Consider $C^0([t_0,t_1])$ equipped with the norm given by $$\Vert x\Vert=\max_{t\in[t_0,t_1]}|x(t)|e^{-\alpha t}$$ for some $\alpha>0$.
Now, consider $T\colon C^0([t_0,t_1])\to C^0([t_0,t_1]),\, (Tx)(t)=x_0+\int_{t_0}^t f(\tau,x(\tau))\operatorname{d}\tau$, for some $x_0\in\mathbb{R}$ and the map $f\colon [t_0,t_1]\times\mathbb{R}\to\mathbb{R}$ which, additionally, has the following property: $$|f(t,x_1)-f(t,x_2)|\leq L|x_1-x_2|,\quad t\in[t_0,t_1],\quad x_1,x_2\in\mathbb{R}$$ for some $L\geq 0$.
I have to show that $$\Vert Tx-Ty\Vert\leq\frac{L}{\alpha}\Vert x-y\Vert,\quad x,y\in C^0([t_0,t_1]).$$
How? I am not able to extrapolate $\alpha$.
Note that for any $t \in [t_0, t_1]$, we have \begin{align*}\lvert Tx(t) - Ty(t)\rvert e^{-\alpha t} &= e^{-\alpha t} \left \lvert\int^t_{t_0} [f(\tau,x(\tau)) - f(\tau,y(\tau))] d \tau \right \rvert\\ &\le e^{-\alpha t} \int^t_{t_0} \lvert f(\tau,x(\tau)) - f(\tau,y(\tau)) \rvert d \tau \\ &\le Le^{-\alpha t}\int^t_{t_0} \lvert x(\tau) - y(\tau) \rvert d\tau \\ &= Le^{-\alpha t} \int^t_{t_0} e^{\alpha \tau} \left(\lvert x(\tau) - y(\tau) \rvert e^{-\alpha \tau} \right) d\tau\\ &\le L \|x - y\| e^{-\alpha t} \int^t_{t_0} e^{\alpha \tau} d\tau\\ &= \frac{L}{\alpha} \| x - y\| e^{-\alpha t} (e^{\alpha t} - e^{\alpha t_0})\\ &= \frac L \alpha \|x - y\| (1- e^{-\alpha(t - t_0)})\\ & \le \frac L \alpha \|x - y \|. \end{align*} Thus taking the supremum over all such $t$, we arrive at $$\| Tx - Ty \| \le \frac L \alpha \|x - y\|.$$