Can the Implicit Function Theorem Solve for y in f(x,y)=0?

[tom_rylex]

Homework Statement

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

Homework 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.

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.

 

[mighty2000]

Hello there! It seems like you are trying to solve a problem involving a real-valued function on a strip. This can be a bit tricky, but don't worry, I'm here to help. Let's break down the problem and see if we can find a solution together.

First, the problem asks if the equation f(x,y) = 0 can be solved for y as a function of x. This means that we need to find a function y = u(x) that satisfies the equation f(x,u(x)) = 0. In order to do this, we can use the fact that f is a continuous function with continuous partial derivatives of first order, and that 0 < m ≤ ∂f/∂y ≤ M. This means that f is a well-behaved function, and we can use it to help us find a solution.

The hint given suggests that we consider a continuous function with the sup norm and define the transformation T:C[a,b] → C[a,b] by (Tu)(x) = u(x) - 2/(m+M)f(x,u(x)). This may seem a bit confusing at first, but essentially what we are doing is using the function f to transform the function u.

Next, we need to show that T is a contraction. This means that for any two functions u and v in the domain of T, the distance between Tu and Tv is always less than the distance between u and v. In other words, we want to show that ||Tu - Tv|| < ||u - v||.

Your attempt at a solution is on the right track, but there are a few mistakes. First, when you apply the Mean Value Theorem, you should get ||u(x) - v(x) + 2f'(c)/(m+M)(v(x) - u(x))||. This is because the derivative of f(x,u(x)) with respect to x is just f'(c), where c is some value between u(x) and v(x).

Next, you took the sup norm of this expression, but you forgot to include the 2f'(c)/(m+M) term. This means that your final expression should be ||u(x) - v(x) + 2f'(c)/(m+M)(v(x) - u(x))|| ≤ ||u(x) - v(x)|| + ||2f'(c)/(m+M)||