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.

  • $\begingroup$ Hint: the dyadic rationals are dense in $[0,1]$. $\endgroup$ – Clement C. May 14 '13 at 19:38
  • $\begingroup$ Ok...so are you saying that I should add/subtract something and get my $\lambda$ using the density of these fractions or...? $\endgroup$ – Nicolo Canali De Rossi May 14 '13 at 20:14
  • $\begingroup$ Fix $x$, $y$; you can first prove the property for all $\lambda$ being dyadic rationals; then try to extend it to all $\lambda\in(0,1)$ by density. $\endgroup$ – Clement C. May 14 '13 at 20:17

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.