THE UN I V E RS I TY OF M I C H I GA N COLLEGE OF LITERATURE, SCIENCE, AND THE ARTS Department of Mathematics Technical Report No. 25 LIPSCHITZ FUNCTIONS AND LIPSCHITZ TRANSFORMATIONS Lamberto Cesari ~,., ORA' Project 024160 submitted for: UNITED STATES AIR FORCE AIR FORCE OFFICE OF SCIENTIFIC RESEARCH GRANT NO. AFOSR-69-1662 ARLINGTON, VIRGINIA administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR May 1971

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ADDENDUM VI. LIPSCHIITZ IFUNCTIONS AND LIPSCHITZ TRANSFORMATIONS Lamberto Cesari VI 1. DIFFERENTIABILITY OF LIPSCHITZ FUNCTIONS If E is any set of the t-space E, t = (t,...,t ), and z(t), t e E, any real-valued function defined on E, we say that z is Lipschitzian on E (or uniformly Lipschitzian on E of constant M), if for some constant M > 0 we have Iz (t)) - Z(t) MIt-tt for all t, t' E E. A function z(t), t E E, is said to be differentiable at a point t e int E, if there is a linear function 0 L(t) = Al(t -t)+...+A (tv-tv) 1 0 V 0 such that z(t) = z(t ) +L(t) + |t-t o e(t), t E N(t ) cE, 1. v where t =(t,...,t ), t =(t,...,t ), e(t) = 0, and e(t) is continuous in a neighborhood N of t. If z(t) is differentiable at t, then the usual first 0 0 order partial derivatives of z at t exist, and zti(t )= Ai, i = l,..,v. 0 0 1 (VI l.i) Theorem (H. Rademacher). If z is Lipschitzian in an open set G c E, then z is differentiable almost everywhere in G. Proof. For the sake of simplicity we assume v = 2, and we write p =.(x,y) instead of t = (t,...t ), p = (x, y ) instead of t = (to,...,t ), and 0 0-0 0 0 1

A(x, y ), B(x, y ) instead of A1...,A. The proof is divided in a number of parts. (a) Let us prove that the partial derivatives A(x,y) = af/ax and B(x,y) = af/ay exist almost everywhere in G. Since f satisfies a uniform Lipschitz condition on G, f(x,y) is AC in each variable x and y, and therefore af/6x exists everywhere in G but in a set E c G whose intersections with each x straight line y = y has measure zero. An analogous property holds for af/6y. Let us consider now the Dini's derivate numbers D f(x,y) = lim sup h [f(x+h,y) - f(x,y)], x h-+O -1 D f(x,y) = lim inf h [f(x-h,y) - f(x,y)]. — x -xK h+O If R = [A,B] is any closed interval R c G and ~ = dist ([A,B], bd G], let m, -1 n denote integers with ~ < m < n, and let H (x,y) = Max h [f(x+h,y) - f(x,y)], (x,y) E R 1/n < I|h < 1/m K (x,y) = lim H (x,y), D f(x,y) = lim K (z,y). m mn x m n-oo m0oo Then H (x,y) is continuous in R and monotone nondecreasing as n inmn creases; K is monotone nonincreasing as m increases, thus the limits above m exist at every (x,y) c R, and D f(x,y) is certainly Borel measurable in R. x Since R is any interval in G, then D f(x,y) is Borel measurable in G. So is x D f(x,y), and hence the set E of the points with D f(x,y) < D f(x,y) is cer— X X -X X tainly measurable. Since E has an intersection of measure zero with each straight x line y = y, EB has 2-measure zero, and thus 6f/ x exist a.e. in G. The same x 2

