Separable Normed Vector Space Isometrically Isomorphic to Linear Subspace of Space of Bounded Sequence

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Theorem

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:

\(\ds \map T {\alpha x + \beta y}\) \(=\) \(\ds \sequence {\map {f_n} {\alpha x + \beta y} }_{n \mathop \in \N}\)
\(\ds \) \(=\) \(\ds \sequence {\alpha \map {f_n} x + \beta \map {f_n} y}_{n \mathop \in \N}\)
\(\ds \) \(=\) \(\ds \sequence {\alpha \map {f_n} x}_{n \mathop \in \N} + \sequence {\beta \map {f_n} y}_{n \mathop \in \N}\)
\(\ds \) \(=\) \(\ds \alpha \sequence {\map {f_n} x}_{n \mathop \in \N} + \beta \sequence {\map {f_n} y}_{n \mathop \in \N}\)
\(\ds \) \(=\) \(\ds \alpha T x + \beta T y\)

so $T$ is a linear transformation.

Also, for $x \in X$:

\(\ds \norm {T x}_\infty\) \(=\) \(\ds \sup_{n \mathop \in \N} \cmod {\map {f_n} x}\)
\(\ds \) \(\le\) \(\ds \norm x\) from the computation above

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.

$\blacksquare$

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