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Inverse Function Theorem
The contraction mapping theorem is a convenient way to prove existence theorems such as the Inverse Function Theorem in multivariable calculus. Recall that a mapf:U!Rn(whereUis open inRn) isdierentiableat a point x2Uif we can write f(x+h) =f(x) +Ah+e(h);(1) whereA:Rn!Rnis a linear transformation (equivalently, annnmatrix) and ke(h)k=khk !0 ash!0. Thederivativeoffatxisf0(x) =A. Iff0(x) is a continuous function ofx,fis said to haveclassC1. ThenormkAkof a linear transformationAis dened to satisfykAxk kAkkxkfor allx.
We need the following standard result.
Lemma 2Supposekf0(x)k Mon some diskD. Then for any pointsx;x+h2D, kf(x+h)f(x)k Mkhk.Theorem 3LetUbe an open set inRn, andf:U!Rna dierentiable function of classC1. Supposex02Uis a point wheref0(x0)is invertible. Then there exist neighborhoodsU0Uofx0andVofy0=f(x0)such thatfjU0:U0!Vis a bijection. Further, the inverseg:V!U0is dierentiable of classC1, with derivative g
0(y) =f0(g(y))1.(That is,fis a localdieomorphismnearx0.)
ProofWe rst simplify. We may assumex0=y0= 0. Further, we may assume f
0(0) =I; otherwise we replacefbyB1f, whereB=f0(0) (in eect, a linear
change of coordinates in one copy ofRn). We restrict attention to the diskD=fx2Rn:kxk g, whereis chosen small enough to makeDUandkf0(x)Ik 12for allx2D(possible because fhas classC1). Putw(x) =f(x)x; thenw0(x) =f0(x)I, and by Lemma 2, kw(x+h)w(x)k 12khk. In terms off, this is kf(x+h)f(x)hk 12khk:(4) We takeVas the open diskfy2Rn:kyk<12g. We assert that given anyy2V, there is a uniquex2Dsuch thatf(x) =y; we then deneg(y) =x. We deneu:D!Dbyu(x) =x+ (yf(x)); thenf(x) =yif and only ifxis a xed point ofu, i.e.u(x) =x. (The motivation for the denition ofuis that ifxis an approximation to the desired value, with erroryf(x),u(x) should be a better approximation, becausefis in some sense close to the identity function.) We must check thatu(x) does in fact lie inD. If we setx= 0 in (4), we obtain kf(h)hk 12khk 12for anyh2D. Replacinghhere byxand using the triangle inequality, we deduce that ku(x)k kyk+kf(x)xk<12+12=: Next, we check thatuis a contraction mapping (and hence automatically contin- uous). From the denition, u(x+h)u(x) =f(x+h) +f(x) +h:
110.413 Intro to Topology JMB File:invfnthm, Rev. A; 2 May 2003; Page 1
2Inverse Function Theorem
Then by (4),
ku(x+h)u(x)k=kf(x+h)f(x)hk 12khk; as required (with contracting factor 12). We tookDclosed in order to make it a complete metric space. Then the Con- traction Mapping Theorem yields a unique xed pointxofu; in fact, we saw that x=u(x) is aninteriorpoint ofD. We takeU0as the open setf1(V)\IntD. To see thatgis continuous, take any two pointsy;y+k2Vand putx=g(y) andx+h=g(y+k). Then (4) becomeskkhk 12khk. By the triangle inequality, khk khkk+kkk 12khk+kkk. So12khk kkkor, more conveniently, khk 2kkk:(5)
We can ensurekhk< simply by takingkkk<12.
As for dierentiability ofg, (1) becomes
y+k=y+A[g(y+k)g(y)] +e(h):(6) NowkAxxk=k(AI)xk 12kxk, from which it is clear thatAhas zero kernel and is therefore invertible. We applyA1to (6) and rearrange, to obtain g(y+k) =g(y) +A1kA1e(h):
With the help of (5), we see that
kA1e(h)kkkk kA1kke(h)kkhkkhkkkk2kA1kke(h)kkhk!0 ask!0. Comparison with (1)) shows thatgis dierentiable atywith derivative g
0(y) =A1. Finally, the formula forg0(y) is visibly continuous iny.RemarkIf we know thatgis dierentiable, the chain rule gives the derivativeg0(y)
easily. The point is to prove thatgisdierentiable. RemarkThe above proof works in any Banach space, except that to see thatA1 exists, one writes it as the sum of the convergent series A
1=IT+T2T3+:::;
whereA=I+TwithkTk 12.
110.413 Intro to Topology JMB File:invfnthm, Rev. A; 2 May 2003; Page 2
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FONCTION INVERSE - maths et tiques