If there are $2$ linearly independent vectors $x,y \in X$ such that $||x+y||=||x||+||y||$, then the unit sphere $S(X)$ contains an interval Let $S(X)= \{x \in X: ||x||=1\}$ be the unit sphere in $X$. Assume that there are $x,y\in X$ linearly independent such that $||x+y||=||x||+||y||$. Prove that $S(X)$ contains the following set:$[x,y]=\{z\in X: z=tx+(1-t)y, t\in [0,1]\}$ for some $x,y$.


So it is obvious that I need to use the $x,y$ that are given to be linearly independent and form a $[x,y]$ in $S(X)$ but I don't know how to start.
This is similar to user1551's answer, but a little simpler.

For the given $x, y$, and all $\lambda,\mu \geqslant 0$, we have
as usual
$$
\|\lambda x + \mu y\| \leqslant \lambda\|x\| + \mu\|y\|,
$$
but also
\begin{align*}
(\lambda + \mu)(\|x\| + \|y\|) &amp; = (\lambda + \mu)(\|x + y\|) \\
&amp; = \|(\lambda + \mu)(x + y)\| \\
&amp; = \|(\lambda x + \mu y) + \mu x + \lambda y\| \\
&amp; \leqslant \|\lambda x + \mu y\| + \mu\|x\| + \lambda\|y\|,
\end{align*}
therefore
$$
\|\lambda x + \mu y\| \geqslant \lambda\|x\| + \mu\|y\|,
$$
therefore
$$
\boxed{\|\lambda x + \mu y\| = \lambda\|x\| + \mu\|y\|}
$$
Putting $\hat{x} = x/\|x\|$, $\hat{y} = y/\|y\|$, we have
$\hat{x} \ne \hat{y}$, $\|\hat{x}\| = \|\hat{y}\| = 1$, and if
$0 \leqslant t \leqslant 1$,
\begin{align*}
\|t\hat{x} + (1 - t)\hat{y}\| &amp; =
\left\lVert\frac{t}{\|x\|}x + \frac{1 - t}{\|y\|}y\right\rVert \\
&amp; = \frac{t}{\|x\|}\|x\| + \frac{1 - t}{\|y\|}\|y\| \\
&amp; = 1.
\end{align*}
From $\|x+y\|=\|x\|+\|y\|$,
$$\|x\|^2+2\|x\|\|y\|+\|y\|^2=\|x+y\|^2=\|x\|^2+\|y\|^2+2\langle x,y\rangle,$$
hence $\langle x,y\rangle =\|x\|\|y\|$. 
Then
$$\|ax+by\|^2=a^2\|x\|^2+b^2\|y\|^2+2ab\langle x,y\rangle =a^2\|x\|^2+b^2\|y\|^2+2ab\|x\|\|y\|=(a\|x\|+b\|y\|)^2,$$
i.e., 
$$ \|ax+by\|=a\|x\|+b\|y\|.$$
In particular, with $a=\frac t{\|x\|}$ and $b=\frac{1-t}{\|y\|}$, 
$$ \left\|t\frac{x}{\|x\|}+(1-t)\frac{y}{\|y\|}\right\|=1.$$