A characterisation of uniform convexivity I want to prove that a Banach space is uniformly convex if and only if
$$\|x_n\| \to 1, \quad \|y_n\| \to 1 \quad \text{and} \quad \left\|\frac{x_n+y_n}{2}\right\|\to1$$
implies
$$ \|x_n - y_n \| \to 0.$$
It's an exercise from Werner's book (German) on functional analysis. Can someone hint to me how this can be done?
The given definition for a Banach space to be uniformly convex is:
For every $\epsilon>0$ there is a $\delta>0$ such that
$$\|x\|\leq1 \text{,} \quad \|y\|\leq1 \quad \text{and} \quad \|x-y\|\geq\epsilon$$ implies
$$\left\|\frac{x+y}{2}\right\|\leq1-\delta.$$
 A: Suppose your Banach space is uniformly convex, and $x_n$ and $y_n$ are sequences with $\|x_n \| \to 1$, $\|y_n \| \to 1$, $\|(x_n+y_n)/2\| \to 1$, but $\|x_n - y_n\|$ does not go to $0$.  By taking a subsequence, we can assume $\|x_n - y_n\| \ge 2 \epsilon$ for some $\epsilon > 0$.  Now take $x = x_n/\|x_n\|$ and $y = y_n/\|y_n\|$ for $n$ sufficiently large, and get a contradiction with the definition of uniformly convex.
A: The other direction is very tricky.
For suppose that the Banach space is not uniformly convex, then there are
\begin{align*}
\|x_{n}\|&\leq 1\\
\|y_{n}\|&\leq 1\\
\|x_{n}-y_{n}\|&\geq\epsilon_{0}\\
\left\|\dfrac{x_{n}+y_{n}}{2}\right\|&>1-\dfrac{1}{n}.
\end{align*}
Consider $u_{n}=x_{n}/\|x_{n}\|$ and $v_{n}=y_{n}/\|y_{n}\|$, then $\|u_{n}\|=\|v_{n}\|=1$ and
\begin{align*}
\left\|\dfrac{u_{n}+v_{n}}{2}\right\|\leq\dfrac{1}{2}(\|u_{n}\|+\|v_{n}\|)=1,
\end{align*}
and
\begin{align*}
\left\|\dfrac{u_{n}+v_{n}}{2}\right\|&\geq\left\|\dfrac{x_{n}+y_{n}}{2}\right\|-\left\|\dfrac{x_{n}}{2\|x_{n}\|}-\dfrac{x_{n}}{2}\right\|-\left\|\dfrac{y_{n}}{2\|y_{n}\|}-\dfrac{y_{n}}{2}\right\|\\
&>1-\dfrac{1}{n}-\dfrac{1}{2}\|x_{n}\|\left(\dfrac{1}{\|x_{n}\|}-1\right)-\dfrac{1}{2}\|y_{n}\|\left(\dfrac{1}{\|y_{n}\|}-1\right)\\
&=\dfrac{1}{2}(\|x_{n}\|+\|y_{n}\|)-\dfrac{1}{n}\\
&\geq\left\|\dfrac{x_{n}+y_{n}}{2}\right\|-\dfrac{1}{n}\\
&>1-\dfrac{2}{n},
\end{align*}
so
\begin{align*}
\left\|\dfrac{u_{n}+v_{n}}{2}\right\|\rightarrow 1.
\end{align*}
By assumption, then $\|u_{n}-v_{n}\|\rightarrow 0$. But then
\begin{align*}
\epsilon_{0}&\leq\|x_{n}-y_{n}\|\\
&\leq\left\|x_{n}-\dfrac{x_{n}}{\|x_{n}\|}\right\|+\left\|y_{n}-\dfrac{y_{n}}{\|y_{n}\|}\right\|+\|u_{n}-v_{n}\|\\
&=2-\|x_{n}\|-\|y_{n}\|+\|u_{n}-v_{n}\|.
\end{align*}
Note that $\|x_{n}\|+\|y_{n}\|>2-2/n$, hence
\begin{align*}
\epsilon_{0}<\dfrac{2}{n}+\|u_{n}-v_{n}\|,
\end{align*}
by taking $n\rightarrow\infty$, we obtain a contradiction.
