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Why is $(1,1,...)$ an extreme point of a closed unit ball in $c$ (the space of all convergent sequences with sup norm)?

Suppose on the contrary that $(1,1,...)= t (x_n)+(1-t) (y_n)$ for some $t$ in $(0,1)$ and $(x_n)\ne (y_n)\in c$. Then, $tx_n+(1-t)y_n=1$ for all $n$, so $t x+(1-t)y=1$, where $x_n\to x, y_n\to y$.

How do I get a contradiction from here to conclude that $(1,1,...)$ is indeed an extreme point?

Koro
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  • Related https://math.stackexchange.com/questions/3475323/extreme-points-of-the-closed-unit-ball-in-space-of-convergent-sequences but it does not explain why $x_n=y_n$. – Koro Feb 26 '23 at 08:18
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    What can you conclude from $1 =tx_n+(1-t)y_n \leq t(1)+(1-t)(1)=1$? – geetha290krm Feb 26 '23 at 08:20
  • @geetha290krm: I don't understand how that helps. I think nothing can be concluded from here. – Koro Feb 26 '23 at 08:23
  • It implies $x_n=y_n=1$. No need to use limits of $(x_n)$ and $(y_n)$. – geetha290krm Feb 26 '23 at 08:24
  • @geetha290krm: Can you please explain how? I don't yet understand how. Thanks. – Koro Feb 26 '23 at 08:25
  • Since you are talking of "a closed unit ball", may be you did not think deeply enough about what the closed unit ball of $c$ is. – Anne Bauval Feb 26 '23 at 08:35
  • @AnneBauval: yeah, I just thought that $|(x_n)|\le 1$, which is true but somehow didn't think that $x_n\le |x_n|\le \sup_n |x_n|=|(x_n)|$ :(. Of course, this assumes that everything is real valued here. – Koro Feb 26 '23 at 08:39
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    Note that if everything is complex valued, the result still holds, with nearly the same argument as in @geetha290krm 's answer. – Anne Bauval Feb 26 '23 at 10:06
  • @AnneBauval: yes, noted. I added that bit about real valued in my last comment so that $x_n\le |x_n|$ makes sense in the comment. :-) – Koro Feb 27 '23 at 11:10

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$1 =tx_n+(1-t)y_n \leq t(1)+(1-t)(1)=1$. If $x_n<1$ or $y_n<1$ then we get $1 =tx_n+(1-t)y_n < t(1)+(1-t)(1)=1$ or $1<1$ which is a contradiction. Hence, the only possibility is $x_n=y_n=1$ for all $n$.

As pointed out by Anne Bauval the conclusion holds even for complex sequences. [If $|c|\leq 1$ and $\Re c=1$ then $\Im c=0$ necessarily, so $c=1$].

geetha290krm
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