f(x)|&&~:~\|x\|<1~&&~{\text{ and }}~&&x\in X\}\\&=\inf &&\{\,c\in [0,\infty )&&~:~|f(x)|\leq c\|x\|~&&~{\text{ for all }}~&&x\in X\}\\&=\sup &&\{\,|f(x)|&&~:~\|x\|=1{\text{ or }}0~&&~{\text{ and }}~&&x\in X\}\\&=\sup &&\{\,|f(x)|&&~:~\|x\|=1~&&~{\text{ and }}~&&x\in X\}\;\;\;{\text{ this equality holds if and only if }}X\neq \{0\}\\&=\sup &&{\bigg \{}\,{\frac {|f(x)|}{\|x\|}}~&&~:~x\neq 0&&~{\text{ and }}~&&x\in X{\bigg \}}\;\;\;{\text{ this equality holds if and only if }}X\neq \{0\}\\\end{alignedat}}}$$

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

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 of a normed linear space [edit]

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.

Examples [edit]

Dual norm for matrices [edit]

The Frobenius norm defined by

$${\displaystyle \|A\|_{\text{F}}={\sqrt {\sum _{i=1}^{m}\sum _{j=1}^{n}\left|a_{ij}\right|^{2}}}={\sqrt {\operatorname {trace} (A^{*}A)}}={\sqrt {\sum _{i=1}^{\min\{m,n\}}\sigma _{i}^{2}}}}$$

is self-dual, i.e., its dual norm is ${\displaystyle \|\cdot \|'_{\text{F}}=\|\cdot \|_{\text{F}}.}$

The spectral norm , a special case of the induced norm when ${\displaystyle p=2}$ , is defined by the maximum singular values of a matrix, that is,

$${\displaystyle \|A\|_{2}=\sigma _{\max }(A),}$$

has the nuclear norm as its dual norm, which is defined by

$${\displaystyle \|B\|'_{2}=\sum _{i}\sigma _{i}(B),}$$

for any matrix ${\displaystyle B}$ where ${\displaystyle \sigma _{i}(B)}$ denote the singular values^{[ citation needed ]}.

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

Finite-dimensional spaces [edit]

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 ${\displaystyle x}$ and ${\displaystyle z.}$ ^[4] The dual of the dual norm is the original norm: we have ${\displaystyle \|x\|_{**}=\|x\|}$ for all ${\displaystyle x.}$ (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 ${\displaystyle z,}$ the value of ${\displaystyle x}$ 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 ${\displaystyle q}$ satisfies ${\displaystyle {\tfrac {1}{p}}+{\tfrac {1}{q}}=1,}$ that is, ${\displaystyle q={\tfrac {p}{p-1}}.}$

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

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

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

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

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

L^p and ℓ ^p spaces [edit]

For ${\displaystyle p\in [1,\infty ],}$ p-norm (also called ${\displaystyle \ell _{p}}$ -norm) of vector ${\displaystyle \mathbf {x} =(x_{n})_{n}}$ is

$${\displaystyle \|\mathbf {x} \|_{p}~:=~\left(\sum _{i=1}^{n}\left|x_{i}\right|^{p}\right)^{1/p}.}$$

If ${\displaystyle p,q\in [1,\infty ]}$ satisfy ${\displaystyle 1/p+1/q=1}$ then the ${\displaystyle \ell ^{q}}$ and ${\displaystyle \ell ^{q}}$ norms are dual to each other and the same is true of the ${\displaystyle L^{q}}$ and ${\displaystyle L^{q}}$ norms, where ${\displaystyle (X,\Sigma ,\mu ),}$ is some measure space. In particular the Euclidean 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 ${\displaystyle \|\,\cdot \,\|_{2}}$ -norm is even induced by a canonical inner product ${\displaystyle \langle \,\cdot ,\,\cdot \rangle ,}$ meaning that ${\displaystyle \|\mathbf {x} \|_{2}={\sqrt {\langle \mathbf {x} ,\mathbf {x} \rangle }}}$ for all vectors ${\displaystyle \mathbf {x} .}$ This inner product can expressed in terms of the norm by using the polarization identity. On ${\displaystyle \ell ^{2},}$ this is the Euclidean inner product defined by

$${\displaystyle \langle \left(x_{n}\right)_{n},\left(y_{n}\right)_{n}\rangle _{\ell ^{2}}~=~\sum _{n}x_{n}{\overline {y_{n}}}}$$

while for the space ${\displaystyle L^{2}(X,\mu )}$ associated with a measure space ${\displaystyle (X,\Sigma ,\mu ),}$ which consists of all square-integrable functions, this inner product is

$${\displaystyle \langle f,g\rangle _{L^{2}}=\int _{X}f(x){\overline {g(x)}}\,\mathrm {d} x.}$$

