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)?
Best Answer
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.