Inverse Function Theorem

The contraction mapping theorem is a convenient way to prove existence theorems such as the Inverse Function Theorem in multivariable calculus.

Recall that a map f: U → ℝ^{n} (where U is open in ℝ^{n}) is differentiable at a point x ∈ U if we can write

f(x + h) = f(x) + Ah + e(h),  (1)

where A: ℝ^{n} → ℝ^{n} is a linear transformation (equivalently, an n × n matrix) and ∥e(h)∥/∥h∥ → 0 as h → 0. The derivative of f at x is f′(x) = A. If f′(x) is a continuous function of x, f is said to have class C^{1}. The norm ∥A∥ of a linear transformation A is defined to satisfy ∥Ax∥ ≤ ∥A∥∥x∥ for all x.

We need the following standard result.

Lemma 2 Suppose ∥f′(x)∥ ≤ M on some disk D. Then for any points x, x + h ∈ D, ∥f(x+h) − f(x)∥ ≤ M∥h∥.

Theorem 3 Let U be an open set in ℝ^{n}, and f: U → ℝ^{n} a differentiable function of class C^{1}. Suppose x_{0} ∈ U is a point where f′(x_{0}) is invertible. Then there exist neighborhoods U′ ⊂ U of x_{0} and V of y_{0} = f(x_{0}) such that f|U′: U′ → V is a bijection. Further, the inverse g: V → U′ is differentiable of class C^{1}, with derivative g′(y) = f′(g(y))^{−1}. (That is, f is a local diffeomorphism near x_{0}.)

Proof We first simplify. We may assume x_{0} = y_{0} = 0. Further, we may assume f′(0) = I; otherwise we replace f by B^{−1} ^{∘} f, where B = f′(0) (in effect, a linear change of coordinates in one copy of ℝ^{n}).

We restrict attention to the disk D = {x ∈ ℝ^{n} : ∥x∥ ≤ δ}, where δ is chosen small enough to make D ⊂ U and ∥f′(x) − I∥ ≤ \frac{1}{2} for all x ∈ D (possible because f has class C^{1}). Put w(x) = f(x) − x; then w′(x) = f′(x) − I, and by Lemma 2, ∥w(x+h) − w(x)∥ ≤ \frac{1}{2}∥h∥. In terms of f, this is

∥f(x + h) − f(x) − h∥ ≤ \frac{1}{2}∥h∥.  (4)

We take V as the open disk \{y ∈ ℝ^{n} : ∥y∥ < \frac{1}{2}δ\}. We assert that given any y ∈ V, there is a unique x ∈ D such that f(x) = y; we then define g(y) = x.

We define u: D → D by u(x) = x + (y − f(x)); then f(x) = y if and only if x is a fixed point of u, i. e. u(x) = x. (The motivation for the definition of u is that if x is an approximation to the desired value, with error y − f(x), u(x) should be a better approximation, because f is in some sense close to the identity function.)

We must check that u(x) does in fact lie in D. If we set x = 0 in (4), we obtain ∥f(h) − h∥ ≤ \frac{1}{2}∥h∥ ≤ \frac{1}{2}δ for any h ∈ D. Replacing h here by x and using the triangle inequality, we deduce that

∥u(x)∥ ≤ ∥y∥ + ∥f(x) − x∥ < \frac{1}{2}δ + \frac{1}{2}δ = δ.

Next, we check that u is a contraction mapping (and hence automatically continuous). From the definition,

u(x+h) − u(x) = −f(x+h) + f(x) + h.

Then by (4),

∥u(x+h) − u(x)∥ = ∥f(x+h) − f(x) − h∥ ≤ \frac{1}{2}∥h∥,

as required (with contracting factor \frac{1}{2}).

We took D closed in order to make it a complete metric space. Then the Contraction Mapping Theorem yields a unique fixed point x of u; in fact, we saw that x = u(x) is an interior point of D. We take U′ as the open set f^{−1}(V) ∩ Int D.

To see that g is continuous, take any two points y, y + k ∈ V and put x = g(y) and x + h = g(y + k). Then (4) becomes ∥k − h∥ ≤ \frac{1}{2}∥h∥. By the triangle inequality, ∥h∥ ≤ ∥h − k∥ + ∥k∥ ≤ \frac{1}{2}∥h∥ + ∥k∥. So \frac{1}{2}∥h∥ ≤ ∥k∥ or, more conveniently,

∥h∥ ≤ 2∥k∥.  (5)

We can ensure ∥h∥ < ϵ simply by taking ∥k∥ < \frac{1}{2}ϵ.

As for differentiability of g, (1) becomes

y + k = y + A[g(y+k) − g(y)] + e(h).  (6)

Now ∥Ax − x∥ = ∥(A − I)x∥ ≤ \frac{1}{2}∥x∥, from which it is clear that A has zero kernel and is therefore invertible. We apply A^{−1} to (6) and rearrange, to obtain

g(y+k) = g(y) + A^{−1}k − A^{−1}e(h).

With the help of (5), we see that

\frac{∥A^{−1}e(h)∥}{∥k∥} ≤ ∥A^{−1}∥\frac{∥e(h)∥}{∥h∥} \frac{∥h∥}{∥k∥} ≤ 2∥A^{−1}∥\frac{∥e(h)∥}{∥h∥} ⟶ 0

as k → 0. Comparison with (1)) shows that g is differentiable at y with derivative g′(y) = A^{−1}. Finally, the formula for g′(y) is visibly continuous in y.

Remark If we know that g is differentiable, the chain rule gives the derivative g′(y) easily. The point is to prove that g is differentiable.

Remark The above proof works in any Banach space, except that to see that A^{−1} exists, one writes it as the sum of the convergent series

A^{−1} = I − T + T^{2} − T^{3} + …,

where A = I + T with ∥T ∥ ≤ \frac{1}{2}.