Let $X$ be a Banach space and let $p\in (1,\infty)$. If $q$ denotes the conjugate exponent to $p$, then $L_q(X^*)$ is easily seen to be isometric to a subspace of $(L_p(X))^*$ via the map $$f\mapsto \int\limits_{[0,1]}\langle \cdot(\omega), f(\omega)\rangle\,{\rm d}\omega\quad (f\in L_q(X^*)).$$ One of the numerous characterisations of the Radon–Nikodym property asserts that $X^*$ has this property with respect to $[0,1]$ endowed with the Lebesgue measure if and only if this map is surjective.

Suppose that $X^*$ fails the Radon–Nikodym property with respect to the Lebesgue measure. Is the range of the above-mentioned map complemented?

My motivation follows from this question of M. González; if the answer were positive, then we would have $\ell_2(X)\not\cong L_2(X)$ for every infinite-dimensional space $X$ with $X^*\cong L_1(\mu)$ and for every infinite-dimensional C*-algebra.

Edit (26.01.2016): This is indeed a known result. It follows from Theorem 9 in:

Z. Hu and B.-L. Lin, Extremal structure of the unit ball of $L_p(\mu, X)^*$, J. Math. Anal. Appl. 200 (1996), 567–590.

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    $\begingroup$ A small correction: $L_p(X)^*=L_q(X^*)$ characterizes the spaces $X$ such that $X^*$ has RNP. So in your question, the hypothesis should be that $X^*$ fails the RNP. $\endgroup$ Dec 16, 2015 at 16:57
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    $\begingroup$ How about doing an example. Taking $({l^1})^* = l^\infty$, we ask: Is $L_q(l^\infty)$ complemented in $(L_p(l^1))^*$? $\endgroup$ Dec 16, 2015 at 17:29

1 Answer 1


Still no answers? Perhaps we should try the following.

Let $g : [0,1] \to X^*$ be weak* measurable, in the sense that, for every $x \in X$, the function $$ \omega \mapsto \langle x, g(\omega)\rangle $$ is measurable. Then there exist weak* measurable functions $g_1, g_2 : [0,1] \to X^*$ such that

(1) $g$ is weak* equivalent to $g_1+g_2$, in the sense that, for every $x \in X$, $$ \langle x,g(\omega)\rangle = \langle x,g_1(\omega)+g_2(\omega)\rangle $$ for almost all $\omega \in [0,1]$. (But the exceptional nullset depends on $x$.)

(2) $g_1$ is Bochner measurable in the sense that the range of $g_1$ is separable.

(3) $g_2$ is totally non-separable in the sense that: for every separable $Y \subseteq X^*$, the outer measure of $g_2^{-1}(Y)$ is zero.

If this works, then perhaps the elements of $(L_p(X))^*$ can be identified as such weak* measurable functions $g$, and the projection is the map sending $g$ to its "Bochner part" $g_1$.

  • $\begingroup$ Gerlad's answer indeed leads to a solution so let me accept it. Full reference is now in the question. $\endgroup$ Jan 26, 2016 at 16:59

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