I have an if and only if, and I am having trouble with one of the arrows! Here it is:
Let $C \subset \mathbb{R}^n$ such that the interior of $C$, $\operatorname{int} C \neq \emptyset$.
$C$ is strictly convex (i.e. $\forall x,y \in C$ then $\lambda x + (1-\lambda)y \in \operatorname{int}C$, $\lambda \in (0,1)$) $\Leftrightarrow \forall x,y \in C$ we have $\frac{1}{2} x + \frac{1}{2} y \in \operatorname{int} C.$
I have shown the $\Rightarrow$ simply by taking $\lambda=\frac{1}{2}$, but I have no idea on how to prove the other implication.
Thanks a lot in advance.
Given: if $x$ and $y$ are contained in $C$, then $\frac12x+\frac12y$ is in the interior of $C$.
Claim 1: If $\lambda=k/2^n$ where $n$ is an integer and $0<k<2^n$, then $$\frac{k}{2^n}x+\frac{2^n-k}{2^n}y \, \in \mathrm{int}\,C$$
Idea of proof: induction on $n$. If $k$ is even, then simplifying the fractions brings us back to $n-1$. If $k$ is odd, apply the inductive step to $$\frac{k-1}{2^n}x+\frac{2^n-(k-1)}{2^n}y, \quad \text{and} \quad \frac{k+1}{2^n}x+\frac{2^n-(k+1)}{2^n}y$$ Then take the average.
Claim 2: If $x \in \mathrm{int}\, C$, $y\in C$, and $0<\lambda<1$, then $\lambda x+(1-\lambda)y\in \mathrm{int}\,C$.
Idea of proof: We already know the result for dyadic rationals $\lambda$. The dyadic rationals are dense. Also, there is $r>0$ such that the ball of radius $r$ centered at $x$ is contained in $\mathrm{int}\,C$. This allows us to move $x$ around and cover other numbers $\lambda$ as well.
To be precise, suppose $\lambda x+(1-\lambda)y\notin \mathrm{int}\,C$. Pick $ n$ so that $1/2^n<r\lambda $. Pick $k$ so that $\lambda\le k/2^n\le \lambda+1/2^n$. Observe that $\mathrm{int}\,C$ contains a ball of radius $rk/2^n$ centered at $$\frac{k}{2^n}x+\frac{2^n-k}{2^n}y$$ Try to show that this ball covers $\lambda x+(1-\lambda)y$.
Claim 3: If $x,y \in C$ and $0<\lambda<1$, then $\lambda x+(1-\lambda)y\in \mathrm{int}\,C$.
This is easy now: replace either $x$ or $y$ (as appropriate) with $(x+y)/2$, which is in the interior of $C$. Then Claim 2 applies.