-2\mu)$, with slope $-\mu/{\lambda}$. Consider now the line from $\pi(p^\infty)$ to any other point $(x,y)=\pi(a_0a_1\dots)\in A_{-{\lambda},\mu}$. This will have a direction of the form $$\begin{aligned}
     \pi(p^\infty) - \pi(a_0 a_1 \dots) &=
    \left(\sum_{i=0}^\infty (1-a_i) (-{\lambda})^i, \sum_{i=0}^\infty (1-a_i) \mu^i\right)\\
   &=
          \left(\sum_{i\text{ even}} (1-a_i) {\lambda}^i, \sum_{i\text{ even}} (1-a_i) \mu^i\right)   \\ &
          + \left(-\sum_{i\text{ odd}} (1-a_i) {\lambda}^i, \sum_{i\text{ odd}} (1-a_i) \mu^i\right).\end{aligned}$$ Clearly, no point in $A_{-{\lambda},\mu}$ can have larger $y$-coordinate that $p^\infty$, whence the second coordinate is always non-negative. If the first coordinate is positive as well, then we are done; so, let us assume that it is negative. We notice that the slope has the form: $$\frac{ \sum_{i\text{ even}} (1-a_i) \mu^i  + \sum_{i\text{ odd}} (1-a_i) \mu^i}
     { \sum_{i\text{ even}} (1-a_i) {\lambda}^i -\sum_{i\text{ odd}} (1-a_i) {\lambda}^i }
\le  -\frac{ \sum_{i\text{ odd}} (1-a_i) \mu^i} { \sum_{i\text{ odd}} (1-a_i) {\lambda}^i }$$ (since ${\lambda}<\mu$). We want $$\frac{ \sum_{i\text{ odd}} (1-a_i) \mu^i} { \sum_{i\text{ odd}} (1-a_i) {\lambda}^i } \le -\frac{\mu}{{\lambda}}.$$ Cross multiplying, this will occur if $${\lambda}\sum_{i\text{ odd}} (1-a_i) \mu^i
  \geq
  \mu \sum_{i\text{ odd}} (1-a_i) {\lambda}^i$$ or, equivalently, $$\sum_{i\text{ odd}} (1-a_i) \mu^{i-1}
  \geq
  \sum_{i\text{ odd}} (1-a_i) {\lambda}^{i-1}.$$ This is clearly true, as ${\lambda}< \mu$. This proves the base case $k=0$.

![Convex hull for $A_{-0.55, 0.8}$[]{data-label="fig:hull-mixed"}](conv-minus055-08.eps){width="275pt" height="275pt"}

Assume the 