Trying to show that $(c_0, \| \cdot \|_s)$ is strictly convex, where $\| x \|_s = \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | x_i |$

Question

I'm trying to show that $ (c_0, \| \cdot \|_s) $ is a strictly convex space, where $$ \| x \|_s = \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | x_i |,$$ $ x = (x_1, x_2, ..., x_i, ...) \in c_0 $.

Let $x, y \in c_0, \| x \|_s = \| y \|_s = 1, x \neq y $. Let $\lambda \in (0 , 1)$.

$$ \| \lambda x + (1 - \lambda) y \|_s = \| (\lambda x_i + (1 - \lambda) y_i)_i \|_s $$ $$ = \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | \lambda x_i + (1 - \lambda) y_i | $$ $$\leq \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} ( \lambda | x_i | + (1 - \lambda) | y_i | ) $$ $$ = \lambda \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | x_i | + (1 - \lambda) \underset{i = 1}{\overset{\infty}{\sum}} \frac{1}{2^i} | y_i | $$ $$ = \lambda \| x \|_s + (1 - \lambda) \| y \|_s = \lambda + 1 - \lambda = 1. $$ So,

$$ \| \lambda x + (1 - \lambda) y \|_s \leq 1. $$ How can I show that $$ \| \lambda x + (1 - \lambda) y \|_s \neq 1 $$?

Than you!

Answer 1

Take $x=(2,0,0\cdots)$ and $y=(0,4,0,0,\cdots)$. Both are in $c_0$, and we have

$$\| \lambda x + (1-\lambda)y \|_s = \sum_{i=1}^{\infty} \frac{|\lambda x_i + (1-\lambda)y_i|}{2^i}$$

$$= \frac{2\lambda}{2} + \frac{4(1-\lambda)}{4} =1$$

Hence your space is not strictly convex