Let $X \equiv (X, \|\cdot\|)$ be a real Banach space. Fix $t \in (0, 2]$. The modulus of convexity for $X$ is $$ \delta(t) = \inf_{x, y \in X} \Big\{\, 1 - \Big\|\frac{x + y}{2}\Big\| : \|x\| \leq 1, \|y\| \leq 1, \|x - y\| \geq t\, \Big\}. $$ Consider the alternative definition, $$ \tilde{\delta}(t) = \inf_{x, y \in X} \Big\{\, 1 - \Big\|\frac{x + y}{2}\Big\| : \|x\| = \|y\| = 1, \|x - y\| = t\, \Big\}. $$ In Lindenstrauss and Tzafriri, Classical Banach spaces, II (pg. 60) it is claimed that $$ \delta(t) = \tilde{\delta}(t). $$ Of course $\delta(t) \leq \tilde{\delta}(t)$ is trivial. However, I do not see why the infima should coincide? Is there a simple explanation?
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$\newcommand\de{\delta}\renewcommand\th{\theta}\newcommand\R{\Bbb R}\newcommand\p\partial$It suffices to show the following: for any $x,y$ in $X$ such that $\|x\|\le1, \|y\|\le1, \|x-y\| \ge t\in(0,2]$ we can find $u,w$ in the linear span of $x,y$ such that $\|u\|=1, \|w\|=1, \|u-w\|=t, \|u+w\|\ge\|x+y\|$.
So, without loss of generality (wlog), $X=\R^2$. Let $B$ be the closed unit ball in $\R^2$ (wrt to the given norm) and let $S=\p B$ be the unit sphere. By continuity and compactness (which follow because all norms on $\R^2$ are equivalent to one another), there is a maximizer $\{x_*,y_*\}$ of $\|x+y\|$ over all subsets $\{x,y\}$ of $B$ such that $\|x-y\|\ge t$. If $s:=\|x_*-y_*\|>t$, then for $\hat x:=(1-a)x_*+ay_*$ and $\hat y:=ax_*+(1-a)y_*$ with $a:=\frac12(1-\frac ts)\in(0,1/2]\subset(0,1)$ we have $\hat x\in B$, $\hat y\in B$, $\hat x+\hat y=x_*+y_*$, and $\|\hat x-\hat y\|=t$.
So, wlog we can maximize $\|x+y\|$ over all subsets $\{x,y\}$ of $B$ such that $\|x-y\|=t$. Letting $v:=\frac{y-x}2$, with $\|v\|=s:=t/2\in(0,1]$ and $z:=\frac{y+x}2$ by $z$, so that $x=z-v$, $y=z+v$, and $z\in (B+v)\cap(B-v)$, we can restate the problem as follows:
Let $M_v$ denote the set of all maximizers of $\|z\|$ over $z\in D_v:=(B+v)\cap(B-v)$. Prove that then there are some $v\in sS$ and $z_v\in M_v$ such that $z_v\in(S+v)\cap(S-v)$.
Indeed, then we can let $u:=z_v-v\in S$ and $w:=z_v+v\in S$, so that $\|u-w\|=2\|v\|=2s=t$.
To prove the highlighted claim, note first that $D_v$ is a symmetric convex compact set. For each real $\th$, let $z(\th)$ be a maximizer $z$ of $\|z\|$ over $z\in D_{v(\th)}$, where \begin{equation*} v(\th):=s\,(\cos\th,\sin\th). \end{equation*} Since the norm is a convex function, for all $\th$ \begin{equation} z(\th)\in\p D_{v(\th)}\subseteq(S+v(\th))\cup(S-v(\th)). \tag{1}\label{1} \end{equation} Moreover, $D_{v(\cdot)}$ is a continuous function (say wrt the Hausdorff metric), which is also periodic with period $\pi$, since $v(\th+\pi)=-v(\th)$. So, we can choose the maximizers $z(\th)$ so that the function $z(\cdot)$ be continuous and periodic with period $\pi$.
Let now \begin{equation*} b(\th):=\|z(\th)+v(\th)\|-\|z(\th)-v(\th)\|, \end{equation*} so that $b$ is a continuous ("balance") function. Since $z(\cdot)$ is periodic with period $\pi$ and $v(\th+\pi)=-v(\th)$, we have $b(\th+\pi)=-b(\th)$. So, for some real $\th$ we have $b(\th)=0$, that is, \begin{equation*} \|z(\th)+v(\th)\|=\|z(\th)-v(\th)\|. \tag{2}\label{2} \end{equation*} But, in view of \eqref{1}, either $\|z(\th)+v(\th)\|=1$ or $\|z(\th)-v(\th)\|=1$. So, by \eqref{2}, $\|z(\th)+v(\th)\|=\|z(\th)-v(\th)\|=1$. That is, $z_{v(\th)}=z(\th)\in(S+v(\th))\cap(S-v(\th))$, and also $v(\th)\in sS$. $\quad\Box$
