Dual Spaces Adjoint Operators and Annihilators — Examples

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The continuous dual of c0 is ell-one, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. With coordinates starting at zero, the map ${\mathrm{\ell }}^1(\mathbb{K})⟶c_0(\mathbb{K}{)}^{\ast },a⟼f_a,f_a(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$ is a linear isometric bijection. The pairing is bilinear, including over $\mathbb{C}$.

From The complex continuous dual of ell-one is ell-infinity, with its stated hypotheses: For complex sequence spaces, with indices starting at zero, ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{C})⟶{\mathrm{\ell }}^1(\mathbb{C}{)}^{\ast },b⟼h_b,h_b(a)={\sum }_{n=0}^{\mathrm{\infty }}{b}_n{a}_n$ is a complex-linear isometric bijection. There is no conjugation in this pairing.

From Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The continuous dual of c0 is ell-one, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. With coordinates starting at zero, the map ${\mathrm{\ell }}^1(\mathbb{K})⟶c_0(\mathbb{K}{)}^{\ast },a⟼f_a,f_a(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$ is a linear isometric bijection. The pairing is bilinear, including over $\mathbb{C}$.

From The complex continuous dual of ell-one is ell-infinity, with its stated hypotheses: For complex sequence spaces, with indices starting at zero, ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{C})⟶{\mathrm{\ell }}^1(\mathbb{C}{)}^{\ast },b⟼h_b,h_b(a)={\sum }_{n=0}^{\mathrm{\infty }}{b}_n{a}_n$ is a complex-linear isometric bijection. There is no conjugation in this pairing.

From Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The bounded complex dual of C_0(X) is regular complex measures, with its stated hypotheses: For an LCH space $X$, every bounded complex linear functional $L$ on $C_0(X;\mathbb{C})$ has a unique representation $L(f)={\int }_{X}fd\mu$ by a finite regular complex Borel measure $\mu$. Conversely each such $\mu$ defines a bounded functional and $\mathrm{\parallel }L\mathrm{\parallel }=\mathrm{\mid }\mu \mathrm{\mid }(X)$.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$ and arbitrary subsets $M\subseteq X$, $N\subseteq X^{\ast }$, define $M^{\perp }=\{f\in X^{\ast }:f(m)=0\text{ for all }m\in M\},{}^{\perp }N=\{x\in X:f(x)=0\text{ for all }f\in N\}\mathrm{.}$ Here $X^{\ast }$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $\mathrm{span}M$, since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$, not in $X^{\ast \ast }$. Empty sets impose no conditions: ${\mathrm{\varnothing }}^{\perp }=X^{\ast }$ and ${}^{\perp }\mathrm{\varnothing }=X$.

From The continuous dual of c0 is ell-one, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. With coordinates starting at zero, the map ${\mathrm{\ell }}^1(\mathbb{K})⟶c_0(\mathbb{K}{)}^{\ast },a⟼f_a,f_a(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$ is a linear isometric bijection. The pairing is bilinear, including over $\mathbb{C}$.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The dual family of a finite basis is a basis of the dual space, with the same dimension, with its stated hypotheses: If $B=(b_1,\dots ,b_n)$ is a basis of a finite-dimensional $F$-vector space $V$, then its dual family $B^{\ast }=(b_1^{\ast },\dots ,b_n^{\ast })$ is a basis of $V^{\ast }$. Consequently $\dim V^{\ast }=\dim V=n$.

From A linear map from a finite-dimensional normed space is bounded, with its stated hypotheses: Let $X$ and $Y$ be normed spaces over the same scalar field, and assume $X$ admits an ordered basis of finite length. Then every linear map $S:X\to Y$ is a bounded linear operator in the sense of def-bounded-linear-operator.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From Transposition reverses composition, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For bounded linear $T:X\to Y$, $S:Y\to Z$ between normed spaces, $(ST{)}^{\ast }=T^{\ast }{S}^{\ast },I_X^{\ast }=I_{X^{\ast }}\mathrm{.}$ For bounded $T,U:X\to Y$ and $a,b\in \mathbb{K}$, $(aT+bU{)}^{\ast }=a{T}^{\ast }+b{U}^{\ast }$.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The continuous dual of c0 is ell-one, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. With coordinates starting at zero, the map ${\mathrm{\ell }}^1(\mathbb{K})⟶c_0(\mathbb{K}{)}^{\ast },a⟼f_a,f_a(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$ is a linear isometric bijection. The pairing is bilinear, including over $\mathbb{C}$.