argument holds for af/6y. Let K denote the set where both A(x,y) = Sf/ax and B(x,y) = af/ay exist. Then G - K = E U E, and IKI = IGI. x y (b) Let us prove that given X > 0 there is a measurable set S c G such that I|S > IGI-A, and the convergence of the limits X [f(x+h,y) - f(x,y)] > A(x,y) as h + 0, k [f(x,y+h) - f(x,y)] + B(x,y) as h> 0, (VI 1.2) is uniform on S. This is trivial consequence of Egoroff's theorem for the convergence of limits ~n[f(x~l/n,y) - f(x,y)] + A(x,y) as n, ~n[f(x,y~l/n) - f(x,y)] > B(x,y) as n +, (VI 1.3) where n = 1,2,..., and we denote by S a corresponding set of uniform convergence of (VI 1.3) with |SI > IGI - A. Now, given ~, 0 < ~ < 1, there is an N such that (x,y) e S, n > N, implies |~n[f(x~l/n,y) - f(x,y)] -A(x,y)| < a, |n[f(x,y+l/n) -f(x,y)] -B(x,y): < ~, and we may well assume N > M/~. If h is any number, 0 < h < (N+1), then 0 O0 for any h, 0 < h < h, there is an integer n such that (n+1) < h < n, hence n+l > h h 1> N > N+l, or n > N > M/~, as well as 1 < (nh) < 0 1+1/n < 2, and M/n <. Thus, for (x,y) S, 0 < h < h, we have "' '*"*. o *0O 3

Ih[f(x+h,y) - f(x,y)] - A(x,y)| Ih-[f(x+l/n,y) - f(x,y)] - A(x,y) + h l[f(x+h,y) - f(x+l/n,y)]| < (nh)-ln[f(x+l/n,y) - f(x,y)] - A(x,y)| + Mh llh-l/nl < 2~ + M/n < 28 + E = 3~. Analogous reasoning holds for -h < h < 0, as well as for the expression 0 1 -h- [f(x,y+h) - f(x,y)] - B(x,y). This proves the uniform convergence of the limits (VI 1.2) on S. Thus, given ~, 0 < E < 1, there is some h (E) > 0 such that (x,y) E S, -. o. -. 0 < Ihl < h (8), implies. ih-[f(x+h,y) - f(x,y)] - A(x,y)| < ~, h- l[f(x,y+h) -f(x,y)] -B(x,y) <~. (c) We may well assume above that S is closed, S c G, where G is bounded, hence S is closed and bounded, that is, compact. Now A(x,y), B(x,y), as uniform limits of continuous functions, are continuous on the compact set S, hence uniformly continuous on S. Thus, given ~ > O, there is some hl(E) > 0 such that p = (x,y) E S, p' = (x',y') S, IP-P'I < h (), implies IA(p) -A(p')|, jB(p) - B(p')l <. (d) From(VI5.iv) we know that S has density one at almost all of its points. Thus, there is a measurable subset S of S, S c S c G, o o IS = ISI = I|G -, such that, if p,y) if K(p,r) is the cir0 0 0 0 0 0 cle of center p and radius r, and if S (p,r) denotes the set S (p,r) = S n K(po,r), then |S (p,r)/IK(p,r)| - 1 as.r + O. Thus, given ~ > O, there is an h (~;p ) > O such that 0 < r < h (~;p ) implies 2 0 -2 0 4

