Isometric embedding of $l_\infty$ to $L_\infty[0;1]$

Question

I'm trying to find an isometric embedding of $l_\infty$ to $L_\infty[0;1]$, i.e such bounded operator $\textsf{T}: l_\infty \to H \subset L_\infty[0;1]$, that $||\textsf{T}(x)||_{L_\infty[0;1]}=||x||_{l_\infty} \forall x \in l_\infty$

I know that isometric operator between $l_\infty$ and entire $L_\infty[0;1]$ doesn't exist, but I want to find an isometric embedding. Could you give me a hint how to build such mapping?

Answer 1

Lets define mapping $\psi : l_\infty \to L_\infty[0;1]$

Let $\xi \in l_\infty, \xi=\{\xi_n\}_{n=1}^\infty$

Then $\xi\overset{\psi}{\mapsto}f_\xi$ where $f_\xi(x)=\xi_n$ on $[S_n; S_{n+1}]$, where $S_n =\begin{cases} 0, n=1 \\ \sum_{k=2}^n1/2^{k-1}, n >1 \end{cases}$

It's easy to understand that $\psi$ is isometry onto it's image.