Question: I have the following function $$\begin{equation} \begin{split} F : \mathbb{R}\times \{\omega \in C^1([-1,1]) \ |\ \omega(0) = 0\} & \longrightarrow C^0([-1,1]) \\ (\alpha, z) & \longmapsto \frac{dz}{d\tau} - \alpha f(\alpha \tau, z(\tau)+\xi) \end{split} \end{equation}$$
With $\xi \in \mathbb{R}$. How can I show it to be continuous in $(0,0)$?
Context: In my advaced analysis course we've introduced Fréchet differentiation on Banach Spaces and stated the implicit function theorem for Banach Spaces as follows:
Given the Banach spaces X, Y, Z, the point $(x_0, y_0) \in X \times Y$ and $U \subset X \times Y$ a nbhd of $(x_0,y_0)$, and an application $F: U \longrightarrow Z$ such that :
1) $F$ is continuous in ($x_0,y_0$)
2) $F(x_0,y_0) = 0$
3) $\exists F_y' (x,y)$ for each $(x,y) \in U$, the application $$\begin{equation} \begin{split} F_y' : U &\longrightarrow \mathcal{B}(Y,Z) \\ (x,y) &\longmapsto F_y'(x,y) \end{split} \end{equation}$$ is continuous in $(x_0,y_0)$ and the operator $F_y'(x_0,y_0)$ is invertible.
Then there exist an open nbhd $I$ of $x_0$, an open nbhd $J$ of $y_0$ and a unique function $\phi: I \rightarrow J$ with $\phi(x_0) = y_0$ such that: $$F(x,y) = 0 \quad (x,y) \in I \times J \iff y = \phi(x)$$
As an application of this theorem we've proved a "weak" version of the local existance and uniqueness theorem for the Cauchy problem:
$$\begin{cases} x' = f(t,x) \qquad f: \Omega \text{ open } \subset \mathbb{R}^2 \rightarrow \mathbb{R} \quad f \in C^1(\Omega)\\ x(0) = \xi \end{cases}$$
and $t \in (0,1)$.
Given that we're searching for a solution $x: (-\alpha, \alpha) \rightarrow \mathbb{R}$, we do the variable change $z(\tau) = x(\alpha \tau) - \xi$ with $z:(-1,1) \rightarrow \mathbb{R} $ and we get the equivalent problem:
$$\begin{cases} z' = \alpha f(\alpha \tau, z(\tau) + \xi) \\ z(0) = 0 \end{cases}$$
Then we've introduced the following function $$\begin{equation} \begin{split} F : \mathbb{R}\times \{\omega \in C^1([-1,1]) \ |\ \omega(0) = 0\} & \longrightarrow C^0([-1,1]) \\ (\alpha, z) & \longmapsto \frac{dz}{d\tau} - \alpha f(\alpha \tau, z(\tau)+\xi) \end{split} \end{equation}$$
A pretty straightforward application of the implicit function theorem for Banach spaces gives the result. What bothers me is that we've not proved that $F$ is continuous in $(0,0)$ (we've not even mentioned it), and it does not seem so trivial to me. Thanks