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Let $X$ be a Banach space. By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.

I am interested in its converse. More precisely,

Question: Let $X$ be a Banach space. If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?

I feel that the statement above is negative. However, I could not produce a counterexample.

In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(\mathbb{R})$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.

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    $\begingroup$ When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point? $\endgroup$
    – Martin Sleziak
    Commented Dec 2, 2018 at 7:11
  • $\begingroup$ I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch. $\endgroup$
    – Martin Sleziak
    Commented Dec 2, 2018 at 7:13
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    $\begingroup$ This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces. $\endgroup$
    – Martin Sleziak
    Commented Dec 2, 2018 at 7:20
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    $\begingroup$ @TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $\beta\mathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $\alpha\mathbb{N}$. These are precisely the totally disconected $K$.) $\endgroup$
    – Dirk Werner
    Commented Dec 2, 2018 at 22:04
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    $\begingroup$ @Idonknow: The question is already answered extensively, but let me add one quick example. The identity of any unital $C^*$-algebra is an extreme point of its closed unit ball, but, of course, not all unital $C^*$-algebras are von Neumann algebras (=$C^*$-algebras with Banach space predual). $\endgroup$
    – Masayoshi Kaneda
    Commented Dec 14, 2018 at 10:15

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Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $\ell_2$ and use $|x| := \|x\|_X + \|Tx\|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.

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No. Let $X$ and $Y$ be Banach spaces, and set $Z=X\oplus Y$, with $\||(x,y)|\|:=\|x\|+\|y\|$. Assume that $x$ is a extreme point of $X$ with $\|x\|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if $$(x,0)=\frac12(a,y)+\frac12(b,z)$$ for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=\|x\|=\|a\|\leq\||(a,y)|\|\leq 1$, so $y=0$, and analogously, $z=0$.

So, $L^2(\mathbb R)\oplus L^1(\mathbb R)$, is not a dual space, but its unit ball has extreme points.

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    $\begingroup$ How can we prove that $L^2(\mathbb R)\oplus L^1(\mathbb R)$ is not a dual space? $\endgroup$
    – Idonknow
    Commented Dec 2, 2018 at 14:56
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    $\begingroup$ @Idonknow This is because, $X\oplus Y$ is dual iff both $X$ and $Y$ are dual. $\endgroup$
    – MSMalekan
    Commented Dec 2, 2018 at 16:03
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    $\begingroup$ @MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] \oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $\ell_1$-direct summand. My argument: If $L_2\oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't. $\endgroup$
    – Dirk Werner
    Commented Dec 2, 2018 at 22:03
  • $\begingroup$ @DirkWerner Thanks for the comment, you are right! $\endgroup$
    – MSMalekan
    Commented Dec 3, 2018 at 3:01

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