Prove that $S=\{x=(x_0,x_1,\dots), x_i\in R; \lim_{i\to\infty}x_i=0\}$ is not strictly convex. There is no given norm, so I assume that it's about the euclidean.
Let's take $x,y\in S$, $x\neq y$ such that $||x||=1=||y||$ and $\frac{1}{2}||x+y||=1$.
$$\sum_{i=0}^{\infty}x_i^2=1=\sum_{i=0}^{\infty}y_i^2$$ $$\sum_{i=0}^{\infty}(x_i+y_i)^2=\sum_{i=0}^{\infty}x_i^2+\sum_{i=0}^{\infty}y_i^2+2\sum_{i=0}^{\infty}y_ix_i=4\implies \sum_{i=0}^{\infty}y_ix_i=1$$ From both those things we can derive to the conclusion that: $$\sum_{i=0}^{\infty}(x_i-y_i)^2=0$$, but this means that $$x_i=y_i$$ Hence this space is strictly convex. Where is my mistake?
Use the $\ell^\infty$ norm $\| \cdot \|$.Consider the sequences $x = (1,1,0, \dots), \, y = (1,0,0,\dots) \in S$. Then $\|x \| = \|y\| = 1$. Let $0 < s < 1$. Then $sx + (1-s)y = (1,s,0, \dots)$ and thus $\|sx + (1-s)y\| = 1$.