Theorem: Normed Dual Space of Space of Zero-Limit Sequences is Isometrically Isomorphic to 1-Sequence Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,} }$ be the space of zero-limit sequences.
Let $\struct {\map {c_0^\ast} \GF, \norm {\, \cdot \,}_{c_0^\ast} }$ be the normed dual space of $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
Let $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$ be the $1$-sequence space.
For each $x \in \map {\ell_1} \GF$, define $f_x : \map {c_0} \GF \to \GF$ by:
:$\ds \map {f_x} y = \sum_{i \mathop = 1}^n x_i y_i$
Then $f_x \in \map {c_0^\ast} \GF$ for each $x \in \map {\ell_1} \GF$.
Further the mapping $T : \map {\ell_1} \GF \to \map {c_0^\ast} \GF$ defined by:
:$\map T x = f_x$ for each $x \in \map {\ell_1} \GF$
is an isometric isomorphism.
Hence, $\struct {\map {c_0^\ast} \GF, \norm {\, \cdot \,}_{c_0^\ast} }$ is isometrically isomorphic to $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$.

Proof:
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis for $\map {c_0} \GF$ given by Space of Zero-Limit Sequences admits Schauder Basis.
We first show that $T$ is a linear isometry.
We firstly argue that $f_x$ is well-defined for each $x \in \map {\ell_1} \GF$.
Let $y \in \map {c_0} \GF$.
We have:
:$\ds \sum_{i \mathop = 1}^\infty \cmod {x_i y_i} \le \norm y_\infty \sum_{i \mathop = 1}^\infty \cmod {x_i} < \infty$
Hence by Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach, we have that:
:$\ds \sum_{i \mathop = 1}^\infty x_i y_i$ is convergent.
Hence $f_x$ is well-defined.
We show that $f_x$ is linear.
Let $y, z \in \map {c_0} \GF$ and $\lambda \in \GF$.
We have:
{{begin-eqn}}
{{eqn	| l = \map {f_x} {y + \lambda z}
		| r = \sum_{i \mathop = 1}^\infty x_i \paren {y_i + \lambda z_i}
}}
{{eqn	| r = \sum_{i \mathop = 1}^\infty x_i y_i + \lambda \sum_{i \mathop = 1}^\infty x_i z_i
		| c = Linear Combination of Convergent Series
}}
{{eqn	| r = \map {f_x} y + \lambda \map {f_x} z
}}
{{end-eqn}}
Hence $f_x$ is linear functional.
We now show that $f_x$ is a bounded linear functional.
Since we have:
:$\ds \cmod {\map {f_x} y} \le \sum_{i \mathop = 1}^\infty \cmod {x_i y_i} \le \norm y_\infty \norm x_1$
we have that $f_x$ is a bounded linear functional satisfying $\norm {f_x}_{c_0^\ast} \le \norm x_1$.
We now show that $T$ is linear.
Let $x, y \in \map {\ell_1} \GF$, $z \in \map {c_0} \GF$ and $\lambda \in \GF$.
We then have:
{{begin-eqn}}
{{eqn	| l = \map {f_{x + \lambda y} } z
		| r = \sum_{i \mathop = 1}^\infty \paren {x_i + \lambda y_i} z_i
}}
{{eqn	| r = \sum_{i \mathop = 1}^\infty x_i z_i + \lambda \sum_{i \mathop = 1}^\infty y_i z_i 
		| c = Linear Combination of Convergent Series
}}
{{eqn	| r = \map {f_x} z + \lambda \map {f_y} z 
}}
{{end-eqn}}
that is:
:$f_{x + \lambda y} = f_x + \lambda f_y$
So that:
:$\map T {x + \lambda y} = \map {T_1} x + \lambda \map {T_1} y$
Hence $T$ is linear transformation.
We now show that if $f \in \map {c_0^\ast} \GF$, then there exists $x \in \map {\ell_1} \GF$ such that $f = f_x$.
That is, $T$ is surjective.
In doing so we will also prove that $T$ is a linear isometry.
We will then be done.
Define $\phi : \N \to \GF$ by:
:$\ds \map \phi n = \begin{cases}\frac {\cmod {\map f {e_n} } } {\map f {e_n} } & \map f {e_n} \ne 0 \\ 0 & \map f {e_n} = 0\end{cases}$
Set:
:$\ds \phi_N = \sum_{n \mathop = 1}^N \map \phi n e_n$
Note that:
:$\ds \norm {\phi_N}_\infty = \sup_{n \mathop \le N} \cmod {\frac {\cmod {\map f {e_n} } } {\map f {e_n} } } = 1$
Hence we have, from Fundamental Property of Norm on Bounded Linear Functional, $\cmod {\map f {\phi_N} } \le \norm f_{c_0^\ast}$.
Meanwhile, we have:
{{begin-eqn}}
{{eqn	| l = \map f {\phi_N}
		| r = \sum_{n \mathop = 1}^N \map \phi n \map f {e_n}
		| c = {{Defof|Linear Functional}}
}}
{{eqn	| r = \sum_{1 \le n \le N, \, \map f {e_n} \ne 0} \frac {\cmod {\map f {e_n} } } {\map f {e_n} } \map f {e_n}
		| c = definition of $\phi$
}}
{{eqn	| r = \sum_{1 \le n \le N, \, \map f {e_n} \ne 0} \cmod {\map f {e_n} }
}}
{{eqn	| r = \sum_{n \mathop = 1}^N \cmod {\map f {e_n} }
}}
{{end-eqn}}
Hence we have:
:$\ds \sum_{n \mathop = 1}^N \cmod {\map f {e_n} } \le \norm f_{c_0^\ast}$
for each $N \in \N$.
From Monotone Convergence Theorem (Real Analysis), we have:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map f {e_n} } \le \norm f_{c_0^\ast} < \infty$
We can therefore define $x \in \map {\ell_1} \GF$ by:
:$\map x n = \map f {e_n}$ for each $n \in \N$.
We note that we then have $\norm x_1 \le \norm f_{c_0^\ast} < \infty$. 
Hence $\norm x_1 = \norm f_{c_0^\ast}$.
Hence $T$ is a linear isometry.
It remains to show that $f = f_x$. 
We have:
:$\ds \map {f_x} y = \sum_{i \mathop = 1}^\infty \map f {e_i} y_i$
We can then write:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^\infty \map f {e_i} y_i
		| r = \lim_{N \mathop \to \infty} \sum_{i \mathop = 1}^N \map f {e_i} y_i
}}
{{eqn	| r = \lim_{N \mathop \to \infty} \map f {\sum_{i \mathop = 1}^N e_i y_i} 
}}
{{eqn	| r = \map f {\sum_{i \mathop = 1}^\infty e_i y_i}
		| c = $f$ is bounded, Continuous Mappings preserve Convergent Sequences
}}
{{eqn	| r = \map f y
		| c = Space of Zero-Limit Sequences admits Schauder Basis
}}
{{end-eqn}}
That is, $\map f y = \map {f_x} y$ for each $y \in \map {c_0} \GF$.
Hence $f = f_x$.
Hence $T$ is surjective.
Hence by Linear Isometry is Injective: Corollary, $T$ is isometric isomorphism.
{{qed}}
Category:Space of Zero-Limit Sequences
Category:P-Sequence Spaces
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