Lectures on Calculus The Inverse Function Theorem

Lectures on Calculus

The Inverse Function Theorem

by William M. Faucette

University of West Georgia

Adapted from Calculus on Manifolds

by Michael Spivak

Lemma One

Lemma: Let ARn be a rectangle and let f:ARn 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



Theorem: Suppose that f:RnRn 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:VW has a continuous inverse f -1:WV which is differentiable and for for all y2W satisfies


Proof: Let be the linear transformation Df(a). Then is non-singular, since det f (a)≠0. Now

is the identity linear transformation.


Proof: If the theorem is true for -1f, it is clearly true for f. Therefore we may assume at the outset that is the identity.


Whenever f(a+h)=f(a), we have

But


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


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


Since

we can apply Lemma One to g(x)=f(x)-x to get


Since

we have


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.


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


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:UR defined by


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.


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


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.


If V=(interior U)f1(W), we have shown that the function f:VW has inverse f 1:WV. We can rewrite

As

This shows that f-1 is continuous.


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.


Since =Df(x), we know that

Setting (x)=f(x+h)-f(x)-(h), we know that


Hence, we have


Therefore,


Since every y12W is of the form f(x1) for some x12V, this can be written

or


It therefore suffices to show that

Since is a linear transformation, it suffices to show that


Recall that

Also, f-1 is continuous, so


Then

where the first factor goes to 0 and the second factor is bounded by 2. This completes the proof. QED

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Lectures on Calculus The Inverse Function Theorem