Le $X$ be a Banach space and $X_{1}$ be its closed unit ball. $x\in X_{1}$ with $\|x\|=1$ is said to be a strongly extreme point if whenever for sequences $\{x_{n}\}, \{y_{n}\}$ in $X_{1}$, $\|\frac{x_{n}+y_{n}}{2}-x\|\rightarrow 0$, then $x_{n}-y_{n}\rightarrow 0$.
An equivalent definition is the following:
If $\|x\pm z_{n}\|\rightarrow 1$, then $\|z_{n}\|\rightarrow 0.$
One direction is clear. Suppose $x$ is a strongly extreme point as in the second definition and suppose $\{x_{n}\}, \{y_{n}\}$ are sequences with norm less than or equal to 1 and $\|\frac{x_{n}+y_{n}}{2}-x\|\rightarrow 0$. It can be shown that lim $\|x_{n}\|,$ lim $\|y_{n}\|=1$. Let $z_{n}=x_{n}-x$. Then $\text{lim}\|z_{n}+x\|=\|x_{n}\|=1$ and $\text{lim}\|z_{n}-x\|=\text{lim }\|y_{n}\|=1$, hence $z_{n}\rightarrow 0$, i.e. $x_{n},y_{n}\rightarrow x$ as required.
For the converse, suppose the first definition holds, and suppose $\|z_{n}\pm x\|\rightarrow 1$. It is to be shown that $z_{n}\rightarrow 0$. We can write $x$ as $x=\frac{x+z_{n}}{2}+\frac{x-z_{n}}{2}$. But we do not know that $\|x\pm z_{n}\|\leq 1$. How else can we use the first definition to prove $z_{n}\rightarrow 0$?
Thanks in advance!