IS (p r) /IK(p r) > 1-E2(1+)) (e) Let p = (x,y) be a point of S, let E be any number, 0 < E < 1, let r > 0 be any number r < h (E), r < hl(E), r < h2(E;p ), and let p = (x,y) be any point of K(p,r ), r = r(l+). If h = - 0 < h <r =r(l+~), 0 0 0 0- 0 then the measure of the set complementary to S in the circle K(p,h(l+E) is o 2 2 2 -2 2 less than ih2 (1+)2 * (1+E) =-h~. On the other hand, the circle K(p,h~) of center p and radius hE lies in K(p,r) since h+hE = h(l+E) < r, and 22 K(p,h~) has area nh. Hence, there is some point p = (x,y) e S n K(p,E), 0 and Ip-poJ < Ip-pj + IP-Po| < hE + h = h(l+E) < r. We have now f(x,y) - f(x,y) = [f(x,y) - f(x,y)] o0o + [f(x,y) - f(x,y)] + [f(x,y) - f(x,y)] A +A+A 0 00 1 2 3 (VI 1.4) Note that Ip-pl < h8, and hence IA = f(p) - f(p)l M|p-pl <MhE. (VI 1.5) Note that p = (x,y) e S, p = (x,y ) E S, with |p-P | r < h (E), and hence 0 00 0 0 I(x-x )- [f(x,y) - f(x,y)]- A(xy)| <E, l(y-yo)-l [f(xo,y) - f(xo,y0)] - B(xo,yo) < E. Finally, since Po' p e S, Ip-p I <r < hl(E), we have IA(x,y) -A(x, y0) < E, ando thus and thus 5

-1 (x-xO) [f(x,y) - f(x,y)] (y-y) [f(x,y) - f(x,y).0 0 0 =A(x y ) + nl 0 0 ] = B(x,y ) + 0 0 2 with IT 1 < 2~, |n21 <E. Noxr 1.2 A2 - (x-x )A(oY) = f(x,y) O~~) OY - f(xy) - O (x-x )A(x,y) I 0 O0 = (x-x ) [A(x,y ) 0 0 o + ]I - (x-x)A(x yo) 1 0o < Ix-x| IA(x,yo) + < MI p-pl + 261 p-Pol Ix-x I 11 0 1l < Mh~ + 26(h+h ) = (M + 2 + 2~)h < (M+4)he. (VI 1.6) Analogously, I1 - (y-o)B(xoyo) f(x,Y) O - f(Xo,Yo) - (y-yo)B(Xoyo) O).. = I(Y-yo)[B(xoYo) + 2] - (y-yo)B(Xoo) < IY-YI IB(x oo) + Y -yo I l21 < Mp-pI + EIP-pIo < Mh + E(h+hE) = (M+l+~)hE < (M+2)h~. (vI 1.7) By (VI 1.4), (VI 1.5), (VI 1.6), (VI 1.7) we have finally 6

f(x,y) - f(x,y) - (x-xo)A(xo,y) - (y-y)B(x, )I < |A l + IA - (x-xo)A(x,y)I + 1A -(y-y)B(x,yo) < Mh~ + (M+4)h~ + (M+2)h~ = (3M+b)h~, for all p = (x,y) E K(p,r ), and where h = Ip-p = ((x-x)2 + (y-yo )/2 This proves that f is differentiable at every point (x,y ) e S, where S c G, S |I > IGI-\, and A is arbitrary. If now we take x = 1/n and we denote by S the corresponding set S, on 0 n = 1,2,..., and we set E = U S, we see that Son c c G, IG-z| < IG-S o < n oon on - l/n; hence Iz| = |GI. On the other hand, f is differentiable at every point (x,y ) S, and hence at every point of E. Theorem (VI l.i) is thereby o o on proved. VI 2. TRANSFORMATIONS OF CLASS K A one-to-one transformation q = Tp, or T: G + U, with inverse p = Tq -1 or T: U + G, of an open set G in E into an open set U of E such that V V both T and T are uniformly Lipschitzian in G and U respectively is said to be a transformation of class K. Thus, if T is such a transformation, there is a constant M > 0 such that, for all points, p, p' e G and q, q' E U we have I Tp ' I < Mp-p', ITq -T- qI <Mlq-q'. (VI 2.1) If x = (x,x) G u = (u then T and T can be rep resented by real-valued functions 7

