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INVERSE FUNCTION THEOREM and SURFACES in Rn Let F ∈ C K(U;Rn)

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INVERSE FUNCTION THEOREM and SURFACES in Rn Let F ∈ C K(U;Rn) INVERSE FUNCTION THEOREM and SURFACES IN Rn Let f 2 Ck(U; Rn), with U ⊂ Rn open (k ≥ 1). Assume df(a) 2 GL(Rn), where a 2 U. The Inverse Function Theorem says there is an open n n neighborhood V ⊂ U of a in R so that fjV : V ! R is a homeomorphism onto its image W = f(V ), an open subset of Rn; the inverse g = f −1 : W ! V is a Ck map. The inverse function theorem as an existence theorem. Given f : X ! Y (X; Y Banach spaces) and y 2 Y , we seek to solve the nonlinear equation: f(x) = y: The idea is that if f is close (in some sense) to a linear operator A 2 L(X; Y ) for which the problem is uniquely solvable, and (for a given b 2 Y ) we happen to know a solution a 2 X to f(x) = b, then for values y close to b the equation should have a unique solution (close to a). For example, suppose f has the form f(x) = Ax + φ(x), where A 2 L(X; Y ) is invertible (with bounded inverse) and φ : X ! Y is Lipschitz, with a small Lipschitz constant Lip(φ). Then to solve Ax + φ(x) = y assuming the linear problem Av = w can be solved uniquely for any w 2 Y −1 (that is, A is invertible, and jvjX ≤ CjwjY , where C = jjA jj) we compute successive approximations, starting from an arbitrary a 2 X and solving the sequence of linear problems with given `right-hand side': Ax1 = y − φ(a); Ax2 = y − φ(x1); Ax3 = y − φ(x2); ··· Convergence of (xn) is established by setting this up as a fixed point problem, for F (x) = A−1[y − φ(x)]: x1 = F (a); x2 = F (x1); ··· Convergence to a fixed point is guaranteed provided F is a contraction of X: jF (x) − F (¯x)j ≤ λjx − x¯j, where 0 < λ < 1. Since jF (x) − F (¯x)j = jA−1[φ(x) − φ(¯x)]j; it suffices to require Lip(φ) < 1=jjA−1jj: Then the estimate (for two solutions of f(x) = y; f(¯x) =y; ¯ y; y¯ 2 Y : jx − x¯j = jA−1[y − y¯] − A−1[φ(x) − φ(¯x)]j ≤ jA−1j(jy − y¯j + Lip(φ)jx − x¯j); 1 (1 − λ)jx − x¯j ≤ jA−1jjy − y¯j; λ = jA−1jLip(φ) < 1 shows that the inverse map g(y) = x is Lipschitz, with constant Lip(g) ≤ jA−1j=(1 − λ). We recall here the standard fixed point theorem for contractions. Contractions have unique fixed points. Let (X; d) be a complete metric space, F : X ! X a λ-contraction, where 0 < λ < 1: d(F (x);F (y)) ≤ λd(x; y); 8x; y 2 X. Then F has a unique fixed point p 2 X, which is globally attracting (F n(x) ! p, 8x 2 X.) n Proof. Let x 2 X. For the sequence of iterates xn = F (x), the contrac- m m tion property easily implies d(xm+1; xm) ≤ λ d(x1; x) := λ δ, and then if n > m ≥ N: λN d(x ; x ) ≤ d(x ; x ) + : : : d(x ; x ) ≤ (λn−1 + : : : λm)δ ≤ δ; n m n n−1 m+1 m 1 − λ so (xn) is a Cauchy sequence, and xn ! p 2 X. Then p is a fixed point: d(F (p); p) ≤ d(F (p);F (xn)) + d(xn+1; p) ≤ λd(xn; p) + d(xn+1; p) ! 0: It follows easily from the contraction property that F can only have one fixed point. Typically the `perturbation' φ of the invertible linear map A will satisfy a Lipschitz condition only in some open set U ⊂ X. Then given a point b 2 Y for which the problem has a unique solution a 2 U (that is, f(a) = b) and a nearby point y 2 Y , we seek solutions x 2 B¯r(a), (closed ball), where r > 0 is small enough that this ball is contained in U. The successive approximations scheme still works, provided we guarantee this ball is invariant under F . So −1 we estimate, for x 2 B¯r(a) (using a = A [b − φ(a)]): jF (x) − aj = jA−1[y − φ(x)] − aj = A−1[y − b] − A−1[φ(x) − φ(a)]j ≤ jA−1j(jy − bj + Lip(φ)jx − aj) ≤ jA−1j(s + Lip(φ)r); and this is bounded above by r provided we pick y 2 Bs(b) (open ball in (1−λ)r −1 Y ), where s ≤ jA−1j (with λ = jA