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MATH 23b, SPRING 2005
THEORETICAL LINEAR ALGEBRA
AND MULTIVARIABLE CALCULUS
The Inverse Function TheoremThe Inverse Function Theorem.Letf:Rn-→Rnbe continuously differentiable on some open set containinga, and suppose detJf(a)?= 0. Then there is some open setVcontainingaand an openWcontainingf(a) such thatf:V→Whas a continuous inversef-1:W→Vwhich is differentiable for ally?W.
Note: As matrices,J(f-1)(y) = [(Jf)(f-1(y))]-1.
Lemma:Let
A?Rnbe an open rectangle, and supposef:A-→Rnis continuously differentiable. If there is someM >0 such that????∂f i∂x
Proof:We write
f i(y)-fi(z) =fi(y1,...,yn)-fi(z
1,...,zn)
n? n? j=1∂f i∂x j(xij)(yj-zj) for somexij= (y1,...,yj-1,cj,zj+1,...,zn) where, for eachj= 1,...,n, we havecjis in the interval (yj,zj), by the single-variable Mean Value
Theorem.
Then i=1||fi(y)-fi(z)|| n? i=1n j=1? ∂fi∂x j(xij)????· |yj-zj| n? i=1n j=1M· ||y-z|| =n2·M· ||y-z|| 1
Proof of the Inverse Function Theorem:
(borrowed principally from Spivak"sCalculus on Manifolds) LetL=Jf(a). Then det(L)?= 0, and soL-1exists. Consider the com- posite functionL-1◦f:Rn→Rn. Then:
J(L-1◦f)(a) =J(L-1)(f(a))◦Jf(a)
=L-1◦Jf(a) =L-1◦L which is the identity. SinceLis invertible, the theorem is equally true or false for bothL-1◦fandfsimultaneously, and hence we prove it in the case when L=I. Supposef(a+h) =f(a). Then|f(a+h)-f(a)-L(h)||h|=|h||h|= 1.
On the other hand, we have have lim
||h||→0f(a+h)-f(a)-L(h)||h||=0, which is a contradiction, and hence there must be some open neighborhood/rectangle
Uaroundain whichf(a+h)?=f(a),?a+h?U,h?=0.
Furthermore, we may choose this neighborhoodUsmall enough so that: •det(Jf(x))?= 0,?x?U ????∂f i∂x j(x)-∂fi∂x j(a)????<12n2,?i,j,?x?U since these are conditions onn2+ 1 continuous functions! Proof of Claim 1:First, we letg(x) =f(x)-x. By construction and the second fact above, we have????∂g i∂x j(x)????=????∂f i∂x j(x)-∂fi∂x and so we apply the Lemma withM=12n2: =||g(x1)-g(x2)|| 12
· ||x1-x2||
and so, combining these inequalities, we have 12 2 Now consider the set∂U, which is compact sinceUis bounded. We know by the reasoning in the second paragraph of the proof that ifx?∂U, then f(x)?=f(a). Hence?d >0 such that||f(x)-f(a)|| ≥d,?x?∂U. (Since bothfand the taking of norms are continuous functions, the expression ||f(x)-f(a)||attains its non-zero minimum on the compact set∂U.) We construct the setW?Rn, thinking of it as a subset of the range off, as follows: W=? y?Rn????||y-f(a)||
Proof of Claim 2: Existence:Considerh:U→Rdefined byh(x) =||y-f(x)||2. A straightfor- ward simplification of this expression givesh(x) =n? i=1(yi-fi(x))2. Note thathis continuous and hence attains its minimum on the compact setU. This minimum doesnotoccur on the boundary, ∂U, by the inequality (1), and hence it must occur on the inte- rior. Sincehis also differentiable, we must have?h(x) = 0 at the minimum, and hence: 0 = ∂h∂x j(x) =n? i=12·(yi-fi(x))·∂fi∂x j(x),?j In other words, collecting this information over the variousiandj, we have 0=Jf(x)·(y-f(x)),
but since we have assumed that detJf(x)?= 0 for anyx?U, it follows thatJf(x) is invertible, and hencey-f(x) =0. Uniqueness:We use Claim 1. Supposey=f(x1) =f(x2).
3 By Claim 2, if we defineV=U∩f-1(W), thenf:V→Uhas an inverse! It remains to show thatf-1is continuous and differentiable. Even though continuity would follow from differentiability, we do this in two steps because we will use the continuity to help prove the differentiability. Claim 3:f-1is continuous.
Proof of Claim 3:
Fory1,y2?W, findx1,x2?Usuch thatf(x1) =y1andf(x2) = y It is now easy to see that givenε >0, we need only chooseδ=ε/2 to guarantee that if||y1-y2||< δ, then||f-1(y1)-f-1(y2)||< ε. Claim 4:f-1is differentiable.
Proof of Claim 4:
Letx?V, letA=Jf(x), and lety=f(x)?W.
We claim thatJf-1(y) =A-1.
Define?(x) =f(x+h)-f(x)-A(h).
Then lim
||h||→0||?(h)||||h||= 0, by the differentiability off. Since det(A) = detJf(x)?= 0 by hypothesis, we know thatA-1 exists, and it is linear sinceAis. Then: A -1(f(x+h)-f(x)) =h+A-1(?(h)) = [(x+h)-x] +A-1(?(h)) Lettingy=f(x) andy1=f(x+h) on both sides yields:
A -1(y1-y) = [f-1(y1)-f-1(y)] +A-1(?(f-1(y1)-f-1(y))) Re-arranging sides:
A -1(?(f-1(y1)-f-1(y))) = [f-1(y1)-f-1(y)]-A-1(y1-y) (2) 4 To show differentiability, we need:
lim ||y1-y||→0||f-1(y1)-f-1(y)-A-1(y1-y)||||y1-y||= 0 but by equation (2) above, this is the same as showing: lim ||y1-y||→0||A-1(?(f-1(y1)-f-1(y)))||||y1-y||= 0. SinceA-1is linear, it suffices to use the Chain Rule and show that: lim ||y1-y||→0||?(f-1(y1)-f-1(y))||||y1-y||= 0,(3) so we factor the expression inside the limit as follows: The first term on the right tends to 0 because of how we defined ?and the fact that the continuity off-1means thatf-1(y1)→ f -1(y). Observing that the second term on the right is less than or equal to 2 (by Claim 1) enables us to use the Squeeze Theorem and conclude
that the product on the right tends to 0, which establishes equation (3). End Proof of Inverse Function Theorem.
(borrowed principally from Spivak"sCalculus on Manifolds)
5quotesdbs_dbs4.pdfusesText_7
FONCTION INVERSE - maths et tiques