Show that a recurrence relation is bounded

Question

I'm studying the following recursive equation as part of my Economics class:

$$k_{t+1}={k_t}^\alpha - c_t$$

For $t=0,1,2,...$, $0\leq \alpha<1$, and $k_0>0$ and $c_t\geq 0$ for every $t$. I need to show that there is a $\hat{K}$ such that $\hat{K}\geq k_t$ for every $t$.

I tried taking $t=0$ and $c_t=0$, but that didn't help cause I only got:

$$k_{t+1} \leq {k_t}^{\alpha} \leq ... \leq {k_1}^{(t-1)\alpha} \leq {k_0}^{t\alpha} $$

What would you suggest?

Answer 1

The case when $c_t=0$ for all $t$ gives $k_t={k_0}^{\alpha^t}$. For $k_0<1$, the sequence is increasing but is bounded by 1. Otherwise the sequence is decreasing and is bounded by $k_0$.

The case when when $c_t \geq 0$ for all $t$ can run into trouble if $k_t$ ever becomes negative since powers can be complex. Assuming $k_t \geq 0$, the sequence can be bounded by a different sequence from the first case. Define $h_0=k_0$ and $h_{t+1}={h_t}^{\alpha}$. Since $c_t \geq 0$, $k_t \leq h_t$ for all $t$.