Isometric Embedding of a separable Banach Space into $\ell^{\infty}$

Question

The problem is:

Let $X$ be a separable Banach space then there is an isometric embedding from $X$ to $\ell^{\infty}$.

My efforts:

I showed that there is an isometry from $X^*$ (topological dual) to $\ell^\infty$ in the following way:

Let $(e_{i})_{i=1}^{\infty}$ be a dense sequence in $B_{X}$ then define $\Phi:X^*\rightarrow\ell^\infty$ by $\Phi(f)=(f(e_{i}))_{i=1}^\infty$.

It is clear that $\Phi$ is an isometry.

An initial idea and a secondary question

Is there any canonical isometry from $X$ to $X^*$ since $X$ is separable (or not)?

Answer 1

Let $(x_n)_{n=0}^\infty$ be a dense sequence in $X$. For each $n$, find $x_n^*\in X^*$ so that $\|x^*_n\|=1$ and $x^*_n(x_n)=\|x_n\|$. It is easy to check that the operator $T: X\to \ell^\infty$ defined by $T(x)=(x_n^*(x))_{n=0}^\infty$ is the desired embedding.

Answer 2

This is almost the same as the other answers, with a bit of detail:

Let X be a separable Banach space. Let $(x_n)$ be a dense sequence in the unit sphere of $X$. For each $n$, use the Hahn-Banach Theorem to find a norm-one functional $f_n\in X^*$ with $f_n^*(x_n)=1$.

Define $\Phi: X\rightarrow\ell_\infty$ via $$ \Phi(x) =(f_n(x) ) $$

$\Phi$ is clearly linear.

Suppose $x\in X$ has norm one and let $1>\epsilon>0$. Choose $n_\epsilon$ so that $\Vert x_{n_\epsilon}-x\Vert<\epsilon$.

Then $$\epsilon>|f_{n_\epsilon}(x_{n_\epsilon}-x)|=|f_{n_\epsilon}(x_{n_\epsilon})-f_{n_\epsilon}(x)| = |1-f_{n_\epsilon}(x)|.$$

As $\epsilon$ was arbitrary, this implies that $\Vert \Phi(x)\Vert=\sup\limits_{n\in \Bbb N}|f_n(x ) |\ge 1$. Also, for any $n$, $$|f_n(x)|\le \Vert f_n \Vert \Vert x\Vert =1.$$ and so $\Vert \Phi(x)\Vert\le 1$.

Thus we have $\Vert \Phi(x)\Vert= 1$, whenever $\Vert x\Vert =1$.

From this it follows that for any non-zero element $x$ of $X$, we have
$$\Vert \Phi(x)\Vert= \Vert x\Vert \sup\limits_{n\in \Bbb N}|f_n(x/\Vert x\Vert) | =\Vert x\Vert;$$ and thus, $\Phi$ is an isometry.