How to prove that this set is convex.

Question

Let $\mathbb{C}$ and $\mathbb{D}$ be two convex sets. Prove that the set $E:=\bigcup\limits_{\lambda \in[0,1]}((1-\lambda)C\cap\lambda D)$ is also convex.

I have two problems here: I don't know how to manage this expresion to prove that it is convex and I can't see this type of set in $\mathbb{R}^2$

Answer 1

Suppose $e_1,e_2\in E$ and let $0\le t\le 1.$ There are $0\le \lambda\le 1;\ c_1,c_2\in C;\ d_1,d_2\in D$ such that $e_1=(1-\lambda) c_1=\lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2\in E.$

Now, it is easy to see that $(1-\lambda)C$ and $\lambda D$ are convex sets.

Then,

$x=(1-t)e_1+te_2=(1-t)(1-\lambda) c_1+t(1-\lambda) c_2\in (1-\lambda)C.$

and

$x=(1-t)e_1+te_2=(1-t)\lambda d_1+t\lambda d_2\in \lambda D.$

Thus, $x\in (1-\lambda)C\cap \lambda D$, so $E$ is convex.

As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $\lambda$ and the other by $\lambda$. Then, try strips. Etc. In short, play with shapes!

Answer 2

Let $x,y\in E$, so that there exists $\lambda, \mu\in [0,1]$ such that $$ x = (1-\lambda) c_x + \lambda d_x, \qquad y = (1-\mu) c_y + \mu d_y, $$ with $c_x, c_y\in C$ and $d_x, d_y\in D$.

Let $t\in (0,1)$ and let us prove that $(1-t)x + ty \in E$. Setting $s := (1-t)\lambda + t \mu$, it is easy to check that $$ \begin{gather*} \frac{(1-t)(1-\lambda)}{1-s} c_x + \frac{t(1-\mu)}{1-s} c_y =: c_z \in C,\\ \frac{(1-t)\lambda}{s} d_x + \frac{t\mu}{s} d_y =: d_z \in D \end{gather*} $$ (by the convexity of $C$ and $D$), and $$ (1-t)x + t y = (1-s) c_z + s d_z \in E. $$ (Here we have implicitly assumed that $s\in (0,1)$; the cases $s=0$ and $s=1$ are trivial.)