I have a question that I think is closely related to the implicit function theorem but I am unsure of how to solve it. Does the following proposition hold?
Let $ a < b$ and $y:(a,b) \to \mathbb R$ be a function. Furthermore, let $F\colon (a,b) \times \mathbb R \to \mathbb R$ be a continuously differentiable function such that $$\forall x \in (a,b): F(x,y(x)) = 0 \\ \forall (x,y) \in (a,b) \times \mathbb R: \frac{\partial}{\partial y}F(x,y) >0.$$ Then $y$ is continuously differentiable.
I think this must be true, but I can't transfer from the local case to the global case and also I don't know if the function that the implicit function theorem gives me can be glued together so that it equals $y$ everywhere. Any help appreciated.
Consider a point $x_0\in(a,b)$, and let $y(x_0)=:y_0$. Then $F(x_0,y_0)=0$, and $${\partial F\over\partial y}(x_0,y_0)>0\ .$$ The implicit function theorem then guarantees the following: There is a window $$W:=[x_0-h,x_0+h]\times[y_0-c,y_0+c]\ \subset \ (a,b)\times{\mathbb R}$$ with center $(x_0,y_0)$ and a $C^1$-function $$\psi:\quad [x_0-h,x_0+h]\to [y_0-c,y_0+c],\qquad x\mapsto y=\psi(x)$$ with $\psi(x_0)=y_0$, such that $$\bigl\{(x,y)\bigm| F(x,y)=0\bigr\}\cap W=\bigl\{(x,\psi(x))\bigm|x_0-h\leq x\leq x_0+h\bigr\}\ .$$ This means that within $W$ the set $F(x,y)=0$ coincides with the graph of $\psi$. This implies that $$y(x)=\psi(x)\qquad(x_0-h\leq x\leq x_0+h)\ ,$$ hence $x\mapsto y(x)$ is $C^1$ in the neighborhood of $x_0$.