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You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ is open endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since the closed unit ball of $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ is open endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ is open endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk-)compact, it admits an extreme point. But, since $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)