I'm trying to solve the following question:
Suppose $V$ is an inner product space and $B$ is the open unit ball in $V$ (thus $B = \{ f \in V : \Vert f \Vert < 1 \}$ ). Prove that if $U$ is a subset of $V$ such that $B \subset U \subset \bar{B}$, then $U$ is convex. Here $\bar{B}$ is the closure of $B$.
Can I get a hint? I have tried to use the fact that it's an inner product space and played around with Cauchy Schwartz inequality, but to no avail. I'm completely stuck.
Let $x,y \in U$ and $0<t<1$. Then $\|tx+(1-t)y\|\leq t\|x\|+(1-t)\|y\| \leq t+(1-t)=1$. If we have strict inequality then $tx+(1-t)y \in B \subset U$. If equality holds then $tx$ and $(1-t)y$ are nonnegative multitples of each other: $tx =s(1-t)y$ for some $s \geq 0$. So $x =ry$ for some $r \geq 0$. The case where $\|x\|<1$ or $\|y\|<1$ is trivial. If you assume that $\|x\|=\|y\|=1$ the we get $r=1$ and $x=y$. This completes the proof.
I have used the fact that $\|x+y\|=\|x\|+\|y\|$ implies $x =ay$ for some $a \geq 0$ (or $y=ax$ for some $a \geq 0$). This is true in any inner product space.