I am studying for my qualifying exams and came across this problem from past exams.
Let $a<b$ be real numbers and $f:\mathbb{R}\times[a,b]\rightarrow\mathbb{R}$ be a continuous function. Assume that $f$ is differentiable with respect to its first variable and that there exists $0 < m < M < \infty$ such that $m \leq \frac{\partial f}{\partial x}(x,y) \leq M$ for all $(x,y) \in \mathbb{R}\times[a,b]$. Prove that there exists a unique continuous function $w:[a,b]\rightarrow\mathbb{R}$ such that $f(w(y),y) = 0$ for all $y\in[a,b]$. (Hint: show that the operator $F$ which maps the function $z(y)$ for $y\in[a,b]$ to $F(z)(y) = z(y) - \frac{1}{M}f(z(y),y)$ is a contraction on a suitable function space.
This problem seems nearly identical to a problem I have done in Carothers aside from it being on the space $C[a,b]$, which motivated this attempt at a solution:
$\frac{\partial F}{\partial z(y)} = 1 - \frac{1}{M}\frac{\partial f(z(y),y)}{\partial z(y)}$ and because of our bound on the partial derivative of $f$ we have that $0 = 1-\frac{M}{M}\leq \frac{\partial F}{\partial z(y)} \leq 1 - \frac{m}{M} < 1$ which gives that $\lvert\frac{\partial F}{\partial z(y)}\rvert \leq \alpha < 1$ which would (maybe?) imply that $F$ has a unique fixed point in $C([a,b])$. Which means there is a unique $w(y) \in C([a,b])$ with $w(y) = w(y) - \frac{1}{M}f(w(y),y)$ which would force $f(w(y),y))=0$ for all $y\in [a,b]$.
I do not know if this method works in this case however. Frankly, I'm not even sure if the derivative of $F$ with respect to the function $z(y)$ even makes sense, and when talking about a regular function $g : \mathbb{R} \rightarrow\mathbb{R}$ you use the mean value theorem to say that $\lvert g'(x)\rvert \leq \alpha < 1$ implies that $g$ is a strict contraction and therefore has a unique fixed point.
Does this method work here, and if not how would I show that $F$ is a strict contraction on $C([a,b])$?
By MVT we get $$F(z_1(y))-F(z_2(Y))=z_1(y)-z_2(y)-\frac 1 M \frac {\partial f} {\partial x} f(t,y) (z_1(y)-z_2(y))$$ for some $t$ between $z_1(y)$ and $z_2(y)$. Hence $F(z_1(y))-F(z_2(Y)) =[z_1(y)-z_2(y)](s)$ where $s$ lies between $1-\frac m M$ and $1-\frac M M$. Thus $0 \leq s <1$ and $F$ is a contraction on $C[a,b]$. Hence it has a fixed point $z$ which gives $f(z(y),y)=0$.