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Given a measurable space $(\Omega,\mu)$ such that $L_\infty(\mu)$ is isomorphic to a dual space, $L_\infty(\mu)$ is an injective Banach space. Indeed, given a subspace $Y$ of $X$ and a norm-one operator $T:Y\to L_\infty(\mu)$, considering a finite partition $P$ of $\Omega$ consisting of measurable subsets, by averaging we get a finite rank operator $T_P:Y\to L_\infty(\mu)$ with $\|T_P\|\leq 1$ that, by the Hahn-Banach theorem, can be extended to an operator $\hat T_P:X\to L_\infty(\mu)$ with $\|\hat T_P\|\leq 1$. So we can define a norm-one extensión $\hat T:X\to L_\infty(\mu)$ of $T$ by taking weak$^*$-limits.

There are examples of $L_\infty(\mu)$ spaces which are not injective. See the comments after Theorem 1.4 in Johnson, Kania and Schechtman (2015)

Suppose that $L_\infty(\mu)$ is non-isomorphic to a dual space. Is it posible for $L_\infty(\mu)$ to be injective?

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  • $\begingroup$ If $L_\infty(\mu)$ is isomorphic to a dual space, must $L_1(\mu)$ be a predual? $\endgroup$
    – Bill Johnson
    Jul 14, 2015 at 14:41
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    $\begingroup$ To Bill Johnson: There are trivial cases in which it is not. Say a finite set in which some of the points have finite measure and some others infinite measure. $\endgroup$
    – M.González
    Jul 14, 2015 at 18:34

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