Show triangle equality implies colinearly Let $X$ be a normed vector space and suppose the following holds;
If $f \in X^*$ is a nonzero bounded linear functional then there exists
at most one element $x \in X$ such that $||x|| = 1$ and $f(x)=||f||$ (This is supposed to be equivalent to strict convexity).
I want to show that if $x \neq 0 \neq y$ and $||x+y||=||x||+||y||$ then $x$ and $y$ are colinear.
It feels like I'm supposed to use Hahn-Banach here (how else would I get a functional?).
I tried the obvious things like $M=\mathbb{F}y$ or $\mathbb{F}x+\mathbb{F}y$ and defining some functional like $f(cy)=c||y||$ but it doesn't work out.
EDIT:
Since it hasn't been commented on yet I guess its not trivial (or maybe so trivial that its not interesting). Ill share some more details;
The most naïve approach is to define $f:\mathbb{F}y \rightarrow \mathbb{F}$ by $f(cy)=c||y||$ which meets H.B conditions (with $p(x)=||x||\geq f(x)$) so we can extend it to a functional on X, call the extension $F$.
set $v=y/||y||$ and we see $||F(x)||<||x||$ (since this was our seminorm) so $||F|| \leq 1$  and $|F(v)|=f(v)=1$ so $||v||=||F||=1$ and we are done if we can show $F(\frac{x}{||x||})=1$ somehow using $||x||+||y||=||x+y||$ (this is where I"m stuck).
The problem is from https://people.math.ethz.ch/~salamon/PREPRINTS/funcana-ams.pdf on page 106 (118 of the PDF).
 A: First, we show that $X$ is strictly convex, that is, for any $x\neq y \in X$ with $||x||=||y||=1$, we get that $||\frac{x+y}{2}|| <1$. Indeed, suppose  that $X$ is not strictly convex and so there exist $x \neq y \in X$ with $||x||=||y||=1 $ and $||\frac{x+y}{2}|| =1$. By the Hahn-Banach theorem, we can find $f \in X^*$ such that $f(\frac{x+y}{2}) = ||\frac{x+y}{2}|| =1 $ and $||f||=1$. Since $f$ is linear, we infer that
$$\frac{f(x)+f(y)}{2} =1 .$$
Since $f(x) \leq 1 $ and $f(y) \leq 1$, it must be true that $f(x)=1$ and $f(y)=1$. By the hypothesis, we get that $ x=y$ which is impossible.
We now show that if $X$ is strictly convex, then for any $x_1,x_2 \in X$, if $||x_1+x_2||=||x_1||+||x_2||$, then $x_1=ax_0$ for some $a>0$. Let $x_1,x_2 \in X$ with $||x_1|| +||x_2||=||x_1+x_2||$ . Without loss of generality, we may assume that $ ||x_1||=1$. Let $ y= x_2/||x_2||$. Then,
\begin{align*}
    2 & \geq  ||{x_1+y}||   \\
    &= ||x_1+x_2 - (1- ||x_2||^{-1} ) x_2|| \\ 
    &\geq ||x_1+x_2|| - (1- ||x_2||^{-1}) ||x_2|| \\
    &= ||{x_1}||+||{x_2}||-||{x_2}||+1\\ 
    &=2
   \end{align*}
and so $|| \frac{x_1+y}{2}=1|| $. Since $X$ is strictly convex, we infer that $x_1=y_1= ||x_2||^{-1} x_2$.
