I have a difference equation $$ p_t^{1-\alpha}=\alpha\sigma(y-p_t-\frac{(\sigma p_{t-1}^\alpha+b)p_t^{1-\alpha}}{\alpha\sigma}) $$ where $\alpha \in [0,1]$ and everything else is >0.
I need to prove that this equation has unique steady state.
This is what I have done so far;
Simplified the expression to write it in the closed form as follows; $$ p_{t-1}=\left[\frac{\alpha y}{p_{t}^{1-\alpha}}-\alpha p_{t}^{\alpha}-\frac{a+1}{\sigma}\right]^{1/\alpha} $$ Substituted $p_{t-1}=p_t=\overline{p}$ in the closed form, this gave. $$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$ I'm stuck here. How can I prove that $\overline{p}$ has a unique solution?
Well, what you want to study is the existence of solutions for $$ f(p;\alpha) = (1+\alpha)p^{\alpha} - \alpha y p^{\alpha-1} = - \frac{a+1}{\sigma}. $$ Where, as you've said, $\alpha \in [0,1]$, $y,\,a,\,\sigma >0$. First, is clear that for $\alpha = 0$, there is no solution for $p>0$ (why?). If $\alpha = 1$, then there is a solution if and only if $\alpha y - \frac{a+1}{\sigma} > 0$$.
Now, let $\alpha \in (0,1)$. $$ \text{What can you say about } \lim_{p\to -\infty} f(p;\alpha) \,\text{ and } \lim_{p\to \infty} f(p;\alpha)\text{?} $$ $$ \text{What about }f'(p;\alpha)\text{?} $$