Is the $L^1$-space dual to a Banach space

Let $(\Omega,\mu)$ be a measure space. It is well known that for $1<p\leq \infty$ one has the duality $$L^p=(L^{p*})^*,$$ where $1/p+1/p^*=1$.

Question. Is it known that the Banach space $L^1$ is not isomorphic to the dual space of any Banach space?

To avoid trivial cases let us assume that $L^1$ is infinite dimensional. I am particularly interested in the case when the total measure of $\Omega$ is finite.

• I just consulted with Ken Davidson, who is a grand master of functional analysis, and his response was: "L^1 is not a dual of any space because its unit ball has no extreme points (see Krein-Milman Theorem)". You might also find this discussion helpful. math.stackexchange.com/questions/210043/… – Anton Jul 5 '16 at 14:14
• See the comment discussion on math.stackexchange.com/a/137683/1543 ; it does depend a bit on which $(\Omega,\mu)$ you are talking about. – Willie Wong Jul 5 '16 at 14:17
• @Anton: Thanks. That proves that $L^1$ is not isometrically isomorphic to a dual space. My question was whether it is isomorphic, but not necessarily isometrically. – makt Jul 5 '16 at 14:18
• @WillieWong: Many thanks. Your link contains the answer to my question. – makt Jul 5 '16 at 14:21
• The Krein-Milman property holds in a separable dual space. It says: any closed bounded convex set is the closed convex hull of its extreme points. So Anton's answer solves this. But in addition, the answer at Willie's reference also solves this. If Willie's reference were in MO, this question would be closed as a duplicate. – Gerald Edgar Jul 5 '16 at 14:22

OP's question was about being isomorphic to a dual space so we need to observe that $L_1$ lacks the Radon–Nikodym property, which is invariant under isomorphisms, and separable dual spaces have this property.

Also, the $\ell_1$-sum of continuum many copies of $L_1[0,1]$ is isometric to $C[0,1]^*$.

Edit of 31.07.2016: It has been pointed out that my answer is incomplete as I do not treat the case when $L_1(\mu)$ is non-separable for a $\sigma$-finite measure $\mu$. By the Radon–Nikodym theorem, we may assume without loss of generality that $\mu$ is actually finite.

The argument is then almost exactly the same as in this case the inclusion $L_2(\mu)\subset L_1(\mu)$ has dense range so $L_1(\mu)$ is weakly compactly generated. It is well known that weakly compactly generated dual spaces have the Radon–Nikodym property, a property that $L_1(\mu)$ is clearly lacking (by Maharam's theorem, $L_1(\mu)$ is isometric to $L_1(\{0,1\}^\lambda)$, where $\{0,1\}^\lambda$ is considered with the product fair-coin-toss (Haar) measure and $\lambda$ is the density character of $L_1(\mu)$).

• A somewhat easier proof: Pelczynski proved via an elementary argument that if $L_1$ is isomorphic to a subspace of $X^*$ then $\ell_1$ is isomorphic to a subspace of $X$ and hence $L_1$ is not isomorphically embeddable into a separable dual space. – Bill Johnson Jul 5 '16 at 18:39

I would like to mention that for general measure spaces the answer is negative: The space $L_1(0,1)^{**}$ (by, e. g., Proposition II.5.3 in Lindenstrauss--Tzafriri (1973)) is a dual $L_1$-space.

Actually for purely atomic (even finite) measures the negative answer is obvious: consider a probability measure on a countable set.

• Your first remark is actually a part of my answer. – Tomasz Kania Jul 29 '16 at 8:58
• @TomekKania Sorry I did not notice that (as it is stated differently). I should mention that I do not understand why your answer is considered as complete: $L_1$ on finite measure spaces does not have to be separable. – August Cleaner Jul 29 '16 at 15:23
• I have added the $\sigma$-finite case too. – Tomasz Kania Jul 31 '16 at 20:02