Let $E$ be a $\mathbb R$-Banach space, $\tau>0$ and $v:[0,\tau]\times E\to E$ be Borel measurable with $$\sup_{t\in[0,\:\tau]}\left\|v(t,x)-v(t,y)\right\|_E\le c_1\left\|x-y\right\|_E\;\;\;\text{for all }x,y\in E\tag1$$ for some $c_1\ge0$.
Question 1: Can we (or which assumption do we need to) show $$\sup_{t\in[0,\:\tau]}\left\|v(t,x)\right\|_E\le c_2(1+\left\|x\right\|_E)\;\;\;\text{for all }x\in E\tag2$$ for some $c_2\ge0$.
Assume $(2)$ holds and let $x\in E$. It can then be shown that $$\Xi(X)(t):=x+\int_0^tv(s,X(s))\:{\rm d}s\;\;\;\text{for }t\in[0,\tau]\tag3,$$ for $X\in C^0([0,\tau],E)$, is a contraction on $C^0([0,\tau],E)$ and hence has a unique fixed point $X$.
Question 2: Assume $v$ is differentiable in the second argument. I would like to know if (or which assumption we need to impose in order to conclude that) the dependence of $X$ on the initial value $x$ is differentiable.
Let $$w(t,A):={\rm D}v(t,X(t))A\;\;\;\text{for }(t,A)\in[0,\tau]\times\mathfrak L(E).$$ In a first step, I would like to show that there is a unique $Y\in C^0([0,\tau],\mathfrak L(E))$ with $$Y(t)=\operatorname{id}_E+\int_0^tw(s,Y(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]\tag4.$$ By the former argumentation, we can show this by verifying that $w$ satisfies $(1)$ and $(2)$ for some suitable constants. Now, clearly, $$\left\|w(t,A)-w(t,B)\right\|_E\le\left\|{\rm D}v(t,X(t))\right\|_{\mathfrak L(E)}\left\|A-B\right\|_{\mathfrak L(E)}\tag5$$ for all $t\in[0,\tau]$ and $A,B\in\mathfrak L(E)$. So, assuming that ${\rm D}v$ is jointly continuous (would separately continuous in each argument be enough?), we could bound the right-hand side of $(5)$ using that $$\sup_{t\in[0,\:\tau]}\left\|{\rm D}v(t,X(t))\right\|_{\mathfrak L(E)}<\infty\tag6.$$
Regarding question 1: Note that if $$c_0:=\sup_{t\in[0,\:\tau]}\left\|v(t,x_0)\right\|_E<\infty\tag7$$ for some $x_0\in E$, then $(2)$ is satisfied by taking $c_2:=\max(c_0+c_1\left\|x_0\right\|_E,c_1)$. So, $(2)$ holds, for example, whenever $v$ is continuous in the first argument.