Lectures on Calculus

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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.

5. Lemma One Proof: We have

6. Lemma One Proof: Applying the Mean Value Theorem we obtain for some zij between xj and yj.

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

8. Lemma One Proof: Then since each |yj-xj|≤|y-x|.

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

10. The Inverse Function Theorem

11. 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

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)|<d/2}. If y2W and x2boundary U, then

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)<g(x). Therefore, the minimum of g does not occur on the boundary of U.

23. 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

24. 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.

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 y12W is of the form f(x1) for some x12V, 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