# show that $\inf\{1-\|\frac 12(x+y)\|: x,y\in S_X,\|x-y\|=\epsilon\}=\inf\{1-\|\frac 12(x+y)\|: x,y\in B_X,\|x-y\|=\epsilon\}$

Here is theorem 5.2.5 in the book Introduction to Banach Space Theory by Megginson:

Let $X$ be a nored space that is not zero-dimensional if the scalar field is $\mathbb C$ and is neither zero- nor one-dimensional if scalar field is $\mathbb R$. Let $\delta_1,\delta_2:[0,2]\to [0,1]$ defined by $$\delta_1(\epsilon)=\inf\{1-\|\frac 12(x+y)\|: x,y\in S_X,\|x-y\|=\epsilon\}\\ \delta_2(\epsilon)=\inf\{1-\|\frac 12(x+y)\|: x,y\in B_X,\|x-y\|=\epsilon\}\\$$ Show that $\delta_1=\delta_2$.

Notation: $B_X$ and $S_X$ are respectively the closed unit ball and unit sphere.

proof of this theorem based on the following lemma:

Let $X$ be a nored space that is not zero-dimensional if the scalar field is $\mathbb C$ and is neither zero- nor one-dimensional if scalar field is $\mathbb R$. If $x_0\in S_X$ and $y_0\in B_X$, then there are members $x_1,y_1\in S_X$ such that $x_1-y_1=x_0-y_0$ and $\|\frac12(x_1+y_1)\|\geq \|\frac 12(x_0+y_0)\|$.

It's clear that $\delta_2\leq \delta_1,$However I dont understand the first part(the following paragraph) of the proof:

suppose that $0<\epsilon\leq 2$, that $x_0,y_0\in B_X$ and that $\|x_0-y_0\|=\epsilon$. to prove the claim it sufices to show that $\delta_1(\epsilon)\leq 1-\|\frac12(x_0+y_0)\|$. Suppose for the moment that $x_0,y_0$ have norm less than one. then there are two closed line segment of length $\epsilon$ on the line through $x_0$ and $y_0$ such that each segment has one endpoint in $S_X$ and the other in $B_X^\circ$. since $\frac12(x_0+y_0)$ is a convex combination of the midpoints of these two segment, the midpoint of at least one of these two segments has norm at least $\|\frac12(x_0+y_0)\|$, so it may be assumed that $x_0\in S_X$.

The rest of the proof is easy by using the lemma.

why it may be assumed $x_0\in S_X$?

why it may be assumed $$x_0\in S_X$$?
Because we just found a segment length $$\epsilon$$ which has one endpoint (denote it by $$x’_0$$) in $$S_X$$, the other (denote it by $$y’_0$$) in $$B_X^\circ$$ such that the norm of its midpoint $$\|\frac12(x’_0+y’_0)\|$$ is at least $$\|\frac12(x_0+y_0)\|$$. So we consider $$x’_0$$ as a new $$x_0$$ and $$y’_0$$ as a new $$y_0$$.