TITLE: Direct proof of injectivity of $L_\infty$
QUESTION [8 upvotes]: I would like to know a simple proof of isometric injectivity of $L_\infty$. The proof I've found in Topics in Banach space theory. F. Albiac, N. Kalton uses two deep result. 

$L_\infty$ as commutative unital $C^*$ algebra is isometrically isomorphic to $C(K)$ for some compact $K$.
Every $C(K)$ space which is a dual space is isometrically injective.

However the proof for $\ell_\infty$ is quite simple. Let $i:X\to Z$ be isometric embedding and $T:X\to \ell_\infty$ be a bounded operator. Let $e_n:\ell_\infty\to\mathbb{C}:x\mapsto x(n)$ be coordiante functionals, then consider bounded functionals $f_n:\mathrm{Im}(i)\to \mathbb{C}:z\mapsto e_n(T(i^{-1}(z)))$ extend them by Hahn-Banach theorem to get functionals $g_n:Z\to\mathbb{C}$. The desired operator is $
\hat{T}:Z\to\ell_\infty: z\mapsto(g_1(z), g_2(z),\ldots)$
My question:
Does there exist a direct proof that $L_\infty$ is isometrically injective, a proof similar to the arguments used for the $\ell_\infty$ space? The problem in mimicking proof for $\ell_\infty$ arose from the fact that I can't find family of functionals $(E_n:n\in\mathbb{N})\subset L_\infty^*$ similar to coordinate functionals $(e_n:n\in\mathbb{N})\subset\ell_\infty^*$. 
Thank you.

REPLY [8 votes]: I am not sure whether the following would meet your requirements but I vaguely remember
having heard it in a course many years ago and I think that it is sufficiently distinct
from the above proofs to justify a brief mention.  The crucial common property of the three spaces in question---the real line, the sequence space and the function space---is that they are 
Dedekind-complete Banach lattices for which the norm is intimately connected to the order
structure---the unit ball coincides with the interval $[-1,1]$.  This fact allows one to mimic directly the standard proof of the classical Hahn-Banach theorem in the two more advanced cases.