# How to prove that this set is convex.

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 $$\mathbb{R}^2$$

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!

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.)