The continuous dual of c0 is ell-one

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 dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ in the literal definition and $\mathbb{K}=\mathbb{C}$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{\ast }:=\mathcal{B}(X,\mathbb{K}),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f\in X^{\ast }$ is in particular a linear functional in the algebraic sense, so $X^{\ast }$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{\ast }$ is the operator norm: $\mathrm{\parallel }f{\mathrm{\parallel }}_{X^{\ast }}:=\mathrm{\parallel }f\mathrm{\parallel }=\sup \{\mathrm{\mid }f(x)\mathrm{\mid }:\mathrm{\parallel }x\mathrm{\parallel }\le 1\}\mathrm{.}$

From The sequence spaces c_0 and ell-infinity, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with absolute value in the real case and modulus $\mathrm{\mid }a+ib\mathrm{\mid }=\sqrt{a^2+b^2}$ in the complex case. A scalar sequence here is a function $x:\mathbb{N}\to \mathbb{K}$, including index zero. Let ${\mathrm{\ell }}^{\mathrm{\infty }}=\{x=(x_n{)}_{n\in \mathbb{N}}:{\sup }_{n\in \mathbb{N}}\mathrm{\mid }{x}_n\mathrm{\mid }<\mathrm{\infty }\},c_0=\{x\in {\mathrm{\ell }}^{\mathrm{\infty }}:x_n\to 0\},$ both equipped with $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n\in \mathbb{N}}\mathrm{\mid }{x}_n\mathrm{\mid }$. Thus $c_0$ is a specified linear subspace of the bounded-sequence space ${\mathrm{\ell }}^{\mathrm{\infty }}$. Here $x_n\to 0$ means that for every real $\epsilon >0$ there is $N\in \mathbb{N}$ such that $\mathrm{\mid }{x}_n\mathrm{\mid }<\epsilon$ for all $n\ge N$. Addition and scalar multiplication are coordinatewise. The scalar triangle inequality makes bounded sequences and null sequences linear spaces and gives the triangle inequality for the displayed supremum norm. Absolute homogeneity follows coordinatewise, and a zero supremum forces every coordinate to vanish.

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 $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0(x\in c_0(\mathbb{K})),\mathrm{\parallel }a-P_{N}a{\mathrm{\parallel }}_1⟶0(a\in {\mathrm{\ell }}^1(\mathbb{K}))\mathrm{.}$