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Lectures on Calculus The Inverse Function Theorem.

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Presentation on theme: "Lectures on Calculus The Inverse Function Theorem."— Presentation transcript:

1 Lectures on Calculus The Inverse Function Theorem

2 by William M. Faucette University of West Georgia

3 Adapted from Calculus on Manifolds by Michael Spivak

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

5 Lemma One Proof: We have

6 Lemma One Proof: Applying the Mean Value Theorem we obtain for some z ij between x j and y j.

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

8 Lemma One Proof: Then since each |y j  x j |≤|y  x|.

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

10 The Inverse Function Theorem

11 Theorem: Suppose that f:R n  R n 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

12 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.

13 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.

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

15 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

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

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

18 The Inverse Function Theorem Since we have

19 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.

20 The Inverse Function Theorem Let W={y:|y  f(a)| { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/8/2330161/slides/slide_20.jpg", "name": "The Inverse Function Theorem Let W={y:|y  f(a)|

21 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

22 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) { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/8/2330161/slides/slide_22.jpg", "name": "The Inverse Function Theorem This function is continuous and therefore has a minimum on U.", "description": "If x2boundary U, then, by the formula on slide 20, we have g(a)

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

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

25 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.

26 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.

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

28 The Inverse Function Theorem Hence, we have

29 The Inverse Function Theorem Therefore,

30 The Inverse Function Theorem Since every y 1 2 W is of the form f(x 1 ) for some x 1 2 V, this can be written or

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

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

33 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


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