jLip(φ) < 1 as before). We summarize as follows: 2 Proposition 1. Perturbations of invertible linear maps are homeomorphisms. Let f : U ! Y (U ⊂ X open) have the form f(x) = Ax + φ(x), where A 2 L(X; Y ) is boundedly invertible and φ is Lipschitz with Lip(φ) < jjA−1jj−1. Then V = f(U) is open in Y , and f : U ! V is a homeomorphism with Lipschitz inverse. If U = X, f is a homeomorphism onto Y (with Lipschitz inverse). Exercise 1. Let f : R ! R have the form f(x) = ax + g(x), where a > 0 and g is Lipschitz on R, with constant M satisfying M < a. (i) Show that f is injective. (ii) Show that f is surjective. Remark use only the definitions, quoting no theorems other than the fixed point theorem for contractions in part (ii). A property of C1 maps. Let f : U ! Rm be C1, U ⊂ Rn open. Given a 2 U, we have the remainder: ra(x) = f(x) − f(a) − df(a)[x − a]; x 2 U: n m Then dra(x) = df(x) − df(a) is continuous from U to L(R ;R ), and van- ishes at x = a. We conclude that, for any > 0, there exists an r > 0 so that jjdra(x)jj ≤ if x 2 Br(a). By the Mean Value Inequality: jjra(x) − ra(y)jj ≤ jjx − yjj; for x; y 2 Br(a): In other words: ra is a Lipschitz map with arbitrarily small Lipschitz con- stant, in a sufficiently small neighborhood of a. Equivalently: r (x) − r (y) lim a a = 0: x!a;y!a jjx − yjj Note that this is equivalent to saying that, for all x; y 2 Br(a)(setting ra(x; y) = ra(x) − ra(y)): r( x; y) f(y) − f(x) = df(a)[y − x] + r(x; y); lim a = 0: x!a;y!a jjx − yjj Remark: Note that only continuity of df at a is used. Conversely, if this condition holds at a 2 U, then df is continuous at a. To see this, combining: f(x) − f(y) = df(a)[x − y] + ra(y; x); f(y) − f(x) = df(x)[y − x] + rx(y); 3 we find: (df(x) − df(a))[y − x] = −[ra(y; x) + rx(y)]. Given > 0, we find ρ > 0 and for each x 2 U, 0 < ηx < ρ so that: jjra(x; y)jj ≤ jjy−xjj for x; y 2 B2ρ(a); jjy−xjj ≤ ηx ) jjrx(y)jj ≤ jjy−xjj: Let x 2 U satisfy jjx−ajj ≤ ρ, and let u 2 Rn be a unit vector. Then letting y = x + ηxu we have: x; y 2 B2ρ(a) and jjy − xjj ≤ ηx, so n ηxjj(df(x)−df(a))[u]jj ≤ jjra(x; y)jj+jjrx(y)jj ≤ 2ηx; 8u 2 R with jjujj = 1: Thus, for the operator norm in L(Rn; Rm): jjdf(x) − df(a)jj ≤ 2, establishing continuity of df at a. Turning to the case m = n, let f 2 C1(U; Rn), with df(a) 2 GL(Rn) (that is, df(a) is an invertible linear map) for all a 2 U. Then: f(x) = f(a)+df(a)[x−a]+ra(x) = df(a)[x]+φ(x); φ(x) = f(a)−df(a)[a]+ra(x); so φ is Lipschitz in U if and only if ra is (with the same Lipschitz constant.) Thus φ has arbitrarily small Lipschitz constant in a sufficiently small ball at a, assuming f 2 C1. This motivates the hypothesis in the statement of the Inverse Mapping Theorem. Inverse Mapping Theorem. Let f : U ! Rn (U ⊂ Rn open) be of class C1 in U. Let a 2 U, and assume df(a) 2 GL(Rn) (that is, df(a) is an isomorphism.) Then there exists a neighborhood V ⊂ U of a so that: (i) The restriction fjV is a bi-Lipschitz homeomorphism onto its image W = f(V ), and W is open in Rn. (ii) The inverse map g = f −1 : W ! V is differentiable in W , and dg(y) = [df(g(y)]−1, for all y 2 W ; (iii) If f 2 Ck(U; Rn) (with k ≥ 1) then g 2 Ck(W ; Rn). 1 Proof. (i): Since f is C , we may find an open ball V = Br(a) so that for x; x¯ in V : f(x) = f(a)+df(a)[x−a]+r(x); where jr(x)−r(¯x)j ≤ λjx−x¯j with λjdf(a)−1j < 1: In other words (since f(a) − df(a)[a] is a constant) in this ball V , f is a per- turbation of the isomorphism df(a). Conclusion (i) follows from Proposition 1. 4 Prior to proving (ii), we consider a general result on differentiability of the inverse (of a homeomorphism). Note that the inverse of the homeomor- phism f(x) = x3 of the real line is not differentiable at x = 0.
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