Given that $\sigma(x)\leq C\Vert x \Vert$ (Lipschitz continuous) I want to show that $\Vert X(1)\Vert<\infty$. We know that $X(t)$ satisfies $$ X'(t) = \sigma(K(t)X(t) + b(t)), $$
where $X: [0,1] \to \mathbb{R}^d, K\in H^1([0,1],\mathbb{R}^{d\times d}), b\in H^1([0,1], \mathbb{R}^d)$ and $X(0)=x\in\mathbb{R}^d$.
I heard about Picard-Lindelöf, and it seems similar, but it doesn't quite fit, since we have that $\sigma(K(t)X(t)+b(t) \leq C\Big\Vert K(t)X(t)+b(t)\Big\Vert$. So if we would rewrite as $X'(t)=f(t, X(t))$, we don't have that the Lipschitz constant only depends on $X(t)$, but also on $t$ (in $K$ and $b$).
I'm also not even sure if I need the differential equation, since I only need to know that $\Vert X(1)\Vert <\infty$.
Integrating the equation gives $$ X(t)-X(0) = \int_0^t\sigma(K(s)X(s) + b(s))ds $$ and hence $$ \|X(t)\|\le\|X(0)\|+C\int_0^t\bigg[\|K(s)\|\|X(s)\|+\|b(s)\|\bigg]ds. \tag{1} $$ Let $$ u(t)=\|X(t)\|, \alpha=\|X(0)\|+\int_0^t \|b(s)\|ds, \beta(t)=\|K(s)\| $$ and then (1) becomes $$ u(t)\le \alpha(t)+\int_0^t\beta(s)u(s)ds. \tag{2}$$ It is easy to see that $\alpha(t)$ is non-decreasing. By Gronwall's Inequality https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality, one has $$ u(t)\le \alpha(t)\exp\left(\int_0^t\beta(s)ds\right) $$ and hence $$ \|X(1)\|\le\left(\|X(0)\|+\int_0^1 \|b(s)\|ds\right) \exp\left(\int_0^1\|K(s)\|(s)ds\right)<\infty.$$