i i T: u = u (x), i = 1,...,v, x C G, T: x = x (u), i = l,...,v, u E U, (VI 2.2) u (x), x (u) uniformly Lipschitzian in G and I respectively with constant M, then certainly T and T are of class K with constant vM. i i Note that, as proved in (VI 1), all functions u (x), x (u) are differentiable almost everywhere in G and U respectively, hence they possess first order partial derivatives (u /8x, au /Su, i,s = l,...,v, almost everywhere in G and U respectively. We shall denote by [du/dx] and [dx/du] the nxn matrices of these partial derivatives, and then the usual Jacobians du/dx = det[du/dx], dx/du = det[dx/du] are defined a.e. in G and U respectively. We shall prove below (VI 2.iii) that, under transformations of class K, sets of measure zero in G and U correspond. We shall need a few lemmas. (VI 2.i) (lemma). If W is any measurable set in G, and W* = T(W) is the image of W under a transformation T: G - Usatisfying ITu-Tv I < M lu-vl for all u,v e G cE, then IW*j < 2VM WI|, where IW*| denotesthe outer measure of W*. Proof. For the sake of simplicity let us assume v = 2. Given E > 0 we may cover W with countably many circles Ki, i = 1,2,..., such that 8

Cx' E K < W + ~. Let pi denote the radius of Ki, and let K' be the image -Z1 i i of K under T. If qi qi denote any two points of K, q Tp, q' = T P., p!i K, then. -p! < 2p, q. -q'1 < 2Mp.; hence K' is contained in the circle, say K', of center any point qi of K and radius 2Mp. Then IW*1 < - 1 K = K i=l p2 2 K 4M2 [-W+]. l-J- 1 3 —JL 1. I~~-L J Since E > 0 is arbitrary, the lemma follows. (VI 2.ii) (lemma). Any measurable set W in E is the countable union of comV pact sets and of a set of measure zero. Proof. As usual assume v = 2 and divide E into the congruent squares 2 Q. = [i < x < i+l, j < y < j+l], i,j = 0, ~1, ~2,.... It is enough to prove the lemma for each of the sets W n Qj. Let us denote by W any of these sets. From [53] we know that there are two sets S and P, S c E c P, with IP-SI = 0, where S is the countable union of closed sets, say S = U, F c W c Q, and J JJ J P is the countable intersection of open sets say P = P., W c P.. Each set F is certainly bounded hence compact, and now J W = (U.F.) U (W- F) where W-U F = W-S c P-S, and Iw-U F.I = |P-S| = 0. (VI 2.iii) Theorem. If T: G - U is any transformation satisfying Tu-Tvl < M u-v| for all u,v E G, then the image W* of any measurable set W c G is measurable, and if |WI = 0 then also IW*1 = O. If T is of class K, 9

then measurable subsets of U and G correspond, and the sets of measure zero also correspond. Proof. That |WI 0 implies IW*| = 0 is a consequence of (VI 2.i). Now, if W is a measurable subset of G, then W = (U.Fj) U Z, where the sets F. are compact, and Z has measure zero. If F* = T(F.), j = 1,2,..., and Z* = T(Z), then Z*I = 0, and all sets F.* are compact as the continuous images of comJ pact sets. Finally W* = T(W) = (U.Fj*) U Z*, and W* is certainly measurable. < J VI 3. THE MAGNIFICATION RATIO AND THE JACOBIAN Let T: G - U be a given transformation of class K. For every point p E G let us consider the hypercube Q = [px -x, i,...,v] of center p and sidelength 2j, where we suppose. sufficiently small so that 1 v 1 v Q c G, and p = (x...x, p = (x,...,x). Let Q' = T(Q) be the image of Q under T. As we know from (VI 2.iii), Q' is a measurable set of measure IQ'I. If the limit exists A(p) = A(xl,...,x ) = lim IT(Q) I/Q| as O0, (VI 3.1) 0.,) =~ O O0 then we say that A(p ) is the magnification ratio of T at the point p. (VI 3i) Theorem. If T: G or T:, x G is ans(VI 5.i) Theorem. If T: G -+ U, or T: u = u (x), i=1,...,v, x~ G, isatransformation of class K, then the magnification ratio exists a.e. in G, and A = Idu/dxl a.e. in G. Proof. For the sake of simplicity we assume v = 2, so as T is given by the real functions u = u(x,y), v = v(x,y), (x,y) E G, and we know already 10