The norms of the continuous dual spaces of ${\displaystyle \ell ^{2}}$ and ${\displaystyle \ell ^{2}}$ satisfy 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 [edit]

More generally, let ${\displaystyle X}$ and ${\displaystyle Y}$ be topological vector spaces and let ${\displaystyle L(X,Y)}$ ^[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 given a canonical norm.

Theorem 1  —Let ${\displaystyle X}$ and ${\displaystyle Y}$ be normed spaces. Assigning to each continuous linear operator ${\displaystyle f\in L(X,Y)}$ the scalar

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

defines a norm ${\displaystyle \|\cdot \|~:~L(X,Y)\to \mathbb {R} }$ on ${\displaystyle L(X,Y)}$ that makes ${\displaystyle L(X,Y)}$ into a normed space. Moreover, if ${\displaystyle Y}$ is a Banach space then 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 $${\displaystyle {\begin{aligned}\|\left(f_{1}+f_{2}\right)x\|~&=~\|f_{1}x+f_{2}x\|\\&\leq ~\|f_{1}x\|+\|f_{2}x\|\\&\leq ~\left(\|f_{1}\|+\|f_{2}\|\right)\|x\|\\&\leq ~\|f_{1}\|+\|f_{2}\|\end{aligned}}}$$ for every ${\displaystyle x\in X}$ satisfying ${\displaystyle \|x\|\leq 1.}$ This fact together with the definition of ${\displaystyle \|\cdot \|~:~L(X,Y)\to \mathbb {R} }$ implies the triangle inequality: $${\displaystyle \|f+g\|\leq \|f\|+\|g\|.}$$ Since ${\displaystyle \{|f(x)|:x\in X,\|x\|\leq 1\}}$ is a non-empty set of non-negative real numbers, ${\displaystyle \|f\|=\sup \left\{|f(x)|:x\in X,\|x\|\leq 1\right\}}$ is a non-negative real number. If ${\displaystyle f\neq 0}$ then ${\displaystyle fx_{0}\neq 0}$ for some ${\displaystyle x_{0}\in X,}$ which implies that ${\displaystyle \left\|fx_{0}\right\|>0}$ and consequently ${\displaystyle \|f\|>0.}$ This shows that ${\displaystyle \left(L(X,Y),\|\cdot \|\right)}$ is a normed space.[8] Assume now that ${\displaystyle Y}$ is complete and we will show that ${\displaystyle (L(X,Y),\|\cdot \|)}$ is complete. Let ${\displaystyle f_{\bullet }=\left(f_{n}\right)_{n=1}^{\infty }}$ be a Cauchy sequence in ${\displaystyle L(X,Y),}$ so by definition ${\displaystyle \left\|f_{n}-f_{m}\right\|\to 0}$ as ${\displaystyle n,m\to \infty .}$ This fact together with the relation $${\displaystyle \left\|f_{n}x-f_{m}x\right\|=\left\|\left(f_{n}-f_{m}\right)x\right\|\leq \left\|f_{n}-f_{m}\right\|\|x\|}$$ implies that ${\displaystyle \left(f_{n}x\right)_{n=1}^{\infty }}$ is a Cauchy sequence in ${\displaystyle Y}$ for every ${\displaystyle x\in X.}$ It follows that for every ${\displaystyle x\in X,}$ the limit ${\displaystyle \lim _{n\to \infty }f_{n}x}$ exists in ${\displaystyle Y}$ and so we will denote this (necessarily unique) limit by ${\displaystyle fx,}$ that is: $${\displaystyle fx~=~\lim _{n\to \infty }f_{n}x.}$$ It can be shown that ${\displaystyle f:X\to Y}$ is linear. If ${\displaystyle \varepsilon >0}$ , then ${\displaystyle \left\|f_{n}-f_{m}\right\|\|x\|~\leq ~\varepsilon \|x\|}$ for all sufficiently large integers n and m. It follows that $${\displaystyle \left\|fx-f_{m}x\right\|~\leq ~\varepsilon \|x\|}$$ for sufficiently all large ${\displaystyle m.}$ Hence ${\displaystyle \|fx\|\leq \left(\left\|f_{m}\right\|+\varepsilon \right)\|x\|,}$ so that ${\displaystyle f\in L(X,Y)}$ and ${\displaystyle \left\|f-f_{m}\right\|\leq \varepsilon .}$ This shows that ${\displaystyle f_{m}\to f}$ in the norm topology 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  —Let ${\displaystyle X}$ be a normed space and for every ${\displaystyle x^{*}\in X^{*}}$ let

