# Implicit Function Theorem

1. Nov 23, 2008

### tom_rylex

1. The problem statement, all variables and given/known data
I'm looking for someone to walk me through a problem, because I keep falling further and further behind ...

Let f be a real-valued function on he strip $R_2: a \leq x \leq b, -\infty < y < \infty$. Can the equation f(x,y) = 0 be solved for y as a function of x;
f is continuous w/ continuous partial derivatives of first order, and
$$0 < m \leq \frac{\partial f}{\partial y} \leq M$$
show a unique function y = u(x) so that f(x,u(x)) = 0

2. Relevant equations
There was s hint to consider a continuous function w/ the sup norm and define the transformation $T:C[a,b] \rightarrow C[a,b]$ by
$$(Tu)(x) = u(x) - \frac{2}{m + M}f(x,u(x))$$
and show that T is a contraction.

3. The attempt at a solution
I'm just not getting it, but here's what I have:
Apply the contraction mapping theorem:

$$\parallel (Tu)(x) - (Tv)(x) \parallel = \parallel u(x) - \frac{2}{m + M}f(x,u(x)) - v(x) + \frac{2}{m + M}f(x,v(x)) \parallel$$
$$= \parallel u(x) - v(x) + \frac{2}{m + M}\left(f(x,v(x)) - f(x,u(x)) \right ) \parallel$$
If I apply the Mean Value theorem, I should be able to get
$$= \parallel u(x) - v(x) - \frac{2}{m + M}f'(c)(v(x) - u(x)) \parallel$$
taking the sup norm, I would state
$$\leq \parallel u(x) -v(x) + \frac{2M}{m + M}(u(x) - v(x)) \parallel_\infty$$
$$\leq \left(1+\frac{2M}{m + M} \right) \parallel u(x) - v(x) \parallel_\infty$$
$$\leq \frac{m+3M}{m+M} \parallel u(x) - v(x) \parallel_\infty$$

I know that's not correct (definitely not a contraction), but I don't know what I'm doing wrong. Please use small words, so I can understand. Thanks.