Consider the following recurrence: $$x_{n+1} = x_n g(x_n), \qquad n \geq 0$$ where $g : [0, \infty) \rightarrow [0, \infty)$ is strictly decreasing. I want to set some necessary and sufficient conditions to $g$ so that the equilibrium point $x = 0$ is asymptotically stable.
I don't know how to start, as I know nothing about $g$. If $g$ were differentiable, then I could be able to get something, but it may not be the case. Some help would be appreciated $:)$
On
I assume that we are in the scalar case. I ignore for now that $g$ is strictly decreasing.
A sufficient condition for global asymptotic stability is that $\sup_{x\ge0} |g(x)|\le1-\epsilon$ for some $\epsilon>0$. Since, $g$ is nonnegative, then one can drop the absolute value.
Define the Lyapunov function $V(x)=|x|$, them we have that $$V(x_{n+1})-V(x_n)=|x_ng(x_n)|-|x_n|=|x_n|\cdot (|g(x_n)|-1)\le-\epsilon|x_n|=-\epsilon V(x_n)$$ which proves global asymptotic (exponential) convergence to 0. Note that if the initial condition is nonnegative, we can drop the absolute value.
It is difficult to give a necessary condition because we would need to know more about the function $g$. For instance, the constraint on $g$ stated above could be violated for some particular values for $x$ and as long as we get back to a stable regime. For instance, we could have $g(x)>1$ for some region of the state-space.
If local asymptotic stability is also fine, then we just need to have $\sup_{x\in[0,r]} |g(x)|\le1-\epsilon$ for some $r>0$. Then, all the trajectories, starting from $x_0\in[0,r]$, will converge to zero.
If we consider now that $g$ is decreasing then we necessarily have $g(x)<g(0)$ and, as a result, we have global convergence to the equilibrium point $0$ if and only if $g(0)\le1$. The limit case where $g(0)=1$ can be treated analogously as in the Marko Karbevski's answer.
Would continuity suffice? If not, simply set $g(0) = \lim_{t \to 0^+}g(t)$. It doesn't change anything in terms of the dynamics.
Assume that $g(0)>0$, as otherwise the problem is trivial.
Take $M= g(0)= \max g$ (or $ = \lim_{t \to 0^+}g(t)$ in case you don't have continuity). Then for any initial condition $0<x_0 < \frac{1}{M}$ we have $0<x_{n} \le x_{n-1} M \le \cdots \le x_0 M^n$.
So a sufficient condition is that $M<1$.
On the other hand, if $M>1$, choose $M_1 = \frac{M+1}{2}$.
There exists a neighborhood $U$ of $0$ such that $\min_U g \ge M_1$. Now if $x_n \to 0$ there exists $N$ such that for all $n\ge N$ we have $x_n \in U$.
Hence $x_{k+N} = x_{k+N-1}g(x_{k+N-1}) \ge \cdots \ge x_{N}M_1^k \to_{k \to \infty} \infty$. A contradiction.
Finally, if $M=1$ it is clear that for any initial condition $x_0$ close to $0$, $x_n$ converges to a limit $l \ge 0$ because it's decreasing and bounded.
Assume that $l >0$. Note that for any $k$, $1>g(l)>g(x_k)$.
Let $N$ be set large enough so that for all $n \ge N$ we have $x_n - l < \frac{l-l \times g(l)}{g(l)}$. Then $x_{n+1} = x_{n} g(x_n) < (l+ \frac{l-l \times g(l)}{g(l)})g(l) = l$. A contradiction!
(Remark: I first arrived at the end and only then did I set the exact upper bound to $x_n - l$: $\frac{l-l \times g(l)}{g(l)}$. Before the end, I only knew that I should set a bound.)
In conclusion: $0$ is asymptotically stable iff $\lim_{t\to 0^+} g(t) \le 1$