Ruston's Theorem Let $X$ be a normed space. The following statement are equivalent:
(i) If $u,v$ are distinct unit vectors differents, then $\|u+v\|<2.$
(ii) If $x,y\in X\setminus\{0\}$ and $\|x+y\|=\|x\|+\|y\|$, then $x=\lambda y$ for some $\lambda>0. $
(iii)  If $\phi\in X^*$ is a nonzero linear functional, then there exists, up to multiplication by a positive scalar, a single $x\in X$ with $\phi(x)=\|x\|.$
To be honest, i do not get in anyone direction. There is here my attepmt.
$i)\rightarrow ii)$ Let us suppose that i) is true and that exist $x,y$ different of the vector zero such that:$||x+y||=||x||+||y||$. Then normalizing the vector $x$ and $y$ we have
$||\frac{x}{||x||}||=1=||\frac{y}{||y||}||$ hence by hypothesis
$$||\frac{x}{||x||}+\frac{y}{||y||}||<2$$ and other thing that i gues i can to use is
$$\tag{$*$}\|x+y\|^2=\|x\|^2+2\|x\|\|y\|+\|y\|^2$$. The other side, if i suppose by contradiction that $x\neq \lambda y$ for all $\lambda>0$ then in particular for $\lambda= \frac{||x||}{||y||}$ then $||x-\frac{||x||}{||y||}y||\neq 0$ and $||x-\frac{||x||}{||y||}y||>|||x||-||\frac{||x||y}{||y||}|||$, well i guest i need to find  some contradiction with the numer $2$ above.
$ii)\rightarrow iii)$ I want to prove by contradiction that exist $x,y$ different vector such that $\phi(x)=||x||$ and $\phi(y)=||y||$. Then $\phi(x+y)=\phi(x)+\phi(y)=||x||+||y||\geq ||x+y||$ but i no know how to guarantee that $\phi(x+y)=||x+y||$. I guest that i can  use too that
$$||\phi||=\sup_{||x||=|}\frac{|\phi(x)||}{||x||}\geq \frac{\phi(\frac{x+y}{||x+y||})}{\frac{x+y}{||x+y||}}$$, i do not how to continue or use other thing with my Hypothesis ii)
I did not try anything with the statement iii). Please if somebody can give me hints i will appreciate so much. Thank you
 A: The key to this question is to recognize that it is about (strict) convexity.
A simple observation is that if $\|x\|\leq1$, $\|y\|\leq1$, and $\|tx+(1-t)y\|=1$ for some $t\in(0,1)$, then $\|x\|=\|y\|=1$. This follows from
$$
1=\|tx+(1-t)y\|\leq t\|x\|+(1-t)\|y\|\leq t+1-t=1. 
$$
Then $t(1-\|x\|)+(1-t)(1-\|y\|)=0$, and as this is sum of nonnegative terms, they both have to be zero and hence $\|x\|=\|y\|=1$.
As Ryszard suggested, $(i)$ (in the form "$\|(u+v)/2\|=1$ for $\|u\|=\|v\|=1$ implies $u=v$") is equivalent to
$$\tag1
\|x\|=\|y\|=1,\ \|tx+(1-t)y\|=1\ \implies \ x=y
$$
(proof at the end).
$(i)\implies(ii)$: if we have $\|x+y\|=\|x\|+\|y\|$, we can rewrite this as
$$
\Big\|\frac{\|x\|}{\|x\|+\|y\|}\,\frac{x}{\|x\|}+\frac{\|y\|}{\|x\|+\|y\|} \,\frac{y}{\|y\|}\Big\|=1.
$$
Now $(1)$ applies and we get that $x/\|x\|=y/\|y\|$.
$(ii)\implies(iii)$: let $\phi\in X^*$ and $x,y\in X$ with $\phi(x)=\|x\|$, $\phi(y)=\|y\|$. By scaling $x$ and $y$ if needed, we may assume that  $\|\phi\|=1$. Then
$$
\|x\|+\|y\|=\phi(x)+\phi(y)=\phi(x+y)\leq\|x+y\|. 
$$
This gives us equality in the triangle inequality, and so by $(ii)$ we have that $y=\lambda x$. Then
$$
\lambda\,\|y\|=\lambda\phi(y)=\phi(\lambda y)=\phi(x)=\|x\|.
$$
This forces $\lambda>0$.
$(iii)\implies(i)$: Suppose that $u,v$ are unit vectors with $\|\frac{u+v}2\|=1$. Use Hahn-Banach to construct $\phi\in X^*$ with $\|\phi\|=1$ and $\phi(u+v)=\|u+v\|$. Then
$$
2=\phi(u+v)=\phi(u)+\phi(v)\leq\|\phi\|\,\|u\|+\|\phi\|\,\|v\|=2.  
$$
As in the observation at the beginning, this forces $\phi(u)=\phi(v)=1$. Hence $\phi(u)=\|u\|$, $\phi(v)=\|v\|$, and then by $(iii)$ we have $u=\lambda v$ for some $\lambda>0$. Since $\|u\|=\|v\|=1$, we get $\lambda=1$ and $u=v$.

$(1)$ is first shown by induction, where the base case is $(i)$, for $t=k/2^n$.  The inductive hypothesis is, for fixed $n$ and $k=1,\ldots,2^n$,
$$\tag2
\|x\|=\|y\|=1,\ \big\|\tfrac{k}{2^n}\,x+\big(1-\tfrac{k}{2^n}\big)\,y\big\|=1\ \implies\ x=y.
$$
If we now assume $(2)$ and we have $\|x\|=\|y\|=1$ and $k\in\{1,\ldots,2^{n+1}\}$,
$$
\big\|\tfrac{k}{2^{n+1}}\,x+\big(1-\tfrac{k}{2^{n+1}}\big)\,y\big\|=1,
$$
we assume without loss of generality that $k\leq 2^n$ (otherwise we switch the roles of $x$ and $y$).
We can write
$$\tag3
1=\big\|\tfrac{k}{2^{n+1}}\,x+\big(1-\tfrac{k}{2^{n+1}}\big)\,y\big\|
=\big\|\tfrac{k}{2^{n}}\big(\frac{x+y}2\big)+\big(1-\tfrac{k}{2^{n}}\big)\,y\big\|.
$$
In principle we only know that $\big\|\frac{x+y}2\big\|\leq1$, but the observation at the beginning gives us that $\big\|\frac{x+y}2\big\|=1$. Then  the inductive hypothesis $(2)$ gives us
$$
\frac{x+y}2=y,\ \text{ which is} \ x=y. 
$$
After having all dyadic $t$, one can get to arbitrary $t\in[0,1]$ by continuity of the norm.
