Consider the first-order non-linear difference equation: \begin{equation} k_{t+1} = sf(k_t) + (1-\delta)k_t \end{equation} where $s$ and $\delta$ are two constants and $s,\delta \in (0,1)$. I assume that $f: \mathbb{R}_+ \to \mathbb{R}_+$ is twice continuously differentiable and $f'(k)>0$ and $f''(k)<0$ for all $k>0$. So, $f(k)$ is assumed to be strictly concave.
At the steady state, $k^{\ast}$ (fixed point) satsifies the following equation: \begin{equation} s f(k^{\ast}) = \delta k^{\ast} \end{equation}
Now, pls, tell me if it's correct how I prove global stability:
First, \begin{equation} k_{t+1} = g(k_t) \end{equation} and for $k^{\ast}$ being a fixed point it must be \begin{equation} k^{\ast} = g(k^{\ast}) \end{equation}
Then, for all $k_t \in (0,k^{\ast})$, \begin{equation} k_{t+1} - k^{\ast} = g(k_t) - g(k^{\ast}) = - \int_{k_t}^{k^{\ast}} g'(k) dk <0 \end{equation} The inequality holds because $g'(k)>0$ for all $k$.
This result shall ensure that for any $k_t \in (0,k^{\ast})$, then $k_{t+1}$ is always lower than $k^{\ast}$, and the convergence of $k \to k^{\ast}$ is ensured from the fact that \begin{equation} \gamma(k_t) = \frac{k_{t+1} -k_t}{k_t} = \frac{s f(k_t) + (1-\delta)k_t - k_t}{k_t} = \frac{sf(k_t)}{k_t} - \delta \end{equation} and since $h(k_t) \equiv k_t/f(k_t)$ is strictly increasing in $k$ and $h(k^{\ast})=s/\delta$, then the growth rate of $k$ is positive but strictly decreasing in $k$, with $\gamma(k)>0$ iff $k< k^{\ast}$ and $\gamma(k)=0$ iff $k= k^{\ast}$.
A similar argument applies to $k_t > k^{\ast}$. Is that correct?