1

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!

Evangelopoulos Foivos
  • 4,429
  • 1
    In the characterization I cited, 1) is equivalent to: For every $x_1,x_2 \in X$ with norm one, either $|x_1+x_2|< 2$ or $x_1=x_2.$ So to prove that 2) implies 1), it suffices to show that whenever $x_1,x_2$ have norm one and $|x_1+x_2| =2 (=|x_1|+|x_2|)$ then $x_1=x_2$ – Evangelopoulos Foivos Oct 16 '24 at 19:40
  • 1
    Btw, to check strict convexity you only need to check that for every $x,y$ in the sphere $S_X$, the middle point is not contained in $S_X$. So you only need to check that $tx+(1-t)y \notin S_X$ only for $t=1/2$. See here for a proof https://math.stackexchange.com/questions/1989904/equivalence-of-the-definition-of-rotund-space – Evangelopoulos Foivos Oct 16 '24 at 19:44
  • @EvangelopoulosFoivos So it's not clear to me why 1) is equivalent to For every $x_1$,$x_2\in X$ with norm one, either $∥x_1+x_2∥<2$ or $x_1=x_2$. Would it be possible for you to demonstrate this fact? It's really not clear to me. –  Oct 16 '24 at 19:52

1 Answers1

1

According to your definition, $X$ is strictly convex iff for all $x,y $ distinct points in the unit sphere $S_X$ and for all $t \in [0,1]$ the point $z_t = tx+(1-t)y \notin S_X$ i.e. $|z_t|< 1$. It turns out that it suffices to only check the $t=1/2$ case. So $X$ is strictly convex if and only if $$\forall x_1,x_2 \in S_X , |x_1+x_2|<2 \equiv |x_1|+|x_2|$$ Now this is equivalent to 1). Indeed, if 1) holds then if $|x_1+x_2| =2 \equiv |x_1|+|x_2|$ then $x_1= \lambda x_2$ but then $\lambda =1$ because $|x_1|=|x_2|=1$. So if 1) holds then for any two distinct $x_1,x_2 \in S_X$ we must have that $|x_1+x_2|<2$ i.e. $X$ is strictly convex.

The converse direction can be found in my answer which I will copy below for convenience. Suppose that $X$ is strictly convex. Let $ x_1,x_2 $ be two points in $X $ such that $ |x_1+x_2|=|x_1|+|x_2| $. We must show that $x_1=ax_2$ for some $a>0$. We may assume that $ |x_1| =1$. Let $ y= x_2/|x_2| $. It then follows that
\begin{align*} 2 & \geq |x_1+y| \\ &= |x_1+x_2 - (1-\|x_2\|^{-1} ) x_2| \\ &\geq |x_1+x_2|- (1- |x_2|^{-1} ) |x_2| \\ &= |x_1|+|x_2|-|x_2|+1\\ &=2 \end{align*} and so $ |\tfrac 12 (x_1+y)| =1$. Therefore we must have that $ x_1=y = |x_2|^{-1} x_2 $.

If you do not feel comfortable with the characterization, you can skip it: To show that 2) implies 1) assume that $|x_1+x_2|=|x_1|+|x_2|$. Set $t= |x_1|/|x_1+x_2|$. Then $$|x_1-0| =|x_1|=t|x_1+y_1|= t|x_1-(-x_2)| $$ and $$|x_2-0| = |x_2| = |x_1+x_2|-|x_1| = (1-t) |x_1+x_2| = (1-t)|x_1-(-x_2)| $$ so that 2) holds for $x=x_1, y =-x_2$ and $z=0.$ Now conclude that $0= tx + (1-t)y$ and thus $x_1=x_2$.

Evangelopoulos Foivos
  • 4,429