Math 125B, Winter 2015. Notes on inverse functions n n Theorem 1 (Inverse Function Theorem). Assume A ⊆ R is open and f : A ! R is a function in Cr(A) for some r ≥ 1. Assume that a 2 A and that det Df(a) 6= 0. Then there exist open sets n −1 r U ⊆ A and V ⊆ R so that a 2 U, f : U ! V is one-to-one and onto, and f : V ! U is in C (V ). Moreover, if y 2 V , and x 2 U with f(x) = y, Df(x) is invertible and (1) Df −1(y) = Df(x)−1: Remark. The exponent −1 on the right of (1) is the inverse of the n × n matrix Df(x), while on the left of (1) it indicates the inverse function. n n Definition. Assume A ⊆ R is open and f : A ! R is a function. We say that f has a differentiable n local inverse at a 2 A if there exist open sets U ⊆ A and V ⊆ R so that a 2 U, f : U ! V is one-to-one and onto, and f −1 : V ! U is differentiable on V . The Inverse Function Theorem gives the sufficient condition for existence of differentiable local −1 inverse. You should note that f may exist without being differentiable. For example, f : R ! R 3 1 −1 1=3 given by f(x) = x is in C (R) and has inverse f (x) = x that is not differentiable at x = 0. The next lemma implies (1) assuming the rest of the Inverse Function Theorem holds. n Lemma 1. Assume A; B ⊆ R is open and f : A ! B is one-to-one and onto. Assume that a 2 A, b = f(a), and that Df(a) and Df −1(b) both exist. Then Df(a) is invertible and Df −1(b) = Df(a)−1: n n Proof. If we denote by I : R ! R the (linear) identity map, then on A f −1 ◦ f = I; and then by the chain rule Df −1(b) · Df(a) = I; which ends the proof. Lemma 1 and Theorem 1 together give the following result, which is the one you should remember for the exams. n n 1 Theorem 2. Assume A ⊆ R is open and f : A ! R is a function in C (A). Then f has a differentiable local inverse at a 2 A if and only if det Df(a) 6= 0 and in this case Df −1(f(a)) = Df(a)−1. We now proceed through the proof of Theorem 1, which does require some effort. 1 n n 1 Lemma 2. Assume A ⊆ R is open and f : A ! R is a function in C (A) for some r ≥ 1. Assume that a 2 A and that det Df(a) 6= 0. Then there exists an open set U ⊆ A with a 2 U and an α > 0 so that (2) jjf(x1) − f(x2)jj ≥ αjjx1 − x2jj for all x1; x2 2 U. In particular f is one-to-one on U. Proof. Let T = Df(a). Then −1 jjx1 − x2jj = jjT (T x1 − T x2)jj ≤ MjjT x1 − T x2jj for some M > 0. Pick α = 1=(2M). Now let g(x) = f(x) − T x. Then, on A, Dg = Df − T and so Dg(a) = 0. As g 2 C1(A), we can 2 choose an > 0 so that the absolute value of all entries of Dg is strictly less than α=n for x 2 B(a). This ball will be our open set U. Fix an i 2 f1; : : : ; ng. By the Mean Value Theorem, for any x1; x2 2 U, there exists a c 2 U so that1 α α jg (x ) − g (x )j = jrg (c) · (x − x )j ≤ n · jjx − x jj = jjx − x jj: i 1 i 2 i 1 2 n2 1 2 n 1 2 Therefore, jjg(x1) − g(x2)jj ≤ αjjx1 − x2jj, and then αjjx1 − x2jj ≥ jjg(x1) − g(x2)jj = jjf(x1) − T x1 − f(x2) + T x2jj ≥ jjT x1 − T x2jj − jjf(x1) − f(x2)jj ≥ 2αjjx1 − x2jj − jjf(x1) − f(x2)jj; from which the claimed inequality follows. The most difficult part of the proof of Inverse Function Theorem is the next preliminary result. n n r Lemma 3. Assume A ⊆ R is open and f : A ! R is a function in C (A) for some r ≥ 1. Let B = f(A). If f is one-to-one on A, and det Df(x) 6= 0 for all x 2 A, then B is open and f −1 : B ! A is in Cr(B). Proof. We divide the proof into several steps. Step 1 : B is open. −1 Pick a b 2 B, and let a = f (B) 2 A. Then pick an r > 0 so that