Equivalent definitions of uniform convexity. I think I'm dumb, but I can't follow a simple proof from "Functional analysis and infinite dimensional geometry". 
They show that two different definitions of modulus of convexity of a norm are the same.
The key step is showing that if $X$ is a 2-dimensional Banach space,
$\sup\lbrace ||x+y||: ||x||\leq 1, ||y||\leq 1, ||x-y||=\epsilon\rbrace=\sup\lbrace ||x+y||: ||x||=1, ||y||= 1, ||x-y||=\epsilon\rbrace$
For this they take $u,v$ a pair that attains the supremum for the left side of the previous equation. The idea is to show that $||u||=1$, and $||v||=1$.
They assume $||v||<1$, then show $||u||<1$, and by an easy argument arrive to a contradiction.
What I don't get is the argument they use to show that $||v||<1$ implies $||u||<1$
They take a functional $\varphi$, with $||\varphi||=1$, and $\varphi(u+v)=||u+v||$. Then they consider $A=\lbrace w:||w||\leq 1, ||w-u||=\epsilon\rbrace.$ Finally, it's easily seen that $\varphi(w)\leq\varphi(v)$ for every $w\in A$. Then.
"This implies that $\varphi$ norms $(v-u)$, so $\varphi(v-u)=||v-u||=\varepsilon$"
(and from here it's easily seen that $||u||<1$)
I really don't see this last bit. 
Thanks in advance.
 A: Assuming that $\varphi (v)\geq \varphi (u)$, I would do it as follows.
Let $S(u,\varepsilon)=\{ w\in X;\; \Vert w-u\Vert=\varepsilon\}$ be the sphere with center $u$ and radius $\varepsilon$. Since $\Vert v\Vert<1$, one can find an open neighbourhood $V$ of $v$ such that $V\cap S$ is contained in the above set $A$. Since $\varphi (w)\leq \varphi (v)$ for all $w\in A$, it follows that the restriction of the linear functional $\varphi$ to the sphere $S(u,\varepsilon)$ has a local maximum at $v$. Translating by $u$ and dividing by $\varepsilon$, we get the following: the restriction of $\varphi$ to the unit sphere $S_X$ attains a local maximum at $h=\frac1\varepsilon\, (v-u)$. 
Assume that we are able to show that in fact the restriction of $\varphi$ to $S_X$ has indeed a $global$ maximum at $h$. Then $\varphi (h)=\sup\{ \varphi (h');\; h'\in S_X\}=\Vert\varphi\Vert =\Vert\varphi\Vert\,\Vert h\Vert$, so $\varphi$ is a norming functional for $h$ and hence for $v-u$. 
Now, let us show that the restriction of $\varphi$ to $S_X$ has a global maximum at $h$. Assume that $\varphi(h')>\varphi (h)$ for some $h'\in S_X$. Then $\varphi (z)>\varphi (h)$ for avery $z\in [h',h)$, because $\varphi$ is concave. Since $[h',h)\subset B_X$, it follows that there are points $z$ in $B_X$ arbitrarily close to $h$ such that $\varphi(z)>\varphi (h)$. For such points $z$ we have $\varphi(z)\geq 0$ (since I'm assuming that $\varphi(v)\geq\varphi(u)$, i.e. $\varphi(h)\geq 0$) so $\varphi (z/\Vert z\Vert)\geq \varphi(z)$  and hence $\varphi (z/\Vert z\Vert) >\varphi(h)$. Therefore, one can find points $y$ in $S_X$ arbitrarily close to $h$ for which $\varphi (y)>\varphi (h)$. This means that there is no local maximum at $h$, a contradiction.
