Chapter 3: Inverse Function Theorem

Chapter 3 Inverse Function Theorem (This lecture was given Thursday, September 16, 2004.) 3.1 Partial Derivatives Definition 3.1.1. If f : Rn → Rm and a ∈ Rn , then the limit f (a1 , . . . , ai + h, . . . , an ) − f (a1 , . . . , an ) h→0 h Di f (a) = lim (3.1) is called the ith partial derivative of f at a, if the limit exists. Denote Dj (Di f (x)) by Di,j (x). This is called a second-order (mixed) partial derivative. Then we have the following theorem (equality of mixed partials) which is given without proof. The proof is given later in Spivak, Problem 3-28. Theorem 3.1.2. If Di,j f and Dj,i f are continuous in an open set containing a, then Di,j f (a) = Dj,i f (a) 19 (3.2) We also have the following theorem about partial derivatives and maxima and minima which follows directly from 1-variable calculus: Theorem 3.1.3. Let A ⊂ Rn . If the maximum (or minimum) of f : A → R occurs at a point a in the interior of A and Di f (a) exists, then Di f (a) = 0. Proof: Let gi (x) = f (a1 , . . . , x, . . . , an ). gi has a maximum (or minimum) at ai , and gi is defined in an open interval containing ai . Hence 0 = gi′ (ai ) = 0. The converse is not true: consider f (x, y) = x2 − y 2 . Then f has a minimum along the x-axis at 0, and a maximum along the y-axis at 0, but (0, 0) is neither a relative minimum nor a relative maximum. 3.2 Derivatives Theorem 3.2.1. If f : Rn → Rm is differentiable at a, then Dj f i (a) exists for 1 ≤ i ≤ m, 1 ≤ j ≤ n and f ′ (a) is the m x n matrix (Dj f i (a)). Proof: First consider m = 1, so f : Rn → R. Define h : R → Rn by h(x) = (a1 , . . . , x, . . . , an ), with x in the j th slot. Then Dj f (a) = (f ◦ h)′ (aj ). Applying the chain rule, ′ j (f ◦ h)′ (aj ) = f ′ (a) · h (a) 0   .   .     .   .   ′ = f (a) ·   1   .     .   .   0 20 (3.3) Thus Dj f (a) exists and is the jth entry of the 1 × n matrix f ′ (a). Spivak 2-3 (3) states that f is differentiable if and only if each f i is. So the theorem holds for arbitrary m, since each f i is differentiable and the ith row of f ′ (a) is (f i )′ (a). The converse of this theorem – that if the partials exists, then the full derivative does – only holds if the partials are continuous. Theorem 3.2.2. If f : Rn → Rm , then Df (a) exists if all Dj f (i) exist in an open set containing a and if each function Dj f (i) is continuous at a. (In this case f is called continuously differentiable.) Proof.: As in the prior proof, it is sufficient to consider m = 1 (i.e., f : Rn → R.) f (a + h) − f (a) = f (a1 + h1 , a2 , . . . , an ) − f (a1 , . . . , an ) +f (a1 + h1 , a2 + h2 , a3 , . . . , an ) − f (a1 + h1 , a2 , . . . , an ) + . . . + f (a1 + h1 , . . . , an + hn ) −f (a1 + h1 , . . . , an−1 + hn−1 , an ). (3.4) D1 f is the derivative of the function g(x) = f (x, a2 , . . . , an ). Apply the mean-value theorem to g : f (a1 + h1 , a2 , . . . , an ) − f (a1 , . . . , an ) = h1 · D1 f (b1 , a2 , . . . , an ). (3.5) for some b1 between a1 and a1 + h1 . Similarly, hi · Di f (a1 + h1 , . . . , ai−1 + hi−1 , bi , . . . , an ) = hi Di f (ci ) 21 (3.6) for some ci . Then P |f (a+h)−f (a)− i Di f (a)·hi | |h| P i i [Di f (ci )−Di f (a) ·h ] limh→0 |h| limh→0 = ≤ limh→0 ≤ limh→0 =0 P i P i |Di f (ci ) − Di f (a)| · |hi | |h| (3.7) |Di f (ci ) − Di f (a)| since Di f is continuous at 0. p Example 3. Let f : R2 → R be the function f (x, y) = xy/( x2 + y 2 if (x, y) 6= (0, 0) and 0 otherwise (when (x, y) = (0, 0)). Find the partial derivatives at (0, 0) and check if the function is differentiable there. 3.3 The Inverse Function Theorem (A sketch of the proof was given in class.) 22