Badly formed question? $\|x\|=1=\|y\|$ and $\|x+y\|=\|x\|+\|y\|$, there is a line segment in the unit sphere Show that if a normed linear space $X$ contains linearly independent vectors $x$ and $y$ such that $\|x\|=1$ and $\|y\|=1$ with $\|x+y\|=\|x\|+\|y\|$, then there is a line segment contained in the unit sphere of $X$.

Now since the unit sphere in $X$ is defined as $X=\{a\in X: \|a\|=1\}$, and this means that $x,y$ both lie on the unit sphere of $X$. But $\|x+y\|=2$ and hence they must lie on opposite ends of the sphere, and hence form a straight line through the origin. But then these two vectors aren't linearly independent?
Is this question badly formed or what am I missing?

You're thinking too Euclidean! It is entirely possible in an arbitrary normed linear space to have points other than antipodal points that satisfy this property. As an example, consider the points $(1, 0)$ and $(0, 1)$ in the unit sphere under the $1$-norm on $\mathbb{R}^2$.
Good question though!
