Plotting some kind of convex hull I have a set of points $x_i \in \mathbb{R}^2$ and want to plot its convex hull with an additional condition. What I want to plot exactly is:
\begin{align}
\left\{\sum\limits_i \alpha_i x_i\middle| \sum\limits_i \alpha_i = 1, \ 0 \leq \alpha_i \leq \mu \right\}
\end{align}
How can I do that? I think it is not possible (in an easy way) in mathematica.
I just need some pictures, so it is ok for me if I can only take a very small set.
 A: Let $x_i = (u_i,v_i)$.  The desired set of points is the set of the $(u,v) \in \mathbb{R}^2$ that satisfy
$$\begin{align}
u &= \sum_i \alpha_i u_i \\
v &= \sum_i \alpha_i v_i \\
1 &= \sum_i \alpha_i \\
0 &\leq \alpha_i \leq \mu \enspace.
\end{align}$$
Splitting each equality into two inequalities, we get a set of inequalities in $u$, $v$, and the $\alpha_i$'s.  If we eliminate the $\alpha_i$'s with, say, Fourier-Motzkin elimination, we obtain constraints on $u$ and $v$ that describe the desired set.
As an example, suppose we are given
$$ x_0 = (0,0), x_1 = (1,0), x_2 = (0,1), x_3 = (1,1) \enspace,$$
and $\mu=0.5$.  Noting that we can do without $\alpha_0$ if we replace $\sum_i \alpha_i = 1$ with $\alpha_1 + \alpha_2 + \alpha_3 \leq 1$, the initial inequalities are:
$$\begin{align}
u &\leq \alpha_1 + \alpha_3 \\
u &\geq \alpha_1 + \alpha_3 \\
v &\leq \alpha_2 + \alpha_3 \\
v &\geq \alpha_2 + \alpha_3 \\
1 &\geq \alpha_1 + \alpha_2 + \alpha_3 \\
0 &\leq \alpha_1 \leq \mu \\
0 &\leq \alpha_2 \leq \mu \\
0 &\leq \alpha_3 \leq \mu
\end{align}$$
For $\mu=0.5$, elimination produces
$$\begin{align}
0 \leq u &\leq 1 \\
0 \leq v &\leq 1 \\
u - v &\leq 0.5 \\
v - u &\leq 0.5 \\
u + v &\leq 1.5 \enspace.
\end{align}$$
These inequalities delimit a pentagon, which is easily seen to be the desired set.  As expected, the pentagon is entirely contained in the convex hull of $x_0,x_1,x_2,x_3$.
