Equivalence of strictly convex Banach spaces

Question

I would like to know if anyone knows how to prove this equivalence that was proven in 1992 in "On strict convexity, LU Zinatniskie Raksti". I ask this because I was unable to access the PDF of this article.

In Banach spaces they are equivalence of strictly convex spaces:

The following conditions are equivalent in the Banach space $X$:

1. $\forall x, y \in X$ : $\|x + y\| = \|x\| + \|y\| \implies \left( (\exists \lambda > 0 : x = \lambda y) \, \vee \, (x = 0) \, \vee \, (y = 0) \right)$;

2. $\forall x, y \in X \, \forall t \in [0, 1] \, \exists! z \in X : \|x - z\| = t \|x - y\|, \text{ and } \|z - y\| = (1 - t) \|x - y\|$.

Thank you to anyone who can help!

Answer 1

Note that the point $z_t = (1-t)x+ty$ always satisfies the condition of 2). So the unique $z$ in 2) must be given by $z=z_t$.

Here is the implication 1) $\implies$ 2): If $|x-z|=t|x-y| $ and $|z-y|= (1-t) |x-y|$ then $$|x-z|+|z-y| = |x-y| = | (x-z)+(z-y)|$$ so that by 1) you have that $x-z=\lambda (z-y)$. In fact, $\lambda = t/(t-1)$. Hence $z$ must be given by $$z=(x+\lambda y)/(\lambda+1)= (1-t)x+ty.$$

For the opposite, we may use the following characterization of strictly convex spaces: For any two distinct points $x_1,x_2$ of norm one, the middle point has norm $ |\frac{x_1+x_2}{2} |<1$. To this end, suppose that $x_1,x_2$ have norm one and $|x_1|+|x_2|=|x_1+x_2|$. We must show that $x_1=x_2$. But $$ |x_1-0| = \frac 12 |x_1+x_2| =|x_2-0| $$ so that 2) holds for $x=x_1, y=-x_2, z=0 $ and $t=1/2$. By the hypothesis $z$ must be given by $z= \frac 12 x + \frac 12 y = \frac 12 x_1 - \frac 12 x_2$ so that $$\frac 12 x_1 - \frac 12 x_2 =0.$$ After an extensive googling, I was able to find the original paper in https://dspace.lu.lv/dspace/handle/7/5330 page 196-197 but I find that the proof there is not very clear.