Handout 4. Inverse and Implicit Function Theorems.
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1 8.95 Hndout 4. Inverse nd Implicit Function Theorems. Theorem (Inverse Function Theorem). Suppose U R n is open, f : U R n is C, x U nd df x is invertible. Then there exists neighborhood V of x in U nd neighborhood W of f(x ) in R n such tht f hs C inverse g = f : W V. (Thus f(g(y)) = y for ll y W nd g(f(x)) = x for ll x V.) Moreover, dg y = (df g(y) ) nd g is smooth whenever f is smooth. for ll y W Remrk. The theorem sys tht continuously differentible function f between regions in R n is loclly invertible ner points where its differentil is invertible. Proof. Without loss of generlity, we my ssume tht x =, f(x ) = nd df x = I. (Otherwise, replce f with f(x) = df x (f(x + x ) f(x )). Note tht if the theorem holds with f,,, I nd function g in plce of f x, f(x ), df x nd g respectively, then it is esily verified tht the theorem s stted holds with g(y) = x + g (df (y f(x ))).) x Since df x is continuous in x t x (see Exercise ), there exists number r > such tht x B r () = df x I. 2 (Recll tht for liner trnsformtion A : R n R m we define the norm of A by A = sup { v } A(v).) Fix y B r/2 (). Define function φ by Note tht dφ x = I df x nd hence φ(x) = x f(x) + y. Thus dφ x /2 if x B r (). d φ(x) φ(x) y + y = φ(tx)dt + y dt = dφ tx xdt + y dφ tx x dt + y r/2 + r/2 = r ()
2 whenever x B r (). i.e. φ is mp from B r () into itself. For ny x, z B r (), φ(z) φ(x) = d φ(x + t(z x))dt dt dφ x+t(z x) (z x) dt dφx+t(z x) z x dt z x. 2 Thus φ : B r () B r () is contrction, nd hence φ hs unique fixed point x y B r (). i.e. there is unique point x y B r () with f(x y ) = y. In fct x y B r () since r 2 > y = f(x y ) x y x y f(x y ) x y 2 x y = 2 x y. Set W = B r/2 () nd V = f (W ) B r (). Note then tht V is open. Define g : W V by g(y) = x y. Then f(g(y)) = y for ll y W nd g(f(x)) = x for ll x V. Next we show tht g is differentible, with dg y = (df g(y) ). First note tht with ψ : B r () R n defined by ψ(x) = x f(x), we hve tht for x, x 2 B r (), x x 2 f(x ) f(x 2 ) (x x 2 ) (f(x ) f(x 2 )) ψ(x ) ψ(x 2 ) x x 2 2 where the lst inequlity follows by estimting s in (), using dψ x = I df x. Hence x x 2 f(x ) f(x 2 ) 2 for ny x, x 2 B r (), which implies g(y ) g(y 2 ) 2 y y 2 (2) for ny y, y 2 W = B r/2 (). In prticulr, g is continuous. 2
3 Now fix y W, nd let A = df g(y). Since W is open, there exists δ > such tht y + k W if k B δ (). Let h = g(y + k) g(y). Then k = y + k y = f(g(y + k)) f(g(y)) = f(g(y) + h) f(g(y)) nd hence, for k B δ () \ {}, g(y + k) g(y) A k A (Ah k) h = k h k A k Ah h h k 2 A f(g(y) + h) f(g(y)) Ah (3) h where the lst estimte follows from (2). Note tht since g(y+k) = g(y) = f(g(y + k)) = f(g(y)) = y + k = y = k =, we hve tht h if k =. Sice A = df g(y), it follows from the definition of differentibility of f tht the right hnd side of (3) tends to h s, nd hence, since h 2 k by (2), it follows tht i.e. g is differentible t y nd g(y + k) g(y) A k lim =. k k dg y = (df g(y) ). (4) Finlly, note tht the function y dg y is the composition of the function y df g(y) nd mtrix inversion A A. Mtrix inversion is smooth mp of the entries, nd the function y df g(y) is continuous since g is continuous nd f is C. Hence we conclude tht y dg y is continuous; i.e. tht g is C. Repetedly differentiting (4) shows tht g is smooth if f is smooth. Exercise. Let L(R n ; R n ) be the set of liner trnsformtions from into itself with the metric d(a, B) = A B. (cf. Exercise of hndout.) Let U R n be open nd f : U R n be C function. Show tht the mp x df x is continuous s mp from U into L(R n ; R n ). Exercise 2. Suppose g : [, b] R n is continuous. Show tht g(t)dt g(t) dt R n 3