that the functions u(x,y), v(x,y) are differentiable at almost every point p = (x,y) e G. It is enough to prove that the statement holds at every point p = (x,y ) G where u and v are both differentiable. Let p = (x,y ) be o O O o such a point. Let A = u, B= u, C = v, D = v where all these derivatives x y x y are computed at (x,yo), and note that u(x +h,y +k) = u(x,y) + Ah + Bk + (h +k ) R(h,k), 0. 0 0 0 v(x +h,y+k) = v(x,yo) + Ch + Dk + (h2+k2)l/2R'(hk) where R, R' + 0 as h, k O. Let u = u(,y), = V( )and let T be the affine transformation T: u = u + Ah + Bk, v = v + Ch = Dk. (VI..2) O 0 Then T maps Q into a parallelogram V of center (u,v ), u = u(x,y ), v = v(x,y), which may be degenerated into a segment or a point. Suppose first L = AD-BC = O. Then V is degenerated to a segment (or a point) of length k\, where \ > 0 is a fixed number which depends on A, B, C, D. Let ~ be an arbitrary number, 0 < ~ < 1. Let t > 0 be sufficiently small so as for h +k < 26 we have IRI, IR'I < ~. Let us assume a <6 and so small that Q c G. As p = (x,y) describes Q, the point Tp describes the 2 2 2 2 segment V. On the other hand, |h|, |k| < I, h +k < 22 < 2 and k0_ _, 11

2 2 1/2. 2 2 1/2 2 1/2 2 1/2 IT -T I (R2+R' 2)1/ (h2+k2) < (22) /(21) ) = 2e2. P P Hence, T(Q) is completely contained in an interval of center (u,v ) and sideO O lengths \X + 4el and 4he. Thus IT(Q)| < (%\ + 4ea)(4aE), QI = 2, and IT(Q) /IQ < (\ + 4E)4E. Thus IT(Q)|I/Q + 0 = L as 2 + 0, and A(p ) = L. Let us suppose now L = IAD-BCI # 0. Then V is a nondegenerated parallelogram of center (u,v ) and area IV| = LIQI. The minimum distance of bd'V from (u,v ) will be pJ where i > 0 is a fixed number which depends on A,B,C, D. Let V1, V2 be the parallelograms of center (u,v ), affine to V, such that O O bd V1 and bd V have minimum distance from (u,v ) equal to (1-~)~1, (1+~)P, 2 o 0 E)4 I respectively. Let ~ be an arbitrary number, 0 < E < 1. Let 5 > 0 be sufficiently small so as for h +k < 25 we have IRI, IR'I _< E/4. Let us assume I < 6 and so small so that Q c G. As p = (x,y) describes bd Q, then 2+2 2 2 |h|, |k| < I, hk 2 < 2 2,and 2 2 1/2 2 2 1/2 2 2 1/2 21/2 ITp-Tpl (R2 +R2)/2(h2+h2)12 < (2pe2/16) 2(2 )1/ 2. bd V Thus, the image of bd Q under T is an oriented curve C which lies between the parallelograms V and V 1 2 Actually, as p describes once bd Q, then Tp describes once the boundary of V, while Tp describes the curve C with | Tp-Tp| < [~s/2. This shows that all points of the interior parallelogram V, are linked by bd V as well as by. C and that V c T(Q) c V; hence 1 2 12

