0$.

Therefore the complement of $\mathrm{supp}(\underline{F})$ is given by (recall that $\underline{F}$ has a point mass at $0$) $$\label{eq:suppc}
\mathrm{supp}(\underline{F})^c=(-\infty,0)\cup(0,\lambda_1)\cup(\lambda_2,\infty).$$ Let us now show that $\lambda_1<\lambda_2$. This would imply the support of $\underline{F}$ is non-empty and consists of one interval.

\[le:1cut\] Let $\lambda_1$ and $\lambda_2$ be the points defined by (\[eq:lambda\]), then $\lambda_1<\lambda_2$ and hence the support of $\underline{F}$ consists of a single interval.

Suppose $\lambda_1>\lambda_2$. Let $z_0\in(\lambda_2,\lambda_1)$. Since both the points $-1$ and $-\frac{1}{a}$ belongs to $(\gamma_1,\gamma_2)$, the function $z(\xi)$ is continuous in $(-\infty,\gamma_1)$ and $(\gamma_2,0)$. By Lemma \[le:image\], these two intervals are mapped onto $(0,\lambda_1)$ and $(\lambda_2,\infty)$ by the map $z(\xi)$. Hence there is at least one point in each of the intervals $(-\infty,\gamma_1)$ and $(\gamma_2,0)$ that is being mapped onto $z_0$. On the other hand, let $s_1$ and $s_2$ be the points in $\{-1,-\frac{1}{a}\}$ that is closer to $\gamma_1$ and $\gamma_2$ respectively, then $z(\xi)$ is continuous in $(\gamma_1,s_1)$ and $(s_2,\gamma_2)$. Since these intervals are mapped onto $(-\infty,\lambda_1,)$ and $(\lambda_2,\infty)$ respectively by $z(\xi)$, there is at least one point in each of these intervals that is mapped onto $z_0$. This would result in 4 distinct points being mapped onto $z_0$ by the map $z(\xi)$, which is not possible. 