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I am looking for a simple proof of the following theorem — wasn't able to come up with one myself. Should be a use of the Bishop–Phelps theorem, in some way:

Let $X$ be a Banach space, $D \subset X^*$ a $w^*$-closed, convex set with nonempty interior. Then the $w^*$-support points of $D$ are dense in $\partial D$.

Note: The set of $w^*$-support points of $D$ is defined as: $\{f \in D : \text{$\exists x_0 \in X$, $x_0\neq0$, s.t $f(x_0)=\sup_{g\in D} g(x_0) $}\}$. Obviously, this set is contained in $\partial D$.

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    $\begingroup$ if $D$ was also symmetric around $0$, it would generate a norm on $X$, which is equivalent to the usual norm, and then you could use Bishop-Phelps theorem $\endgroup$
    – erz
    Commented Apr 24, 2021 at 18:26
  • $\begingroup$ Why do we even need $D$ to be symmetric? I think its enough to know that 0 is in the interior of $D$.. (It wont induce a norm, but we may use Bishop-Phelps on the pre-polar set) $\endgroup$
    – Tomer
    Commented Apr 24, 2021 at 20:14
  • $\begingroup$ yeah, I think you're right, perhaps you could answer your own question with some details? $\endgroup$
    – erz
    Commented Apr 25, 2021 at 2:13
  • $\begingroup$ We apply Bishop-Phelps to the pre-polar set, $C$, and we get that $A(C)$ (the functionals which attain sup on $C$) is dense in $X^*$ ; In particular, $A(C)$ is dense in $D$. Let's fix some $f \in \partial D$, and choose a sequence $g_n$ from $D \cap A(C)$ s.t $g_n \to f$. we may 'normalize' each $g_n$ s.t it would be in $\partial D$ (i.e, multiply by some appropriate scalar) s.t the new normalized sequence would converge to $f$ as well, but from $\partial D$, and it is still a sequence of functionals which attain their norm on $C$. $\endgroup$
    – Tomer
    Commented Apr 25, 2021 at 7:42
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    $\begingroup$ The missing detail i referred to is why attaining the maximum on $C$ is the same as being a w* support? I suggest you write it up as an answer and accept it so that the question was considered answered $\endgroup$
    – erz
    Commented Apr 25, 2021 at 8:39

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Here is my proof. Let $C$ be the pre-polar of $D$:

$$C = \{x: \sup\nolimits_{f \in D} \lvert f(x)\rvert \leq 1\}.$$

By applying Bishop–Phelps to $C$, we get that $A(C):=\{f \in X^*:\text{$\exists x_0 \in C$ s.t $f(x_0)=\sup_{x \in C} f(x) $}\}$ is dense in $D$. Fix some $f_0 \in \partial D$. We may choose a sequence $(g_n) \subset D \cap A(C) $ s.t $g_n \to f_0$. We may normalize $g_n$ (i.e, multiply by appropriate scalars, while keeping the same name) s.t $g_n \to f_0$ and now $g_n \in \partial D \cap A(C) $ (as $A(C)$ is invariant under positive scalar multiplication).

Now, i claim that if $g \in A(C) \cap \partial D$, then $g$ is a $w^*$-support point for $D$. Indeed, by Hahn–Banach, there exists some $\phi_0 \in X^{**}$ s.t $ \phi_0(g)=\max \phi_0(D) = 1$. Let $C^{**} := \{ \phi \in X^{**}:\sup_{f \in D} \lvert\phi(f)\rvert \leq1 \}$. So, we have $\phi_0 \in C^{**}$, $C = C^{**} \cap X $, and by Goldstein, $C$ is $w^*-$dense in $C^{**}$. Therefore, we may choose $(x_n) \subset C$ s.t $x_n \to \phi_0$ ($w^*$ convergence). Thus, $ x_n(g) = g(x_n) \to \phi_0(g) = 1$. But, since $g \in A(C)$, $\exists x_0 \in C$ s.t $g(x_0)=\sup_{x \in C} g(x) = 1$. But by the definition of $C$, $\lvert x_0(D)\rvert \leq 1$. Hence $g(x_0)=1=\sup_{f \in D}f(x_0)$, i.e, $g$ is a $w^*$-support point for $D$.

We fixed $ f_0 \in \partial D$ and found a sequence $(g_n)$ of $w^*$-support points for $D$ which convegre to $f_0$. Hence the set of $w^*$-support points for $D$ is dense in $\partial D$.

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  • $\begingroup$ Edit: I believe there is an error in this proof - we do not know, necessarily, that $\phi_0 \in C^{**}$, because it might be that $\phi_0$ attains smaller values then -1 on $C$. However, i think there's in easy fix: We simply take $C$ to be the set {$\phi \in X^{**} : m \leq \phi(f) \leq 1$, for all $ f \in D $}, where $m = inf_{f \in D} \phi_0 (f) $ $\endgroup$
    – Tomer
    Commented Apr 29, 2021 at 9:58
  • $\begingroup$ What does $x_0(D)$ mean? $\endgroup$
    – LSpice
    Commented Oct 26, 2021 at 20:26

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