I am facing the problem of showing the existence of a solution to the following integral equation: \begin{equation}\label{Eq: IE} z(t) = e^{\lambda(T-t)-\beta\int_{t}^{T}z(s)^{\frac{1}{\beta-1}}ds}\rho(t) + \int_{t}^{T}e^{\lambda(\tau-t)-\beta\int_{t}^{\tau}z(s)^{\frac{1}{\beta-1}}ds}z(\tau)^{\frac{\beta}{\beta-1}}\nu(t,\tau)d\tau, \quad t\in[0,T] \end{equation} where $\nu:(t,\tau)\in \{[0,T]^2: 0 \leq t \leq \tau \leq T\} \mapsto (0,\beta], \quad \beta\in(0,1)$ and $\rho: [0,T] \mapsto (0,\infty)$ are given continuous functions.
It is proved in the reference article that the following bounds on the solution $z:[0,T]\mapsto(0,\infty)$ of the above integral equation hold: $$e^{\lambda-\bar{\lambda}(T-t)}\min_{t\in[0,T]}\rho(t) \leq z(t) \leq e^{\lambda(T-t)}\max_{t\in[0,T]\rho(t)}\rho(t), \qquad t\in[0,T]$$ with $\bar{\lambda}=\sup_{0 \leq t \leq s \leq T}\frac{-ln(\nu(t,s))}{s-t} < \infty$.
The author states that we can obtain the existence of a solution $z(t)$ by means of the contraction mapping theorem, which gives us the local solvability and then exploiting a continuation argument.
Therefore, following his reasoning, I should show that the operator $T: (C[0,T], \|\cdot\|_{\infty}) \mapsto (C[0,T], \|\cdot\|_{\infty}) $ which defines the integral equation, i.e. $$T(w(t)) = e^{\lambda(T-t)-\beta\int_{t}^{T}w(s)^{\frac{1}{\beta-1}}ds}\rho(t) + \int_{t}^{T}e^{\lambda(\tau-t)-\beta\int_{t}^{\tau}w(s)^{\frac{1}{\beta-1}}ds}w(\tau)^{\frac{\beta}{\beta-1}}\nu(t,\tau)d\tau$$ is actually a local contraction on $[T-\epsilon, T]$, namely that for some $L\in(0,1)$ it holds: $$\|T(z_1(t))-T(z_2(t))\|_{\infty} = \sup_{t\in[T-\epsilon,T]}|T(z_1(t))-T(z_2(t))| \leq L\sup_{t\in[T-\epsilon,T]}|z_1(t)-z_2(t)| = L \|z_1(t)-z_2(t)\|_{\infty}$$
However, I am having some problems to show the above. Can someone provide the necessary steps to show that it actually holds ?