I would like to check if the proof of the following result is true.
Proposition: Let us note $y_u:[0,T]\mapsto \mathbb{R}^n$ the solution of the following ODE $$y_u'(t) := f(y_u(t), u(t))$$ where $y_u(0) = y^0$ is fixed and $u\in L^\infty([0,T];U)$ with $U$ a closed convex bounded subset of $\mathbb{R}^m$ and $f$ is differentiable and uniformly Lipschitz. Let $g:\mathbb{R}^n\times \mathbb{R}^m\mapsto \mathbb{R}$ a differentiable uniformly Lipschitz function. Let $(u_n)_n$ be a sequence converging to $\bar{u}$ in the weak $L^2$ topology. Define the sequence $(G_n)_n$ as follows $$G_n(t) := \int_0^t g(y_{u_n}(s),u_n(s))ds;\;\; G_n(0) = 0$$ Then $(G_n)_n$ uniformly converges to $G$ solution of $$G(t) := \int_0^t g(y_{\bar{u}}(s),\bar{u}(s))ds;\;\; G(0) = 0$$
Proof: First, $\forall t_1,t_2\in[0,T]$, From Hölder inequality, we have
$$\parallel G_n(t_1) - G_n(t_2)\parallel \leq \parallel g(y_{u_n} ,u_n)\parallel_{L^1([t_1,t_2])}\leq\sup_n \parallel g(y_{u_n}, u_n)\parallel_{L^2}\sqrt{\vert t_1 - t_2\vert}$$
Therefore, the sequence $(G_n)_n$ is bounded and equicontinuous. From Arzela-Ascoli, it contains a uniformly converging subsequence to some $L$. Let us prove that the whole sequence $(G_n)_n$ uniformly converges to $G$ by contradiction. Assume that, there exists a subsequence $(G_{n_k})_k$ such that $\exists K>0$ and $\epsilon >0$ satisfying
$$\Vert G_{n_k}-L \Vert_{L^\infty}\geq \epsilon\;\;\forall k\geq K$$
One can extract a sub-subsequence $(G_{n_{k_j}})_j$ uniformly converging to some $G_1$ with $\parallel G_1 - L\parallel_{L^\infty}>0$. However, $L^2([0;T];U)$ being weakly compact, one can extract from $(u_{n_{k_j}})_j$ a weakly convergent subsequence. By definition, this sequence weakly converges to $\bar{u}$ and proves that $(G_{n_{k_j}})_j$ contains a subsequence converging to $L=G(y_{\bar{u}}, \bar{u})$ which contradicts the initial assumption and proves the result.