From The complex continuous dual of ell-one is ell-infinity, with its stated hypotheses: For complex sequence spaces, with indices starting at zero, ${\mathrm{\ell }}^{\mathrm{\infty }}(\mathbb{C})⟶{\mathrm{\ell }}^1(\mathbb{C}{)}^{\ast },b⟼h_b,h_b(a)={\sum }_{n=0}^{\mathrm{\infty }}{b}_n{a}_n$ is a complex-linear isometric bijection. There is no conjugation in this pairing.

From Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

From The canonical evaluation map into the bidual, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$, define $J_X:X\to X^{\ast \ast },(J_{X}x)(f)=f(x)(f\in X^{\ast })\mathrm{.}$ With the dual norm from def-dual-space-of-a-normed-space, evaluation is linear in $f$ and $\mathrm{\mid }(J_{X}x)(f)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$, so $J_{X}x$ is a bounded functional on $X^{\ast }$. The map is canonical and uses no chosen basis or conjugation.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The continuous dual of c0 is ell-one, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. With coordinates starting at zero, the map ${\mathrm{\ell }}^1(\mathbb{K})⟶c_0(\mathbb{K}{)}^{\ast },a⟼f_a,f_a(x)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{x}_n$ is a linear isometric bijection. The pairing is bilinear, including over $\mathbb{C}$.

From Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

From Finite truncations approximate null and summable sequences, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Use coordinates indexed by $\mathbb{N}=\{0,1,\dots \}$. Define ${\mathrm{\ell }}^1(\mathbb{K})=\{a:{\sum }_{n=0}^{\mathrm{\infty }}\mathrm{\mid }{a}_n\mathrm{\mid }<\mathrm{\infty }\}$, with coordinatewise operations and norm $\mathrm{\parallel }a{\mathrm{\parallel }}_1={\sum }_n\mathrm{\mid }{a}_n\mathrm{\mid }$. Let $P_N$ retain coordinates $0,\dots ,N$ and set all others to zero. Then $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0(x\in c_0(\mathbb{K})),\mathrm{\parallel }a-P_{N}a{\mathrm{\parallel }}_1⟶0(a\in {\mathrm{\ell }}^1(\mathbb{K}))\mathrm{.}$

From c_0 is a closed subspace of ell-infinity, with its stated hypotheses: $c_0$ is a closed linear subspace of ${\mathrm{\ell }}^{\mathrm{\infty }}$ in the sup norm.

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$ and arbitrary subsets $M\subseteq X$, $N\subseteq X^{\ast }$, define $M^{\perp }=\{f\in X^{\ast }:f(m)=0\text{ for all }m\in M\},{}^{\perp }N=\{x\in X:f(x)=0\text{ for all }f\in N\}\mathrm{.}$ Here $X^{\ast }$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $\mathrm{span}M$, since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$, not in $X^{\ast \ast }$. Empty sets impose no conditions: ${\mathrm{\varnothing }}^{\perp }=X^{\ast }$ and ${}^{\perp }\mathrm{\varnothing }=X$.

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From Counting measure specializes the representation theorem to ${\mathrm{\ell }}^p$ and ${\mathrm{\ell }}^q$, with its stated hypotheses: Let $1\le p<\mathrm{\infty }$ and let $q$ be conjugate to $p$. Every bounded linear functional $\mathrm{\Lambda }:{\mathrm{\ell }}^p\to \mathbb{R}$ is of the form $\mathrm{\Lambda }(a)={\sum }_{n=0}^{\mathrm{\infty }}{a}_n{b}_n$ for a unique sequence $b\in {\mathrm{\ell }}^q$. Moreover, $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }b{\mathrm{\parallel }}_{{\mathrm{\ell }}^q}\mathrm{.}$

From ${\mathrm{\ell }}^p$ is the $L^p$ space of counting measure, with its stated hypotheses: On $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mathrm{\#})$ with counting measure, every function $f:\mathbb{N}\to \mathbb{R}$ is measurable. Writing $a_k:=f(k)$, one has $\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mathrm{\#}={\sum }_{k=0}^{\mathrm{\infty }}\mathrm{\mid }{a}_k{\mathrm{\mid }}^p(0<p<\mathrm{\infty }),$ by the counting-measure integral dictionary, so ${\mathcal{L}}^p(\mathrm{\#})$ is exactly the usual sequence class ${\mathrm{\ell }}^p$. Also $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{k\in \mathbb{N}}\mathrm{\mid }{a}_k\mathrm{\mid },$ because a subset of $\mathbb{N}$ has counting measure zero only when it is empty. Hence the quotient by almost-everywhere equality does nothing: for counting measure on $\mathbb{N}$, equality almost everywhere means equality everywhere.