Definition of extreme point which involve two norm $1$ points I migrate the following post from Maths Overflow.
Let $E$ be a Banach space. 
Denote $B_E$ and $S_E$ to be the unit ball and sphere of $E$ respectively.
A common definition of extreme point of $B_E$ that I have came across is the following: 

Definition $1$  $x$ is an extreme point of $B_E$ if $$x = \frac{1}{2}(y+z)$$ for some $y,z\in B_E,$ then $x=y=z.$

I have seen another definition of extreme point which involves $y$ and $z$ to be norm $1.$ For example, Lei Li et al. paper, page $549,$ paragraph starts with 'On the other hand,...' and Botelho's et al. paper, page $823,$ first sentence of the proof of Theorem $3$.
They use the following definition of extreme point: 

Definition $2$ $x$ is an extreme point of $B_E$ if 
  $$x = \frac{1}{2} (y+z)$$ for some $y,z\in S_E,$ then $x=y=z.$

I can prove that Definition $1$ implies Definition $2.$
But I am not able to prove its converse. 
So here is my question:

Question: Does Definition $2$ imply Definition $1?$ If yes, can prove a proof or reference? Otherwise, can provide counterexample?

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Based on the discussion at Maths Overflow, it seems that Definition $2$ implies Definition $1.$
To prove it, it suffices to prove that Definition $2$ implies that $\|x\|=1.$
 A: This answer is essentially the one from Jan-Christoph Schlage-Puchta in a MO comment, we show that every extreme point according to Definition 2 is an extreme point according to Definition 1. 
Since the norm is convex, a point lying in the middle between two points of the unit Ball must be an interior point unless it lies in the middle between two boundary points. So it suffices to show that no interior point can satisfy Definition 2.
So let $x$ be an interior point of the unit sphere, we show there are $y,z\in S_E$ such that $x=1/2 y+1/2 z$. Clearly $y\neq x\neq z$. The argument actually shows us something stronger, we can find $y,z$ in any plane containing $x$ (this means we have to deal with the trivial one-dimensional case separately). The norm restricted to a plane is again a norm and the unit ball and the unit sphere are the intersections of the original ones with the plane. So we reduce the problem from general Banach spaces to two-dimensional one. Now for each point $r$ in the two-dimensional Euclidean sphere, let $f(r)$ be the  vector $x+\alpha_r r$ where $\alpha_r\geq 0$ is chosen so that $\|x+\alpha_r r\|=1$. It is not hard to show that $f$ is a continuous function of the unit sphere and so is $r\mapsto\alpha_r$. So $r\mapsto |\alpha_r|$ is a continuous function from the two dimensional Euclidean sphere into $\mathbb{R}$. By the Borsuk-Ulam theorem, there must exist antipodal points $r$ and $-r$ in the Euclidean sphere such that $|\alpha_r|=|\alpha_{-r}|$. But then $x=1/2 f(r)+1/2 f(-r)$ and so $x$ is not an extreme point according to Definition 2.
