4 Inverse function theorem

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1 Tel Aviv Universit, 2013/14 Analsis-III,IV 53 4 Inverse function theorem 4a What is the problem b Simple observations before the theorem c The theorem d Iterations The solution = g() of an equation f() = near a given nondegenerate point is an eas matter in dimension one, but for f : R n R n it means a sstem of n (nonlinear) equations in n unknowns. Still, under appropriate conditions, the inverse mapping to a continuousl differentiable mapping is continuousl differentiable. An iterative process converges to the solution. 4a What is the problem Recall the mapping f : R 2 R 2, f(r, θ) = (r cos θ, r sin θ), treated in 2e6. It is not one-to-one, since f(r, θ + 2π) = f(r, θ) and f( r, θ + π) = f(r, θ). However, its restriction to the open set U = (0, ) ( π, π) is one-to-one, and f(u) is the open set V = R 2 \ (, 0] {0}. Thus, (f U ) 1 : V U. B 2e6, f is differentiable on U. We wonder, is (f U ) 1 differentiable on V? π 0 π U f U V The first coordinate r = of (f U ) 1 (, ) evidentl is differentiable on V. The second coordinate θ is differentiable on V b the argument used in 2b18(b): θ=arcsin θ=arccos θ= arccos However, this is just good luck. In general, the inverse mapping is not a combination of well-known functions. (Not even in dimension one; tr for

2 Tel Aviv Universit, 2013/14 Analsis-III,IV 54 instance to find from 5 + =, or + e =.) Can we deduce differentiabilit of f 1 from differentiabilit of f? Of course, we need a multidimensional theor; R 2 is onl the simplest case. 4b Simple observations before the theorem It is not a problem to differentiate the inverse mapping assuming that it is differentiable. B the chain rule, therefore (Df 1 ) f(0 ) (Df) 0 = ( D(f 1 f) ) 0 = I, (Df 1 ) 0 = ( (Df) f 1 ( 0 )) 1. The same argument shows that f 1 cannot be differentiable at f( 0 ) if the operator (Df) 0 is not invertible. (Recall also 2b13(b): and must be of the same dimension.) It can happen that (Df) 0 is not invertible and nevertheless f is invertible. Eample: f : R R, f() = 3, 0 = 0. f()= 3 f 1 ()= 1/3 If f : R R is such that the operator (Df) is invertible (that is, f () 0) for all then f is one-to-one (think, wh). This is not the case for f : R 2 R 2. Eample: f(, ) = (e cos, e sin ). Thus we turn to the local problem: the germ of f at 0 is given, and we eamine the germ of f 1 at 0 = f( 0 ). It can happen that f is differentiable near 0 and (Df) 0 is invertible, but f is not one-to-one near 0. Eample: f : R R, f() = sin 1 for 0, f(0) = 0, 0 = Hint: consider f () for 0.

