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 $T:X\to Y$ be a bounded operator. Let $e_n:\ell_\infty\to\mathbb{C}:x\mapsto x(n)$ be coordiante functionals, then consider bounded functionals $f_n:\mathrm{Im}(i)\to \ell_\infty:z\mapsto e_n(T(i^{-1}(z)))$ 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.