A $\mathcal{K}$ function $\alpha: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ obeys $\alpha(0) = 0$ and $\alpha(a) > \alpha(b)$ for $a > b$. An identity function $\mathrm{id}: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ obeys $\mathrm{id}(s) = s$.
Suppose we have some $\mathcal{K}$ function $\alpha < \mathrm{id}$, i.e., $\alpha(s) < \mathrm{id}(s)$ for all $s > 0$. Given an $s \in \mathbb{R}_+$, do we have $\lim_{k \rightarrow \infty}\underbrace{\alpha\circ\alpha\circ \cdots \circ \alpha}_{k}(s) = 0$ (here $\circ$ denotes composition of functions)?
My intuition is that the above claim may not be true, and the answer could be $s - c$ with any constant $c < s$. But I have no idea how to proceed the proof, and I can not come up with some example.
The infinite composition of functions is equivalent to the discrete time system
$$ s_{k+1}=\alpha(s_k) $$
, $\mathcal{K}$-function $\alpha$ and $\alpha(s)<\text{id}(s), s>0$. So you are practically asking if this discrete system (with unique equilbrium at $0$) is asymptotically stable for all $s_0 \in \mathbb{R}_+$.
Since we consider non-negative $s$, use the Lyapunov function $V(s)=s$.
Then $V(s_{k+1})-V(s_k)=\alpha(s_k)-s_k=\alpha(s_k)-\text{id}(s_k)<0$ for $s_k>0$ so your limit should always converge to zero.