4 where denotes the Eucliden norm. You my use without proof tht h(t)dt h(t) dt for sclr vlued function h. Exercise 3. Define f : R R by f(x) = 2 + x 2 sin x if x nd f() =. Compute f (x) for ll x R. Show tht f () >, yet f is not one to one in ny neighborhood of This. exmple shows tht in the Inverse Function Theorem, the hypothesis tht f is C cnnot be wekened to the hypothesis tht f is differentible. Exercise 4. Define f : R 2 R 2 by f(x, y) = (e x cos y, e x sin y). Show tht f is C nd tht df (x,y) is invertible for ll (x, ) y R 2 nd yet f is not one to one function globlly. Why doesn t this contrdict the Inverse Function Theorem? Next we prove the Implicit Function Theorem. This theorem gives conditions under which one cn solve, loclly, system of equtions f i (x, y) =, where x R m nd y R n, for y in terms of x. (Thus, y = (y,..., y n ) where y,..., y n re regrded s n unknowns, stisfying the n equtions f i (x, y) =, i =,..., n.) Geometriclly, the set of solutions (x, ) yto the system of equtions is the grph of function y = g(x). Note tht we hve from liner lgebr tht if for ech i, the function f i is liner with constnt coefficients ( ) in the vribles y j, then whenever the (constnt) n n mtrix f i is invertible, the system of equtions is solvble yfor in terms y j i,jn We shll use the following nottion: For Rn n vlued function f(x, y) = (f (x, y), f 2 (x, y),..., f n (x, y)) in domin U R m+n R m R n, where x R m, y R n, we ( shll ) denote by dx f the prtil differentil represented f by the n m mtrix i x j nd by d y f the prtil differentil in,jm ( ) represented by the n n mtrix x i =, 2,... n of x. Implicit function theorem sys tht whenever f i re C nd this mtrix is invertible t point (, b), then the system is solvble y in for terms of x loclly in neighborhood of (, b). f i y j i,jn. Theorem 2 (Implicit Function Theorem). Let U R m+n R m R n be n open set, f : U R n C function, (, b) U point such tht f(, b) = nd d y f (,b) invertible. Then there exists neighborhood V of 4
5 (, b) in U, neighborhood W of in R m nd C function g : W such tht R n {(x, y) V : f(x, y) = } = {(x, g(x)) : x W }. Moreover, dg x = (d y f) d x f (x,g(x)) (x,g(x)) nd g is smooth if f is smooth. Proof. Define F : U by F (x, y) = (x, f(x, y)). Then F is C in U, F(, b) = (, ) nd det df (,b) = det d y f (,b) =. Hence by the Inverse Function Theorem, F hs C inverse F : W V for neighborhoods V of (, b) ndw of (, ) in R m R n. Set W = {x R m : (x, ) W }. Then W is open in R m. Note then tht if x W, then (x, ) W so tht (x, ) = F (x, y ) where (x, y ) V is uniquely determined by x. (In fct, by the definition of F, x = x.) Define g : W R n by setting y = g(x). Thus g(x) is defined by F (x, ) = (x, g(x)); i.e. by g(x) = π F (x, ) where π : R m R n R n is the projection mp π(x, y) = y. Then {(x, y) V : f(x, y) = } = {(x, y) V : F (x, y) = (x, )} = {(x, g(x)) : x W }. Since π is smooth mp nd F is C, it follows tht g is C. The formul for dg x follows by differentiting the identity R m+n f(x, g(x)) on W using the chin rule. By repetedly differentiting this identity, it follows tht g is smooth if f is smooth. 5
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