Vrftjkry

Let $S(X)= {x in X: ||x||=1}$ be the unit sphere in $X$. Assume that there are $x,yin X$ linearly independent such that $||x+y||=||x||+||y||$. Prove that $S(X)$ contains the following set:$[x,y]={zin X: z=tx+(1-t)y, tin [0,1]}$ 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|) & = (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*}
|that{x} + (1 - t)hat{y}| & =
leftlVertfrac{t}{|x|}x + frac{1 - t}{|y|}yrightrVert \
& = frac{t}{|x|}|x| + frac{1 - t}{|y|}|y| \
& = 1.
end{align*}

Define
$$
phi(alpha)=|x+alpha y|-|x|-alpha|y|,quad alphage 0.
$$
It is a convex function, $phi(alpha)le 0$ and $phi(0)=phi(1)=0$. Then from convexity $phi(alpha)=0$, $forallalphage 0$. Hence,
$$
|x+alpha y|=|x|+alpha|y|,quad forallalphage 0.
$$
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
$$
|that x+(1-t)hat y|=1.
$$

Pick an arbitrary $pin[frac12,1]$ and let $q=1-pin[0,frac12]$. 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 $pin[0,1]$.

Consequently, $|ax+by|=a|x|+b|y|$ for every $a,bge0$. 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 $a,bge0$. In particular, when $a=tin[0,1]$ and $b=1-t$, we have $[u,v]subset S(X)$.

From $|x+y|=|x|+|y|$,
$$|x|^2+2|x||y|+|y|^2=|x+y|^2=|x|^2+|y|^2+2langle x,yrangle,$$
hence $langle x,yrangle =|x||y|$.
Then
$$|ax+by|^2=a^2|x|^2+b^2|y|^2+2ablangle x,yrangle =a^2|x|^2+b^2|y|^2+2ab|x||y|=(a|x|+b|y|)^2,$$
i.e.,
$$ |ax+by|=a|x|+b|y|.$$
In particular, with $a=frac t{|x|}$ and $b=frac{1-t}{|y|}$,
$$ left|tfrac{x}{|x|}+(1-t)frac{y}{|y|}right|=1.$$

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 $(x+y)/lVert x + yrVert$, and $R$ is the restriction of the closed unit ball to the span of $x$ and $y$.