Theorem: Separable Normed Vector Space Isometrically Isomorphic to Linear Subspace of Space of Bounded Sequences

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a separable normed vector space over $\Bbb F$.
Let $\struct {\map {\ell^\infty} {\Bbb F}, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.
Then there exists a linear subspace $Y$ of $\map {\ell^\infty} {\Bbb F}$ such that: 
:$X$ is isometrically isomorphic to $Y$.

Proof:
Let $\mathcal S = \set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$. 
By Existence of Support Functional, for each $n \in \N$ there exists $f_n \in X^\ast$ such that $\norm {f_n}_{X^\ast} = 1$ and $\map {f_n} {x_n} = \norm {x_n}$.
Then, for each $x \in X$, we have: 
:$\cmod {\map {f_n} x} \le \norm x$ 
from Supremum Operator Norm as Universal Upper Bound.
So, we have: 
:$\ds \sup_{n \in \N} \cmod {\map {f_n} x} \le \norm x$ 
So we can define $T : X \to \map {\ell^\infty} {\Bbb F}$ by: 
:$T x = \sequence {\map {f_n} x}_{n \mathop \in \N} = \tuple {\map {f_1} x, \map {f_2} x, \ldots}$
From Image of Vector Subspace under Linear Transformation is Vector Subspace, $T \sqbrk U$ is a vector subspace of $\map {\ell^\infty} {\Bbb F}$.
It remains to show that $T$ is a linear isometry. 
Let $\alpha, \beta \in \Bbb F$ and $x, y \in X$. 
We have:
{{begin-eqn}}
{{eqn	| l = \map T {\alpha x + \beta y}
	| r = \sequence {\map {f_n} {\alpha x + \beta y} }_{n \mathop \in \N}
}}
{{eqn	| r = \sequence {\alpha \map {f_n} x + \beta \map {f_n} y}_{n \mathop \in \N}
}}
{{eqn	| r = \sequence {\alpha \map {f_n} x}_{n \mathop \in \N} + \sequence {\beta \map {f_n} y}_{n \mathop \in \N}
}}
{{eqn	| r = \alpha \sequence {\map {f_n} x}_{n \mathop \in \N} + \beta \sequence {\map {f_n} y}_{n \mathop \in \N}
}}
{{eqn	| r = \alpha T x + \beta T y
}}
{{end-eqn}}
so $T$ is a linear transformation.
Also, for $x \in X$:
{{begin-eqn}}
{{eqn	| l = \norm {T x}_\infty
		| r = \sup_{n \mathop \in \N} \cmod {\map {f_n} x}
}}
{{eqn	| o = \le
		| r = \norm x
		| c = from the computation above
}}
{{end-eqn}}
So $T$ is a bounded linear transformation.
So by Continuity of Linear Transformations, $T$ is continuous. 
Let $x_k \in \mathcal S$.
Note that while: 
:$\ds \sup_{n \mathop \in \N} \cmod {\map {f_n} {x_k} } \le \norm {x_k}$
we have: 
:$\map {f_k} {x_k} = \norm {x_k}$
So we actually have: 
:$\ds \norm {T x_k}_\infty = \sup_{n \mathop \in \N} \cmod {\map {f_n} {x_k} } = \norm {x_k}$
for each $k \in \N$.
Now take general $x \in X$.
Then there exists a sequence $\sequence {n_k}_{k \mathop \in \N}$ such that: 
:$x_{n_k} \to x$ 
as $k \to \infty$.
Since $T$ is continuous , we have: 
:$T x_{n_k} \to T x$
as $k \to \infty$, by Sequential Continuity is Equivalent to Continuity in Metric Space.
So taking $k \to \infty$ in: 
:$\norm {T x_{n_k} }_\infty = \norm {x_{n_k} }$
we have: 
:$\norm {T x}_\infty = \norm x$
for each $x \in X$.
So $T : X \to Y$ is a linear isometry.
Hence $X$ and $Y$ are isometrically isomorphic.
{{qed}}
Category:Isometric Isomorphisms (Normed Vector Spaces)
Category:Separable Spaces
Category:Normed Vector Spaces
Category:Space of Bounded Sequences
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