Can we always find $f_1$ and $f_2$ in $F$ such that $\lVert f_1 \rVert= \lVert f_2 \rVert=1$ and $\lVert f_1 + f_2 \rVert=2$

Question

Let $F$ be a normed linear space with a norm that is not strictly convex. Is it true then that we can always find different $f_1$ and $f_2$ in $F$ such that $\lVert f_1 \rVert= \lVert f_2 \rVert=1$ and $\lVert f_1 + f_2 \rVert=2$. Ofcourse, when $F$ is not just the space containing only the zero element.

Answer 1

As $\|\cdot\|$ is not strictly convex, there exist distinct points $f_1,f_2$ on the unit sphere such that some convex combination $tf_1+(1-t)f_2$, $0<t<1$, of them is not an interior point of the unit ball. By convexity of the ball, $\|f_1\|=\|f_2\|=\|tf_1+(1-t)f_2\|=1$. In fact this implies (using convexity again) that $\|uf_1+(1-u)f_2\|=1$ for all $u\in[0,1]$. In particular $\|\frac12 f_1+\frac12f_2\|=1$ as desired.