The transpose of an injective map need not have norm-dense range

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 ${}^{\perp }\mathrm{\varnothing }=X$.