on non-strictly convex norms The 131 problem of the book  "Finite-Dimensional Linear Analysis: A Systematic Presentation in Problem Form"(Russian version) states that if $(X,\|\cdot\|)$ non-strictly convex space then there are linearly independent $x,y\in X$ such that $\|\alpha x+\beta y\| = |\alpha|+|\beta|$ for all $\alpha,\beta\in\mathbb R$. I only could prove that  if $(X,\|\cdot\|)$ non-strictly convex space then there are linearly independent $x,y\in X$ such that $\|\alpha x+\beta y\| = \alpha+\beta$ for all $\alpha\geq 0$ and $\beta\geq 0$ in other words statement of problem only for nonegative real numbers. Does anybody know how to solve the problem in general? Thank's in advance. 
 A: This assertion is in general false when $\alpha,\beta$ have different signs. Here's a counterexample: we know that to every origin-symmetric convex body $C$ in $\mathbb{R}^d$ containing the origin as an interior point there corresponds a norm on $\mathbb{R}^d$ having $C$ as its unit ball. 
Now consider in 2 dimensions the convex body which is an axis-aligned rectangle $R$ centered at the origin, where the top and bottom side of $R$ were bent out in a symmetric manner to obtain arcs. Denote the top-right and bottom-right vertices of $R$ by $x,y$. Then we see that the norm is not strictly convex because the left and right sides are still straight line segments where all points have the same norm 1. 
However, with $\alpha = -1/2$ and $\beta = 1/2$, the point $-x/2+y/2$ is strictly inside $R$, and thus has norm $< 1 = |\alpha| + |\beta|$.
EDIT: as pointed out in the comments, we're not quite done yet but we're almost there; the points $x,y$ we considered above were two specific points, but they are in a sense worst-case. 
Observe that any $x,y$ that satisfy $\|\alpha x +\beta y\| = 1$ whenever $\alpha,\beta\geq 0$ and $\alpha+\beta=1$ must be some two points on either the left or right side of $R$ (the straight sides). Now, take any two $x,y$ on the same straight side, and consider the point $(y-x)/2$ again. Observe that each such points falls inside the rectangular region determined by the straight sides of $R$: indeed, $y-x$ ranges over all vectors of the form $(0,l)$ where $|l|\leq L$ and $L$ is the sidelength of the straight sides of $R$. Thus $(y-x)/2$ ranges over vectors $(0,l)$ with $|l|\leq L/2$, and all of these lie strictly inside the unit ball of our norm, showing the contradiction.