$${\displaystyle \left\|x^{*}\right\|~:=~\sup \left\{|\langle x,x^{*}\rangle |~:~x\in X{\text{ with }}\|x\|\leq 1\right\}}$$

where by definition ${\displaystyle \langle x,x^{*}\rangle ~:=~x^{*}(x)}$ is a scalar. Then

1. ${\displaystyle \|\,\cdot \,\|:X^{*}\to \mathbb {R} }$ is a norm that makes ${\displaystyle X^{*}}$ a Banach space.^[10]
2. If ${\displaystyle B^{*}}$ is the closed unit ball of ${\displaystyle X^{*}}$ then for every ${\displaystyle x\in X,}$

   $${\displaystyle {\begin{alignedat}{4}\|x\|~&=~\sup \left\{|\langle x,x^{*}\rangle |~:~x^{*}\in B^{*}\right\}\\&=~\sup \left\{\left|x^{*}(x)\right|~:~\left\|x^{*}\right\|\leq 1{\text{ with }}x^{*}\in X^{*}\right\}.\\\end{alignedat}}}$$

   

   Consequently, ${\displaystyle x^{*}\mapsto \langle x,x^{*}\rangle }$ is a bounded linear functional on ${\displaystyle X^{*}}$ with norm ${\displaystyle \|x^{*}\|~=~\|x\|.}$
3. ${\displaystyle B^{*}}$ is weak*-compact.

Proof

Let ${\displaystyle B~=~\sup\{x\in X~:~\|x\|\leq 1\}}$ denote the closed unit ball of a normed space ${\displaystyle X.}$ When ${\displaystyle Y}$ is the scalar field then ${\displaystyle L(X,Y)=X^{*}}$ so part (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]

See also [edit]

• Convex conjugate
• 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

Notes [edit]

1. ^ Rudin 1991, p. 87
2. ^ Rudin 1991, section 4.5, p. 95
3. ^ Rudin 1991, p. 95
4. ^ This inequality is tight, in the following sense: for any ${\displaystyle x}$ there is a ${\displaystyle z}$ for which the inequality holds with equality. (Similarly, for any ${\displaystyle z}$ there is an ${\displaystyle x}$ that gives equality.)
5. ^ Boyd & Vandenberghe 2004, p. 637
6. ^ Each ${\displaystyle L(X,Y)}$ is a vector space, with the usual definitions of addition and scalar multiplication of functions; this only depends on the vector space structure of ${\displaystyle Y}$ , not ${\displaystyle X}$ .
7. ^ Rudin 1991, p. 92
8. ^ Rudin 1991, p. 93
9. ^ Rudin 1991, p. 93
10. ^ Aliprantis 2006, p. 230 harvnb error: no target: CITEREFAliprantis2006 (help)
11. ^ Rudin 1991, Theorem 3.3 Corollary, p. 59
12. ^ Rudin 1991, Theorem 3.15 The Banach–Alaoglu theorem algorithm, p. 68
13. ^ Rudin 1991, p. 94

References [edit]

• Aliprantis, Charalambos D.; Border, Kim C. (2006). Infinite Dimensional Analysis: A Hitchhiker's Guide (3rd ed.). Springer. ISBN9783540326960.
• Boyd, Stephen; Vandenberghe, Lieven (2004). Convex Optimization. Cambridge University Press. ISBN9780521833783.
• 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.
• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN978-1584888666. OCLC 144216834.
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN978-0-07-054236-5. OCLC 21163277.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN978-1-4612-7155-0. OCLC 840278135.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN978-0-486-45352-1. OCLC 853623322.

External links [edit]

• Notes on the proximal mapping by Lieven Vandenberge

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Source: https://en.wikipedia.org/wiki/Dual_norm

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