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.
 A: 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\|) & = (\lambda + \mu)(\|x + y\|) \\
& = \|(\lambda + \mu)(x + y)\| \\
& = \|(\lambda x + \mu y) + \mu x + \lambda y\| \\
& \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}\| & =
\left\lVert\frac{t}{\|x\|}x + \frac{1 - t}{\|y\|}y\right\rVert \\
& = \frac{t}{\|x\|}\|x\| + \frac{1 - t}{\|y\|}\|y\| \\
& = 1.
\end{align*}
A: Define
$$
\phi(\alpha)=\|x+\alpha y\|-\|x\|-\alpha\|y\|,\quad \alpha\ge 0.
$$
It is a convex function, $\phi(\alpha)\le 0$ and $\phi(0)=\phi(1)=0$. Then from convexity $\phi(\alpha)=0$, $\forall\alpha\ge 0$. Hence,
$$
\|x+\alpha y\|=\|x\|+\alpha\|y\|,\quad \forall\alpha\ge 0.
$$
Now define
$$
\hat x=\frac{x}{\|x\|},\quad \hat y=\frac{y}{\|y\|},\quad 
t=\frac{\|x\|}{\|x\|+\alpha\|y\|}\in(0,1].
$$
We have
$$
\|t\hat x+(1-t)\hat y\|=1.
$$
A: Let's say we have a closed, bounded, convex region $R$ in the plane. Suppose that 


*

*distinct points $O$, $A$, $B$, and $C$ lie in $R$,

*$A$, $B$, and $C$ lie on the boundary and $O$ lies on the interior, and 

*$C$ lies on line segment $AB$.


Then the whole of line segment $AB$ lies on the boundary $R$. For suppose to the contrary that some point $D$ on $AB$ does not: then ray $OD$ strikes the boundary of $R$ at some distinct point $E$, since $R$ is bounded, and all of segments $BE$ and $EA$ lie in $R$. Assume without loss of generality that $D$ is between $B$ and $C$. Then ray $OC$ strikes segment $EA$ at some point $F$ and $C$ lies between $O$ and $F$. But that's impossible, since that would put $C$ on the interior of $R$.
To map back to the original problem: the plane in question is the span of $x$ and $y$, $O$ is the zero vector, $A$ is $x/\lVert x \rVert$, $B$ is $y/\lVert y \rVert$, $C$ is $(x+y)/\lVert x + y\rVert$, and $R$ is the restriction of the closed unit ball to the span of $x$ and $y$.

A: Pick an arbitrary $p\in[\frac12,1]$ and let $q=1-p\in[0,\frac12]$. Then
\begin{align}
\|x\|+\|y\| = \|x+y\|
&= \|(px+qy)+(qx+py)\|\\
&\le \|px+qy\|+\|qx+py\|\\
&\le(\|px\|+\|qy\|)+(\|py\|+\|qx\|)\tag{1}\\
&=(p\|x\|+q\|y\|)+(p\|y\|+q\|x\|)\\
&=\|x\|+\|y\|
\end{align}
and hence equalities must hold in $(1)$. Therefore, $\|px+(1-p)y\|=\|px\|+\|(1-p)y\|=p\|x\|+(1-p)\|y\|$ for every $p\in[0,1]$.
Consequently, $\|ax+by\|=a\|x\|+b\|y\|$ for every $a,b\ge0$. As $x,y$ are linearly independent, they are nonzero and we may normalise them to unit vectors $u=\frac{x}{\|x\|}$ and $v=\frac{y}{\|y\|}$. By absorbing $\|x\|,\|y\|$ into $a,b$ respectively, we obtain $\|au+bv\|=a+b$ for every $a,b\ge0$. In particular, when $a=t\in[0,1]$ and $b=1-t$, we have $[u,v]\subset S(X)$.
A: 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 $b=\frac{1-t}{\|y\|}$, 
$$ \left\|t\frac{x}{\|x\|}+(1-t)\frac{y}{\|y\|}\right\|=1.$$
