Proof Of The Inverse Function Theorem

Proof of the Inverse Function Theorem: A Comprehensive Guide

The Inverse Function Theorem is a cornerstone result in multivariable calculus, providing conditions under which a function has a locally invertible inverse. Understanding its proof not only solidifies your grasp of the theorem itself but also deepens your understanding of concepts like differentiability, linear approximations, and the Contraction Mapping Theorem. This article will present a detailed, step-by-step proof, breaking down the complexities into digestible chunks.

Understanding the Theorem

Before diving into the proof, let's formally state the Inverse Function Theorem:

Theorem: Let f: U ⊂ Rⁿ → Rⁿ be a continuously differentiable function, where U is an open subset of Rⁿ. If, at a point a ∈ U, the Jacobian matrix Df(a) is invertible (i.e., its determinant is non-zero), then there exists an open set V containing a and an open set W containing f(a) such that f: V → W is a bijection, and its inverse function f⁻¹: W → V is continuously differentiable. Furthermore, the derivative of the inverse function at f(a) is given by:

Df⁻¹(f(a)) = [Df(a)]⁻¹

The Proof: A Step-by-Step Approach

The proof relies heavily on the Contraction Mapping Theorem, a powerful tool in analysis. We'll break down the proof into several key steps:

Step 1: Reformulating the Problem Using Newton's Method

The core idea is to show that the equation f(x) = y has a unique solution x near a for each y near f(a). We can rewrite this equation as:

x = x - f(x) + y

This resembles a fixed-point iteration, a common strategy in numerical analysis. Define a function g(x) = x - f(x) + y. Finding a solution to f(x) = y is equivalent to finding a fixed point of g(x), i.e., a point x such that g(x) = x.

Step 2: Applying the Mean Value Theorem for Vector-Valued Functions

Crucially, we're not directly applying Newton's method iteratively. Instead, we utilize the Mean Value Theorem to analyze the behavior of g(x). The Mean Value Theorem, extended to vector-valued functions, guarantees that for any x, x' ∈ U, there exists a point c on the line segment connecting x and x' such that:

f(x) - f(x') = Df(c)(x - x')

We'll use this to bound the difference g(x) - g(x').

Step 3: The Role of the Jacobian Matrix

The invertibility of Df(a) plays a crucial role. Since the derivative Df is continuous, the invertibility of Df(a) implies that Df(x) is invertible in some neighborhood of a. This allows us to bound the norm of the inverse Jacobian:

||[Df(x)]⁻¹|| ≤ M for some constant M > 0, and all x in a small enough neighborhood of 'a'. This boundedness is essential in the next step.

Step 4: Showing that g(x) is a Contraction Mapping

Now, we leverage the Mean Value Theorem and the bounded inverse of the Jacobian to demonstrate that g(x) is a contraction mapping in a sufficiently small neighborhood of a. We need to show that there exists a constant k ∈ (0, 1) such that:

||g(x) - g(x')|| ≤ k||x - x'||

for all x and x' in this neighborhood. The following steps demonstrate this:

1. Start with g(x) - g(x') = x - f(x) + y - (x' - f(x') + y) = x - x' - (f(x) - f(x'))
2. Apply the Mean Value Theorem: g(x) - g(x') = x - x' - Df(c)(x - x') = (I - Df(c))(x - x') where I is the identity matrix.
3. Bound the norm: ||g(x) - g(x')|| ≤ ||I - Df(c)|| ||x - x'||
4. By choosing a sufficiently small neighborhood around 'a', we can make ||I - Df(c)|| arbitrarily close to 0 because Df(c) is close to Df(a). This ensures we can select a k < 1.

Step 5: Applying the Contraction Mapping Theorem

Because we've established that g(x) is a contraction mapping in a neighborhood of a, we can now apply the Contraction Mapping Theorem. This theorem guarantees that g(x) has a unique fixed point x in this neighborhood. This fixed point is the solution to f(x) = y.

Step 6: Establishing the Inverse Function and its Differentiability

The existence of a unique solution x for each y in a neighborhood of f(a) proves the existence of a local inverse function f⁻¹.

To show that f⁻¹ is differentiable, we can use the implicit function theorem. This theorem states that if we have a continuously differentiable function F(x,y) = f(x) - y = 0, and the partial derivative with respect to x is invertible at a point, then there exists a continuously differentiable function x = g(y) that satisfies the equation. In our case, F(x,y) = f(x) - y, and the partial derivative with respect to x is simply Df(x), which is invertible near 'a'. Therefore, f⁻¹ is differentiable. The derivative is found by differentiating both sides of f(f⁻¹(y)) = y with respect to y and using the chain rule, leading to Df(f⁻¹(y))Df⁻¹(y) = I, which gives us the formula for Df⁻¹(f(a)).

Step 7: Continuous Differentiability of the Inverse

Finally, the continuity of Df⁻¹ follows from the continuity of Df and the formula Df⁻¹(y) = [Df(f⁻¹(y))]⁻¹. Since matrix inversion is a continuous operation, and Df and f⁻¹ are continuous, the continuity of Df⁻¹ is guaranteed.

Conclusion

The proof of the Inverse Function Theorem is a beautiful example of how seemingly disparate concepts from analysis and linear algebra intertwine to produce a powerful result. The careful use of the Mean Value Theorem, the Contraction Mapping Theorem, and the properties of the Jacobian matrix allows us to conclude the existence of a locally invertible inverse function with continuous derivatives. This theorem forms the bedrock for understanding many advanced concepts in analysis and differential geometry, highlighting its significance in various branches of mathematics. A solid understanding of this proof empowers you to confidently apply the Inverse Function Theorem in various applications, from solving systems of nonlinear equations to understanding the behavior of differentiable manifolds.