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.


2 Answers 2


Write $L_\infty$ as the closure of a net (directed by inclusion) of finite dimensional $\ell_\infty$ spaces. Compose the operator into $L_\infty$ with norm one projections onto these subspaces and extend. Use weak$^*$ compactness of the unit ball of $L_\infty$ to pass to a limit of a subnet of these operators. $$ $$ Basically the same argument works for any dual space that is $\mathcal{L}_{\infty,\lambda}$ for all $\lambda > 1$.

  • $\begingroup$ Wow, really nice! $\endgroup$
    – Nik Weaver
    Oct 23, 2012 at 21:25
  • $\begingroup$ What is the easiest way to prove the existence of such a net? Do we order by inclusion all the finite subfields of the domain of $\mu$ ($L_\infty = L_\infty(\mu)$) and consider subspaces of functions from L∞ which are measurable with respect to these fields? $\endgroup$ Nov 1, 2012 at 14:18
  • $\begingroup$ @SlavojZizek For more details see Banach space theory. The basis for linear and non-linear analysis. M. Fabian, P. Habala, P. Hajek, V. Montesinos, V. Zizler page 287 exercise 5.91 $\endgroup$
    – Norbert
    Nov 5, 2012 at 19:42

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.

  • $\begingroup$ I like this proof too. $\endgroup$
    – Nik Weaver
    Feb 12, 2013 at 17:51

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