(1-) 21I< I|V I IT(Q)I < V I = (1+~)21 V with IVI = LlQ|, and finally (1-E) L < IT(Q)I/IQI < (1+)2 L, for every ~ with 0 < I < 6 and Q c G. This proves that IT(Q) //IQ +' L as I 0, or A(p ) = L. O We have proved that A(p ) = L = IAC-BDI at every point p = (x,y ) c G 0 0 00 where both u(x,y), v(x,y) are differentiable, that is, almost everywhere in G. (VI 3.ii) Let T: G U, or T: u = u (x), i = l,...,v, x E G, be a transformation of class K. Let A be a measurable subset of G where D(x) <_ for some constant 1, and A' = T(A). Then A' is measurable and IA' I < p |A. Proof. As usual we may assume v = 2, and IAI <. Let A be the set of 0O all points p E A of density one and at which A(p) exists. Then |A I = IAI. Given ~ > 0, for every point p e A there are squares K of diameter as small O as we want with |AnK|I/IK = IA nKI/IKi > 1-~, IT(K)| C (P+e)IKI. O 0 By Vitali's covering theorem there is a countable system of points p e A s O0 and corresponding disjoint squares K, s = 1,2,..., such that S p K, lAnK I/IK I = A AK 1/IK I > 1-8, IT(K ) < (P+~)JK ), s S s s o s s s S s = 1,2,..., IA-U K l = 0. S S 13

Then |T(A-U K )| == 0, and S IA'1 = I T(A)1 = T[(u A K ) U (A-UK )] = s IT(AnK )I + T(A-U K )| S S S s < T(K ) < (P+~))K I < (P+)(1-~) Z JAnK I S S S S < (+-~)(1-~)- |A|. Since ~ > 0 is arbitrary, we have proved that IA't < PIAI. i i (VI 3.ii) Let T: G - U, or T: u = u (x), i = l,...,v, x c G, be a transformation of class K. Let A be a measurable subset of G where D(x) > c for some constant a > 0, and A' = T(A). Then A' is measurable, and IA' > a IA|. Proof. As usual we assume v = 2. Since the statement is trivial for a = O, we may well assume a > O. As in the proof of (VI 3.ii) we assume |A| < + o. Let A be the set of all points p E A of density one and at which A(p) exists. Then IA I = IA. Given ~, 0 < ~ <, for every. point p c A there are squares K of diameter as small as we want with p E K, IAnK|/IKI = |A nK|/I|K > 1-~, IT(K) > (a-~)IKI. Then IK-Al = IK- lAnKI < IKi (1-~)IKI = ~IKI, and by (VII 2.i) also 14

T(K-A): 4M_ 21|K|. We shall assume ~ > 0 sufficiently small so that ~+4M~ < a. By Vitali's covering theorem there is a countable system of points p c A and corresponding disjoint squares K, s = 1,2,..., such that s p c K, lAnK |I/IK = IA AnK /|K I > 1-~, s s s o S 0s s T(Ks) > (a-e)IK1, T(K -A) < 4M |K, S = 1,2,..., I|A-U Kl = 0. s s Then IT(A-U K )I = 0 and S S IA'I = |T(A)j = T[(U AnK ) U (A-U K )] = I T(AnK )I + IT(A-U K )I S S S S ZT(K ) - ZT(K -A) s s S s > ( Z-E~) zS I S - 4M2 z |IK -- / s s s S > (a-&-4M 2 ) z |AnK | = (Ca —4M 2)|A|. S S Since ~ > 0 is arbitrary, we have proved that IA' > A|. VI 4. THE RULE OF MULTIPLICATION OF JACOBIANS (VI 4.i) Let T: U - G, or T: x = x(u), u c U, x = (x,...,x ), u = (u,...,u), i i be a transformation of class K. Let z = z (x), i = 1,...,v, x E G, be Lip 15

