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A Banach Space $X$ is said to be strictly convex or rotund if for all $x,y\in X$ we have $\lVert x+y\rVert<\lVert x\rVert+\lVert y\rVert$ unless $x,y$ are multiple of each other.

We have to prove the equivalence $$\text{$X$ is strictly convex (rotund) $\iff $ for all $x,y\in S(X)$ with $x\ne y$ we have $\lVert x+y\rVert <2$}$$

Here $S(X)=\{x\in X:\ \lVert x\rVert=1\}$. $\implies$ is obvious. But I cannot prove the other direction. There is another equivalence here, but I'm not getting any hint from this.

Can anyone help me with a hint or way out? Thanks for help in advance.

DeltaEpsilon
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1 Answers1

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For the converse, suppose that for any two distinct points $x_1,x_2 \in S(X)$ one has that $\| \tfrac 12 (x_1+x_2)\| <1$. 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 $.

Evangelopoulos Foivos
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