3 Tel Aviv Universit, 2013/14 Analsis-III,IV 55 Thus, we assume that f is continuousl differentiable near 0. That is, (Df) is continuous near 0. It follows that ( ) (Df) 1 is continuous near 0, see Eercise 4b1 below. Now, assuming again that the inverse mapping is differentiable (and therefore continuous) we see that it must be continuousl differentiable, since ( (Df) 1 f ()) 1 is continuous. 4b1 Eercise. If A, A n M n,n (R), A n A, and A is invertible then A n is invertible for all n large enough, and A 1 n A 1. Prove it. 1 4c The theorem 4c1 Theorem. Assume that a mapping f : R n R n is continuousl differentiable near 0, and the operator (Df) 0 is invertible. Then there eists an open neighborhood U of 0 and an open neighborhood V of 0 = f( 0 ) such that f U is a homeomorphism U V, continuousl differentiable on U, and the inverse mapping (f U ) 1 : V U is continuousl differentiable on V. 4c2 Remark. The equalit (Dg) 0 = ( (Df) 0 ) 1 for g = (f U ) 1 is often included into this theorem. However, it is just an immediate implication of the chain rule, as noted in Sect. 4b. Moreover, (Dg) = ( ) (Df) 1 whenever U, V, = f(). 4c3 Remark. R n ma be replaced with an arbitrar n-dimensional vector or affine space. Likewise, the theorem applies to f : S 1 S 2 for two n-dimensional affine spaces S 1, S 2 ; in this case 0 S 1, 0 S 2, (Df) 0 : S 1 S 2 and (Dg) 0 = ( (Df) 0 ) 1 : S 2 S 1. 4c4 Remark. Onl the germ of f at 0 is relevant. Thus, the theorem ma be applied to a function defined on a neighborhood of 0 (rather than the whole R n ). But never forget: U is generall smaller than the given neighborhood. In contrast, the net result applies to the whole given U. 4c5 Theorem. Assume that U, V R n are open, f : U V is a homeomorphism, continuousl differentiable, and the operator (Df) is invertible for all U. Then the inverse mapping f 1 : V U is continuousl differentiable. 1 Hint. One wa: use determinants. Another wa: first, reduce the general case to the special case A = I (via A 1 A n I); second, prove that A 1 I A I 1 A I whenever A I < 1 (via the triangle inequalit).

4 Tel Aviv Universit, 2013/14 Analsis-III,IV 56 Proof of Theorem 4c1 given Theorem 4c5. Prop. 3b9 and Lemma 3b8 (or Prop. 3b7) provide open sets U 0 and V f( 0 ) satisfing the conditions of Theorem 4c5. B Theorem 4c5 these U, V satisf the conclusion of Theorem 4c1. Proof of Theorem 4c5. Let 0 U, 0 = f( 0 ) V ; it is sufficient to prove that the mapping g = f 1 is differentiable at 0. (Continuit of Dg follows, see the end of Sect. 4b.) Similarl to Sect. 3e we reduce the general case to a special case: 0 = 0, 0 = 0, and (Df) 0 = id. Similarl to the proof of Prop. 3e2 we have U ε such that (f() f()) ( ) ε, (1 ε) f() f() (1 + ε) for all, U ε, and in particular (for = 0), f() ε, (1 ε) f() (1 + ε) for all U ε. The set V ε = f(u ε ) is an open neighborhood of 0 (recall 3b3), and g() ε g(), (1 ε) g() (1 + ε) g() for all V ε. Therefore g() ε 1 ε for all V ε. We see that g() = + o( ), that is, id is the derivative of g at 0. 4c6 Remark. The equalit (Dg) 0 also follows from the proof above. = ( (Df) 0 ) 1 was known before, but 4c7 Remark. Continuit of Dg was known before, but also follows readil from the arguments of the proof above, as follows. For all 1, 2 U ε, (f( 1 ) f( 2 )) ( 1 2 ) ε 1 2, (1 ε) 1 2 f( 1 ) f( 2 ) (1 + ε) 1 2 ; (f( 1 ) f( 2 )) ( 1 2 ) ε 1 ε f( 1) f( 2 ) ; therefore for all 1, 2 V ε (g( 2 ) g( 1 )) ( 2 1 ) ε 1 ε 2 1.