Q = Br(a) ⊆ A. Then @Q is compact and the so is f(@Q), which is therefore closed. Moreover, as f is one-to-one, b2 = f(@Q). Thus there is a δ > 0 so that B2δ(b) \ f(@Q) = ;. Pick an arbitrary y 2 Bδ(b) we will show that there is an x0 2 A so that f(x0) = y, which will 2 r show that Bδ(b) ⊆ f(A) and establish Step 1. Let '(x) = jjf(x) − yjj . Then ' : A ! R is in C (A). As Q is compact, ' achieves minimum on Q, say at x0 2 Q. Note that '(a) = jjf(a) − yjj2 = jjb − yjj2 < δ2; 2 and so minQ ' < δ . On the other hand, if x 2 @Q, then jjf(x) − yjj ≥ jjf(x) − bjj − jjb − yjj ≥ 2δ − δ = δ; 1 n p Recall that for x 2 R , jjxjj1 = max jxij and jjxjj1 ≤ jjxjj2 ≤ njjxjj1. 2 2 which implies that x0 2= @Q, and thus is a local minimum in the interior of Q. This implies that f(x0) = y. Step 2 : g = f −1 is continuous. We need to show that for every open set U ⊆ A, g−1(U) is open. But g−1(U) = f(U), which is open by Step 1. Step 3 : g = f −1 is differentiable on B. Let T = Df(a). We need to show that g(b + k) − g(b) − T −1k (3) ! 0 jjkjj as k ! 0. By Lemma 2, there is an open set U containing a, and some α > 0, such that jjf(x2) − f(x1)jj ≥ αjjx2 − x1jj for all x1; x2 2 U. By Step 1, there exists an > 0 so that b + k 2 f(U) whenever jjkjj < . Thus, when jjkjj < , jjkjj = jjb + k − bjj ≥ αjjg(b + k) − g(b)jj which we rewrite jjg(b + k) − g(b)jj 1 (4) ≤ : jjkjj α We rewrite the expression in (3) g(b + k) − g(b) − T −1k k − T (g(b + k) − g(b)) jjg(b + k) − g(b)jj = −T −1 ; · jjkjj jjg(b + k) − g(b)jj jjkjj and we use (4); also, because −T −1 is linear we obtain an M > 0 so that jjg(b + k) − g(b) − T −1kjj M jjk − T (g(b + k) − g(b))jj ≤ · : jjkjj α jjg(b + k) − g(b)jj Now we let h = g(b + k) − g(b) = g(b + k) − a. By continuity of g (Step 2), h ! 0 as k ! 0. Observe that b + k = f(a + h) and so k = f(a + h) − f(a). Therefore, jjg(b + k) − g(b) − T −1kjj M jjf(a + h) − f(a) − T hjj (5) ≤ · ! 0 jjkjj α jjhjj as h ! 0, because T = Df(a). Clearly, (5) implies (3). Step 4 : g = f −1 2 Cr(B). We first assume that r = 1 and show that g 2 C1(B). We know that Dg(x) = Df(g(x))−1. n;n 1 Moreover, the entries of the inverse of an invertible matrix T 2 R are rational, thus C , functions of entries of T . As g is continuous, and the entries of Df are continuous, this implies that the entries of Dg are continuous. Now we proceed by induction on r. Assume the claim holds for functions in Cr−1. If f is in Cr, then certainly f 2 Cr−1, and g 2 Cr−1 by the induction hypothesis. Also, Df is in Cr−1 (or more precisely its entries are) and so Dg, as a composite function of two Cr−1 functions and a C1 function (the inverse) is also in Cr−1. Thus g is in Cr. 2See Problem 7 in Discussion Problems 9. 3 Proof of Theorem 1. By Lemma 2, there is an open set U1 ⊆ A, with a 2 U1, on which f is one-to-one. As det Df(x) is a continuous function of x on A (as a polynomial in the entries of Df(x)), there is a open set U2, with a 2 U2, so that det Df(x) 6= 0 for x 2 U2. Take U = U1 \ U2. Lemma 3 implies that this set has the propertied claimed in Theorem 1. Finally, as already remarked, (1) follows by Lemma 1. m Next is the most famous consequence of the Inverse Function Theorem. In it, we view an R - n+m n m valued function f defined on a subset of R as a function on R × R . That is, f is given by n m f(x; y), where x 2 R and y 2 R . The next theorem tells us when can we the relation f(x; y) = 0 defines a differentiable function y = g(x) from n variables to m variables.