Recent content by Kokuhaku

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

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 $\|g\|_p,\|h\|_p \leq 1$...

I see that Opalg, but that works only for $a+\varepsilon <0$, that is $\varepsilon < -a$. But $\varepsilon$ can be arbitrary. Am I missing something?

If $\{x_n\}_{n \ge 1}$ is real sequence and $\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n|<0$, prove that $\{x_n\}$ is Cauchy sequence. My work: Let $a=\limsup\limits_{n \to \infty} \frac{1}{n} \log |x_{n+1}-x_n| <0$. Then, for every $\varepsilon >0$ there exist $N \in...