schitzian functions in G, write z = z(x), x c G, z = (z,...,z ) and let Z(u) = z(x(u)) = (Z,...,Z ), u U. Then the functions Z are differentiable a.e. in T, (VI 4.1) z1/auj = (azi/ax )(axr/auj), i,j = l,..,v, and the r=l rule of multiplication of Jacobians hold: dZ/du = (dz/dx)(dx/du), a.e. in U. (VI 4.2) Proof. Each function Z (u) = z (x(u)) is uniformly Lipschitzian in U as the composition of Lipschitzian functions. Hence, each function Z (u) is certainly differentiable a.e. in U by (VI l.i). Actually, if H c G is the O set of points where some z are not differentiable in G, then meas H = 0, 0 and we know from (VI 2.iii) that the set HI = T (H ) c U also has measure zero in U. Let H be the set of points in U where some of the functions 2 x (u) are not differentiable. Then meas H2 = 0, and if H = H U H2, also meas H = O. If u c U - H, and x = Tu, then all z are differentiable at o O O x, and all xj are differentiable at u. Now the proof that all Z (u) = O O z (x(u)) are differentiable at u and that formulas (VI 4.1) hold is the same as in calculus. Formula (VI 4.2) follows by the composition rule of matrices. (VI 4.ii) If T: U + G is a transformation of class K, say T: x = x(u), -1 u c U, and T:u = u(x), x E G, then [dx/du][du/dx] ( = I, X=X(u) [du/dx] U [dx/du] = I, x=x(u)

a.e. in U, where I is the unitary matrix, and hence the products of the relative determinants (jacobians) is equal to one a.e. in U. An analogous statement holds by exchanging x with u. A consequence of (VI 4.i) (VI 4.iii) If T: U + G is a transformation of class K, and U is connected, say T: x = x(u), u e U, T: u = u(x), x G, then either dx/du > 0 a.e. in U, or dx/du < 0 a.e. in U. If dx/du > 0 a.e. in U, then du/dx > 0 a.e. in G, and T is said to be positive. Proof. By (VI 4.i) and (VI 4.ii) we see that (dx/du)(du/dx) = 1 x(X(u) a.e. in U, hence dx/du i 0 a.e. in U, and dx/du and du/dx have the same sign in corresponding points (a.e. in U). It remains to prove that dx/du has the same sign wherever it is different from zero in U. To prove this we need the concept of topological degree [3] and a few of its properties. I. U is an point of U, c any cell in U, (c closed), T any continuous map from U into E, and x any point x C E - T(bd c), then the topological degree O(x;c,T) is an V integer (~ 0), and (a) O(x;c,T), as a function of x in E, is constant on V each component of the open set E - T(bd c), and (x) O(x,c,T) # 0 implies V that x = Tu for some u e int c. Also, (c) if T is 1-1 then O(x;T,c) = +1 for every point x c T(int c). Thus, if T is 1-1, since T(int c) is a single component of E - T(bd c), we deduce from (a) that either O(x;c,T) = 1 for all V x c T(int c), or O(x;c,T) = -1 for all x E T(intc ). Now, if u c U, and c,c' are any two cells in U, with u e c cin t c, c int cc' cU, then we can 17

prove that (6) O(x;c,T) = O(x;c',T). Indeed, if we divide 6(c'-c) into finitely many not overlapping cells cl,...,CN, then by an additive property N of the topological degree we have O(x;c',T) = O(x,c,T) + E. O(x;c.,T). J=l j J Then O(x;c,T) # O(x;c',T) would imply O(x;cj,T) # 0 for at least one j, -1 hence by (D), there would be a counter image u' c T x inthe interior of c, hence T(u) = T(u') = x, u ~ u', a contradiction, since T is one-one. Thus O(x;c,T) = O(x;c',T)., C. C 1 Finally, if ul, u2 are any two points of U and U is connected, then we can take a line I in U joining ul and u2, and a cell c containing ~ and hence u and u2 in its interior. If cl, c2 are two cells with u e int ci, Ci c int c, i = 1,2, then by (6), O(xi,c,T) = (xi;c,T), i = 1,2, and by (a), 0(xl,c,T) = (x2,c,T). Hence (E) 0(xl;cl,T) = 0(xl,c,T) = 0(x2,c,T) 0(x2,c2,T). Let us assume now that T be as in the statement. We have seen in the proof of (VI 3.i) that for any u e U with dx/du 0, we can take a cube c (cell) as small as we want, with u e int c, c c U, and 0(x,c,T) = sgn(dx/dt). Thus, if ul, u are any two points of U where the Jacobian is f 0, and cl, c2 corresponding cubes, and xi = Tui, i = 1,2, then, by (E), sgn1 2/dt) O(X2, C21 U="2 sgn(dx/dt) = 0(x;cl,T) = 0(x2,c,T)= sgn(dx/dt) Statement (VI 4.iii) is thereby proved. 18

