- By
**juan** - Follow User

- 60 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Lectures on Calculus' - juan

**An Image/Link below is provided (as is) to download presentation**

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Lectures on Calculus

### by William M. Faucette

### Adapted from Calculus on Manifolds

The Inverse Function Theorem

University of West Georgia

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

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

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

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

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:UR 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)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.

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

Then

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

Download Presentation

Connecting to Server..