- 46 Views
- Uploaded on
- Presentation posted in: General

Lectures on Calculus

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Lectures on Calculus

The Inverse Function Theorem

by William M. Faucette

University of West Georgia

Adapted from Calculus on Manifolds

by Michael Spivak

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.

Proof: We have

Proof: Applying the Mean Value Theorem we obtain

for some zij between xj and yj.

Proof: The expression

has absolute value less than or equal to

Proof: Then

since each |yj-xj|≤|y-x|.

Proof: Finally,

which concludes the proof. QED

The Inverse Function Theorem

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 f1: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