5 Tel Aviv Universit, 2013/14 Analsis-III,IV 57 On the other hand, g( 1 + h) g( 1 ) = (Dg) 1 (h) + o( h ). It follows that (Dg) 1 (h) h ε h + o( h ) ; 1 ε (Dg) 1 (h) h 2ε h for all h near 0, therefore for all h; 1 ε (Dg) 1 id (Dg) 1 (Dg) 0 as ε 1 ε for all 1 V ε ; 4c8 Remark. We see that continuit of the map A A 1 is not necessaril used when proving the inverse function theorem. Curiousl enough, the former can be deduced from the latter. To this end, consider the inverse to a mapping (A, ) (A, A) from M n,n (R) R n to itself. It gives not onl continuit of the map A A 1 (on the open set of all invertible matrices) but also its continuous differentiabilit. But this is not a revelation: elements of A 1 are just rational functions (that is, fractions of polnomials) of the elements of A. (See also 2e7.) 4c9 Eercise. (a) Let f : U V be as in Theorem 4c5 and in addition f C 2 (U) (recall Sect. 2g). Prove that f 1 C 2 (V ). 1 (b) The same for C k (... ) where k = 3, 4,... 4c10 Remark. Now we see that the paths γ used in Sect. 3f, 3g are not just continuous, the are continuousl differentiable, which resolves the doubt of 3f3(a). In relation to 3f3(b) one ma guess that a small ball must contain a connected portion of the path. This need not hold for a continuous path in general, not even for a differentiable path. However, it holds for a continuousl differentiable path (see 4c11). 2 4c11 Eercise. Let γ : ( ε, ε) R n be continuousl differentiable. Prove that the set {t ( ε, ε) : γ(t) γ(0) < r} is an interval provided that r is small enough. 3 4c12 Eercise. Let ψ : U V be as in 3e3(b). Prove that ψ is continuousl differentiable. 1 Hint: (Dg) = ((Df) g() ) 1 where g = f 1. 2 A similar fact for surfaces is beond our course. 3 d Hint: dt γ(t) γ(0) 2 = 2 γ (t), γ(t) γ(0) t γ (0) 2.

6 Tel Aviv Universit, 2013/14 Analsis-III,IV 58 4d Iterations We know that (under appropriate conditions) the solution of the equation f() = eists and is unique. How to compute numericall? Taking into account that is close to 0 = f( 0 ), must be close to 0, and the operator T = (Df) 0 is invertible, we guess that and hopefull, We iterate this operation, = f() = f( 0 + ( 0 )) 0 + T ( 0 ), 0 + T 1 ( 0 ) = 0 + T 1 ( f( 0 )). n+1 = n + T 1( f( n ) ) for n = 0, 1, 2,... and hope that n. These iterations are well-defined for a mapping f : S 1 S 2 between affine spaces (as in 4c3). Choosing appropriate coordinates we return to the special case treated in the proof of 4c5: f : R n R n, 0 = 0, 0 = 0, and T = id. Now we ma use neighborhoods U ε and the related inequalities. Also, the iterations become just n+1 = n + f( n ), that is, n+1 n = f( n ). In particular, 1 0 = 0. That is, 1 =. We have (f( 1 ) f( 2 )) ( 1 2 ) ε 1 2 for all 1, 2 U ε. Assuming (for now) that n U ε for all n we get for all n > 0 n+1 n = ( n + f( n )) ( n 1 + f( n 1 )) = therefore for all n 0, = ( n n 1 ) (f( n ) f( n 1 )) ε n n 1, n+1 n ε n 1 = ε n ; n+k n (ε n + ε n ) 1 = εn 1 ε ; we see that n are a Cauch sequence, thus the limit = lim n n must eist, and n εn 1 ε.

7 Tel Aviv Universit, 2013/14 Analsis-III,IV 59 Also, f( n ) = n+1 n ε n, which implies f() =, provided that U ε. There eists r > 0 such that U ε contains the open r-ball centered at 0. Assuming < (1 ε)r we get n 1 1 ε < r (since k = 0+k 0 ε0 1 ε ), which ensures that U ε contains 1, 2,... and. We summarize. 4d1 Proposition. Assume that 0 R n, f : R n R n is differentiable near 0, Df is continuous at 0, and the operator T = (Df) 0 is invertible. Then for ever near 0 = f( 0 ) the iterative process n+1 = n + T 1( f( n ) ) for n = 0, 1, 2,... is well-defined and converges to a solution of the equation f() =. In addition, 0 = O( 0 ). 4d2 Remark. The proof of Prop. 4d1 does not use results of Sect. 3c, 3d. Thus, Prop. 4d1 is an alternative wa 1 toward Lemma 3b8 and Prop. 3b7. 1 The third wa, if we count also the invariance of domain mentioned in Sect. 3b.