For me a Banach space $X$ is strictly convex if all points of unit sphere are not inner points of straight lines in a unit ball.

Here is the equivalence in Banach spaces:

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

I want to prove that (2) implies (1).

Below is a demonstration of the Evangelopoulos Foivos:

Proof. 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 $$\frac 12 x_1 - \frac 12 x_2 =0.$$

My doubt: I wanted to understand in detail, because to prove this implication we can assume that $x_1$ and $x_2$ have norms equal to 1. This is not at all clear to me.

Thank you to anyone who can explain it to me!

For me a Banach space $X$ is strictly convex if all points of unit sphere are not inner points of straight lines in a unit ball.

Here is the equivalence in Banach spaces:

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

I want to prove that (2) implies (1).

Below is a demonstration of an answer by Evangelopoulos Foivos:

Proof. 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 $$\frac 12 x_1 - \frac 12 x_2 =0.$$

My doubt: I wanted to understand in detail, because to prove this implication we can assume that $x_1$ and $x_2$ have norms equal to 1. This is not at all clear to me.

Thank you to anyone who can explain it to me!

For me a Banach space $X$ is strictly convex if x if all points of unit sphere are not inner points of straight lines in a unit ball.

Here is the equivalence in Banach spaces:

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

I want to prove that (2) implies (1).

Below is a demonstration of the Evangelopoulos Foivos:

Proof. 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 $$\frac 12 x_1 - \frac 12 x_2 =0.$$

My doubt: I wanted to understand in detail, because to prove this implication we can assume that $x_1$ and $x_2$ have norms equal to 1. This is not at all clear to me.

Thank you to anyone who can explain it to me!

For me a Banach space $X$ is strictly convex if all points of unit sphere are not inner points of straight lines in a unit ball.

Here is the equivalence in Banach spaces:

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

I want to prove that (2) implies (1).

Below is a demonstration of the Evangelopoulos Foivos:

Proof. 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 $$\frac 12 x_1 - \frac 12 x_2 =0.$$

My doubt: I wanted to understand in detail, because to prove this implication we can assume that $x_1$ and $x_2$ have norms equal to 1. This is not at all clear to me.

Thank you to anyone who can explain it to me!