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I am trying to show that the closed unit ball in $C_0(R)$ has no extreme points. This is what I got so far and I am stuck. Please help me.

Suppose that $f \in C_0(R)$ and is an extreme point of the closed unit ball. Let $g(s) = f(s) + |f(s)| + 1$ and $h(s) = f(s) + |f(s)j| - 1$. Then $\|g\|_\infty \le 1$ and $\|h\|_\infty \le 1$ since $\|f\|_\infty \le 1$.

Is that the right way to start this problem?

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    Hint from a different direction: what are the extreme points of $\mathbb{R}^n$ with the $\infty$-norm? Why can the equivalent not occur in $C_0$ with the $\infty$-norm?2012-04-02

1 Answers 1

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I'm not sure what you are doing will lead anywhere.

Here's what you need to show:

If $f$ is in the unit ball of $C_0$, then there exists two functions $g$ and $h$ in the unit ball of $C_0$ such that $g\ne h$ and $f$ is a nontrivial convex combination of $g$ and $h$.

To show this, you could use the following

Hint: for $f$ in the unit ball of $C_0$, eventually $|f(x)|<1/2$. Using this, find two functions $g$ and $h$ in the unit ball of $C_0$ with $f={1\over2}(h+g)$.

Very informally, you can add a "bump" and subtract the same bump to $f$ over an interval on which the inequality first mentioned holds. $g$ and $h$ will have the forms $f+b$ and $f-b$, where $b$ is the "bump". Do this in such a way that the resulting functions are in the unit ball of $C_0$. Draw a picture here...

I hope that made sense...




enter image description here

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