In $l_2^2$, for any two linearly independent extreme points of unit ball $x_p, x_q$, $\|ax_p+bx_q\|= 1\implies |a|\leq 1, |b|\leq 1$

Question

I was studying about the extreme points of unit ball of a Banach space. I have a confusion in the following viewpoint:

Is it true that for any two linearly independent extreme points $x_p, x_q$ of the unit ball of $l_2^2$, $\|ax_p+bx_q\|= 1\implies |a|\leq 1, |b|\leq 1? $

I think it is true because for any parallelogram inside a unit circle, the length of the diagonal connecting to the origin will be $1$ only when each of the corresponding two sides connecting to the origin are of length less than or equal to $1$. Is my thinking true? If anyone can explain it mathematically it will be very much appreciated. Thank you in advance.

Answer 1

This answer summarizes my comments on $x_p\approx -x_q$. Fix $\epsilon$ such that $0<\epsilon\leq 1/100$. Define \begin{align*} x_p &= (1,0,0,0,...)\\ x_q &= (-1+\epsilon, \delta, 0, 0, ...) \end{align*} where $\delta = \sqrt{\epsilon(2-\epsilon)}$. Then $||x_p||=||x_q||=1$. Using $a=1$ we search for $b\neq 0$ such that $$||x_p+bx_q||=1$$ Equivalently \begin{align*} (1-b(1-\epsilon))^2 + b^2\delta^2 = 1 \end{align*} $$-2b(1-\epsilon) + b^2(1-\epsilon)^2 + b^2\delta^2=0$$ Since $b\neq 0$ we factor it out to obtain $$ -2(1-\epsilon) + b(1-\epsilon)^2 + b\delta^2 = 0$$ so $$ \boxed{b = \frac{2(1-\epsilon)}{(1-\epsilon)^2 + \delta^2}}$$ For $\epsilon\approx 0$ we get $\delta \approx 0$ and $b \approx 2$. For $\epsilon = 1/100$ we get $b=1.98$.

Answer 2

Let $x$ be an extreme point with $x_1=0$ and let $y=\alpha x+\sqrt{1-\alpha^2}e_1$ with some $\alpha \in (0,1)$. Then $||y||=1$, and $||x-\frac{1}{\alpha}y||=\frac{\sqrt{1-\alpha^2}}{\alpha}$. Taking $\alpha$ sufficiently large (larger than $\frac{1}{\sqrt{2}}$ to be precise) this can be made less than 1, so we have a counterexample.