show that $E$ is not uniform convex space

Question

If $E=ℝ^2 $ with norm $||x||=|x_1|+|x_2|$ where $x=(x_1,x_2)$ i need to proof by definition that $E$ is not uniform convex space. Then by definition i have, for all $0<\epsilon\leq2$ exist $\delta>0$: $x,y\in B_1(0)$ unitary ball, and $\epsilon< ||x-y||$ imply $||\frac{x+y}{2}||<1-\delta$. So, using the negation of the definition i pick many couples $x,y\in B_1(0)$, for example $x=(1,0)$ and $y=(-1,0)$ then $||x-y||=2=\epsilon_0$ i need to proof that for all $\delta>0$, $1-\delta \leq||\frac{x+y}{2}||=0$. I did the same with many couples $x, y$ but i donf go anyplace. I will appreciate the hints for prove the affirmation.

Answer 1

How about $x=(1,0)$ and $y=(0,1)?$ Then, $\|x-y\|=2$ but $\frac{\|x+y\|}{2}=1$, whch contradicts the definition of uniform convexity. The geometric idea is that for a space to be uniformly convex, if points in the closed unit ball are far apart, then the interior of the line segment connecting them must be strictly interior to this ball. Drawing the ball in the $\ell^1$ norm then shows how to get a counterexample. Compare the following balls and see which induce uniformly convex spaces:

Answer 2

Hint: the unit sphere $\{x : \|x\|=1\}$ consists of line segments. What if $x$ and $y$ are on the same segment?