# If the closed unit ball of Banach space has at least one extreme point, must the Banach space the be a dual space?

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.

• 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? – Martin Sleziak Dec 2 '18 at 7:11
• 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. – Martin Sleziak Dec 2 '18 at 7:13
• This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces. – Martin Sleziak Dec 2 '18 at 7:20
• @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 $K$.) – Dirk Werner Dec 2 '18 at 22:04
• @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). – Masayoshi Kaneda Dec 14 '18 at 10:15

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.
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.
• How can we prove that $L^2(\mathbb R)\oplus L^1(\mathbb R)$ is not a dual space? – Idonknow Dec 2 '18 at 14:56
• @Idonknow This is because, $X\oplus Y$ is dual iff both $X$ and $Y$ are dual. – Meisam Soleimani Malekan Dec 2 '18 at 16:03
• @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. – Dirk Werner Dec 2 '18 at 22:03