Lectures on Calculus

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# Lectures on Calculus - PowerPoint PPT Presentation

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.

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### 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 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

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

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