Exercise: Let $(M, d_M)$ be a metric space and $F: \mathbb{R}^{1+d} \times M \rightarrow~\mathbb{R}^{d}$ a limited function, lipschitz on the second variable and uniformly continuous on the following distance:
$$d((t, x, \lambda), (s, y, \mu)) = |t-s| + ||x-y|| + d_M(\lambda, \mu),$$
i.e, given $\epsilon > 0$, there exists $\delta >0$ such that
$$||F(t,x, \mu) - F(s,y,\lambda)|| < \epsilon, \text{whenever} \hspace{0.15cm} d((t,x,\mu), (s,y,\lambda)) < \delta.$$
Fixe $(t_0, x_0) \in \mathbb{R}^{1 + d}$ and let $t \mapsto \gamma_\lambda(t)$ be the maximal solution of $x'~=~F(t,x,\lambda)$ with initial condition $\gamma_\lambda(t_0) = x_0$. Show that $\gamma_\lambda(t)$ is continuous on the parameter $\lambda$: given $t \in \mathbb{R}$ and $\epsilon > 0$, there exists $\delta > 0$ such that
$$||\gamma_\lambda (t) - \gamma_\mu (t) || < \epsilon, \text{whenever} \hspace{0.15cm} d_M (\lambda, \mu) < \delta.$$
Attempt:
Consider $(t_0, x_0) \in \mathbb{R}^{1+d}$ and let $t \mapsto \gamma_{\lambda}(t)$ be the maximal solution of the diferential equation $x' = F(t,x, \mu)$, with initial condition $\gamma_{\lambda}(t_0) = x_0$.
Let $t \geq t_0$ and $\epsilon > 0$. The following proof is completely analogous for $t \leq t_0$.
Considering $\lambda, \mu$ in the metric space $(M, d_M)$, due to Picard's Theorem, we have
$$\gamma_{\lambda} (t) = x_0 + \displaystyle\int_{t_0}^t F(s, \gamma_{\lambda}(s), \lambda) ds \hspace{0.15cm} \text{and} \hspace{0.15cm} \gamma_{\mu} (t) = x_0 + \displaystyle\int_{t_0}^t F(s, \gamma_{\mu}(s), \mu) ds,$$
with $\gamma_{\mu}(t)$ solution of $x' = F(t,x, \mu)$, with initial condition $\gamma_{\mu} (t_0) = x_0$.
Therefore,
$ \begin{array}{ccl} || \gamma_{\lambda}(t) - \gamma_{\mu}(t)|| & = & || \displaystyle\int_{t_0}^t \left[ F(s, \gamma_{\lambda}(s), \lambda) - F(s, \gamma_{\mu}(s), \mu) \right] ds || \\ & \leq & \displaystyle\int_{t_0}^t || F(s, \gamma_{\lambda}(s), \lambda) - F(s, \gamma_{\mu}(s), \mu) || ds \\ & = & \displaystyle\int_{t_0}^t ||F(s, \gamma_{\lambda}(s), \lambda) - F(s, \gamma_{\mu} (s), \lambda) + F(s, \gamma_{\mu} (s), \lambda) - F(s, \gamma_{\mu}(s), \mu) || ds. \end{array}$
It follows that
$ \begin{array}{ccl} || \gamma_{\lambda}(t) - \gamma_{\mu}(t)|| & \leq & \displaystyle\int_{t_0}^t ||F(s, \gamma_{\lambda}(s), \lambda) - F(s, \gamma_{\mu} (s), \lambda) || ds + \displaystyle\int_{t_0}^t ||F(s, \gamma_{\mu} (s), \lambda) - F(s, \gamma_{\mu}(s), \mu) || ds.\end{array}$
Note that, since $F$ is lipschit on the second variable, there exists $C > 0$ such that
$$ || F(t, x_1, \mu) - F(t, x_2, \mu)|| < C ||x_1 - x_2||, \hspace{0.1cm} \forall \hspace{0.1cm} (t, x_1, \mu), (t, x_2, \mu) \in \mathbb{R}^{1+d} \times M. $$
We also know that $F$ is uniformly continuous on the distance $d$, so given $\epsilon'$, there exists $\delta > 0$ such that
$$||F(t,x, \mu) - F(s, y, \lambda)|| < \epsilon',$$
whenever $d((t,x,\mu), (s,y,\lambda)) < \delta$.
For $t=s$ e $x=y$, by the definition of the distance $d$, we get that
$$d_M(\lambda, \mu) < \delta \Rightarrow ||F(t,x,\mu) - F(t,x,\lambda)|| < \epsilon'.$$
From the information above mentioned, whenever $d_M(\lambda, \mu) < \delta$, we have
$ \begin{array}{ccl} || \gamma_{\lambda}(t) - \gamma_{\mu}(t)|| & \leq & \displaystyle\int_{t_0}^t ||F(s, \gamma_{\lambda}(s), \lambda) - F(s, \gamma_{\mu} (s), \lambda) || ds + \displaystyle\int_{t_0}^t ||F(s, \gamma_{\mu} (s), \lambda) - F(s, \gamma_{\mu}(s), \mu) || ds \\ & < & \displaystyle\int_{t_0}^t C || \gamma_{\lambda} (s) - \gamma_{\mu}(s) || ds + \displaystyle\int_{t_0}^t \epsilon' ds \\ & = & \displaystyle\int_{t_0}^t C ||\gamma_{\lambda}(s) - \gamma_{\mu}(s)|| ds + \epsilon' (t-t_0). \end{array}$
Since $t \geq t_0$, the function $\epsilon' (t-t_0)$ is continuous and not decreasing. Therefore, we can apply Gronwall's Lemma for the function $||\gamma_{\lambda}(t) - \gamma_{\mu}(t)||$. So,
$$ || \gamma_{\lambda}(t) - \gamma_{\mu}(t)|| < \epsilon' (t-t_0) e^{\displaystyle\int_{t_0}^t C ds} = \epsilon' (t-t_0)e^{C(t-t_0)}.$$
How do I take a proper $\epsilon'$ in order to make $||\gamma_{\lambda}(t) - \gamma_{\mu}(t)|| < \epsilon$? Does it somehow follow from the hypotesis that F is limited? I didn't use it so far. Any help would be appreciated!