Extreme points of L^p[0,1] on unit closed ball

[Kokuhaku]

I would like to prove that if $\|f\|_p = 1$ then $f$ is extreme point of unit closed ball in $L^p[0,1]$. [here $1<p<\infty$]

I suppose we should try to prove it by contradiction. That is, if $f \in L^p$, with $\|f\|_p=1$, is not extreme, then exists $g,h \in L^p$ with $\|g\|_p,\|h\|_p \leq 1$ and $\lambda \in (0,1)$ such that $f=(1-\lambda)g+\lambda h$.

Then we can use that $$1=\|f\|_p^p = \int_{[0,1]} |(1-\lambda) g(t) + \lambda h(t)|^p \, d\mu(t),$$ but I don't know how to proceed from this point. I suppose we can show that last term is $<1$, using some integral inequalities and $\|g\|_p,\|h\|_p \leq 1$, but I don't see how.

Or, maybe, to use Minkowski inequality to obtain $1=\|f\|_p \le (1-\lambda) \|g\|_p + \lambda \|h\|_p$ and to prove that it can't be equality in Minkowski inequality. We know that we have equality if and only if there exist $\alpha,\beta \ge 0$ such that $\alpha g = \beta h$ almost everywhere.

 

[Opalg]

Re: Extreme points of \$L^p[0,1]\$ on unit closed ball

... use Minkowski inequality to obtain $1=\|f\|_p \le (1-\lambda) \|g\|_p + \lambda \|h\|_p$ and to prove that it can't be equality in Minkowski inequality. We know that we have equality if and only if there exist $\alpha,\beta \ge 0$ such that $\alpha g = \beta h$ almost everywhere.

Yes, that is exactly how to do it. Minkowski's inequality is strict unless $g$ and $h$ are positive scalar multiples of each other (almost everywhere). In that case, $f$ is also a scalar multiple of $g$ (or $h$). Since $f$, $g$ and $h$ all have norm $1$, it follows that they must all be equal (almost everywhere).

 

[Kokuhaku]

Re: Extreme points of \$L^p[0,1]\$ on unit closed ball

Since $f$, $g$ and $h$ all have norm $1$, ...

I don't see how to prove this.

If $g=\alpha h$, then we have $f=((1-\lambda)\alpha+\lambda)h$ and from that I can see that $\|h\|=((1-\lambda)\alpha+\lambda)^{-1}$, but why that is equal to one, I don't see.

edit: Oh, wait, maybe I see it. If we would have $\|h\|_p <1$, then we would have $1=\|f\|_p = (1-\lambda) \|g\|_p + \lambda \|h\|_p < (1-\lambda)\|g\|_p + \lambda \le (1-\lambda) + \lambda = 1$ (because $\|g\| \le 1$). Contradiction. Also, from $\|h\|_p=1$ we see that $\|g\|_p=1$.

And now we know that $\alpha = 1$, and then $f=g=h$ a.e. Great!Also, it's kinda funny, but I see many things when I am writing post :)

edit 2: I am just wondering, but are this only extreme points in $L^p$ on unit closed ball? That is, is this if and only if case?