# The Inverse Function Theorem

## Presentation on theme: "The Inverse Function Theorem"— Presentation transcript:

The Inverse Function Theorem
Lectures on Calculus The Inverse Function Theorem

University of West Georgia
by William M. Faucette University of West Georgia

by Michael Spivak

Lemma One Lemma: Let ARn be a rectangle and let f:ARn be continuously differentiable. If there is a number M such that |Djf i(x)|≤M for all x in the interior of A, then for all x, y2A.

Lemma One Proof: We have

Lemma One Proof: Applying the Mean Value Theorem we obtain
for some zij between xj and yj.

Lemma One Proof: The expression
has absolute value less than or equal to

Lemma One Proof: Then since each |yj-xj|≤|y-x|.

Lemma One Proof: Finally, which concludes the proof. QED

The Inverse Function Theorem

The Inverse Function Theorem
Theorem: Suppose that f:RnRn is continuously differentiable in an open set containing a, and det f (a)≠0. Then there is an open set V containing a and an open set W containing f(a) such that f:VW has a continuous inverse f -1:WV which is differentiable and for for all y2W satisfies

The Inverse Function Theorem
Proof: Let  be the linear transformation Df(a). Then  is non-singular, since det f (a)≠0. Now is the identity linear transformation.

The Inverse Function Theorem
Proof: If the theorem is true for -1f, it is clearly true for f. Therefore we may assume at the outset that  is the identity.

The Inverse Function Theorem
Whenever f(a+h)=f(a), we have But

The Inverse Function Theorem
This means that we cannot have f(x)=f(a) for x arbitrarily close to, but unequal to, a. Therefore, there is a closed rectangle U containing a in its interior such that

The Inverse Function Theorem
Since f is continuously differentiable in an open set containing a, we can also assume that

The Inverse Function Theorem
Since we can apply Lemma One to g(x)=f(x)-x to get

The Inverse Function Theorem
Since we have

The Inverse Function Theorem
Now f(boundary U) is a compact set which does not contain f(a). Therefore, there is a number d>0 such that |f(a)-f(x)|≥d for x2boundary U.

The Inverse Function Theorem
Let W={y:|y-f(a)|<d/2}. If y2W and x2boundary U, then

The Inverse Function Theorem
We will show that for any y2W there is a unique x in interior U such that f(x)=y. To prove this consider the function g:UR defined by

The Inverse Function Theorem
This function is continuous and therefore has a minimum on U. If x2boundary U, then, by the formula on slide 20, we have g(a)<g(x). Therefore, the minimum of g does not occur on the boundary of U.

The Inverse Function Theorem
Since the minimum occurs on the interior of U, there must exist a point x2U so that Djg(x)=0 for all j, that is

The Inverse Function Theorem
Since the Jacobian [Djf i(x)] is non-singular, we must have That is, y=f(x). This proves the existence of x. Uniqueness follows from slide 18.

The Inverse Function Theorem
If V=(interior U)f1(W), we have shown that the function f:VW has inverse f 1:WV. We can rewrite As This shows that f-1 is continuous.

The Inverse Function Theorem
We only need to show that f-1 if differentiable. Let =Df(x). We will show that f-1 is differentiable at y=f(x) with derivative -1.

The Inverse Function Theorem
Since =Df(x), we know that Setting (x)=f(x+h)-f(x)-(h), we know that

The Inverse Function Theorem
Hence, we have

The Inverse Function Theorem
Therefore,

The Inverse Function Theorem
Since every y12W is of the form f(x1) for some x12V, this can be written or

The Inverse Function Theorem
It therefore suffices to show that Since  is a linear transformation, it suffices to show that

The Inverse Function Theorem
Recall that Also, f-1 is continuous, so

The Inverse Function Theorem
Then where the first factor goes to 0 and the second factor is bounded by 2. This completes the proof. QED