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Lectures on Calculus The Inverse Function Theorem

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by William M. Faucette University of West Georgia

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Adapted from Calculus on Manifolds by Michael Spivak

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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.

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Lemma One Proof: We have

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

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

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

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

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The Inverse Function Theorem

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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

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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.

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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.

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

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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

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

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

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The Inverse Function Theorem Since we have

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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.

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The Inverse Function Theorem Let W={y:|y f(a)|

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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

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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)

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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

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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.

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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.

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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.

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

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The Inverse Function Theorem Hence, we have

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The Inverse Function Theorem Therefore,

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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

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

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

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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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