VI 5- TRANSFORMATION OF MULTIPLE INTEGRALS (VI 5.i) Theorem. Let T: G -> U be a transformation of class K. Thus, T is one to one, and T: u = u(x), x e G, T x = x(u), u E U, are uniformly Lipschitzian. Let f(x), x C G, be any measurable function in G, and let F(u) = f(x(u)), u c U. Then F(u) is measurable in U, and f F(u)du = / f(x)ldu/dxldx, (VI 5.1) U G whenever either F(u) or f(x)Idu/dxl is known to be L-integralbe, and then both F(u) and f(x)ldu/dx| are L-integrability and (VI 5.1) holds. Proof. As usual we assume v = 2 and we use the notation T: u = u(x,y), v = v(x,y), (x,y) c G, for T, and we write f(x,y), F(u,v) = f(x(u,v), y(u,v)), and L(x,y) = ju v - u v |. x y y x (a) Let us assume first that A is any measurable subset of G with I A < + oo, and that f(x,y), (x,y) c A, is any bounded measurable nonnegative function. Then A' = T(A) is also measurable with IA'I < o by force of (VI 2.i), F and fL are bounded and measurable, as we shall prove that A' F(q)dq = [A f(p)L(p)dp. (VI 5.2) For every 6 > 0 and k = 0,1,2,..., let e = [pJk6 < f(p) < k+l]. Then k each set ek c G is measurable, and so is the set e' = T(e ). Then ke stk k = k lekl '< A F(q)dq k= (k+)61e I, (VI 5-3) where the left sum converges as 6 + 0 since f F(q)dq < + oo. Further At 19

k |e' = IA'I < + X, and hence the right sum also converges as 6S 0, and the k k two limits coincide with fA F(q)dq. For every j - 0,1,2,..., let s. = [p c Ajb6 < A(p) < (j+l)6]. Then s is measurable, the sets s' = Tsj are measurable, and by force of (VI 3.ii), (VI 3.iii) we have j6|eks J le s j < (j+l)f|es. (V 5.4) By summation on j we have j jleks l < le' 1 < E (j+l)bleks.1, k i where the sums converge as 6 + 0 since lek1 < + o, and jle s I = le I < + oo. kj j k From (VI 5.3) and (VI 5.4) we have 2k k 'jEleks < I A CF(q)dq < k(k+l).(j+l) eks. (VI 5.5) Again, the left side converge as 6 + 0 since fA,F(q)dq < + oo, and the right side also converges if we note that 6, j kle sjl = k E kle kl f(p)dp EkEj jleksj = 2.jle.l < IA't kjk3 Zk Vj leksj = IAL and we can rearrange the sums since all terms are nonnegative. Consequently, 2 2 2kCj (k+l)(j+l)le - k e i < () (VI 56) Eky'ij k kjj k 20

where 2 ri() 6' FA f(p)dp + 6^A' + 6 2Al, and rq(6) + 0 as -+ 0. On the other hand, for all p e eksj we have kj 62 e sI| < f(p)A(p) < (k+l)(j+1)2 leks, k k J hence and from Since rq( s kj kile s < f(p)A(p)dp < S Z k+ j +l) k j kj - j (VI 5.5) and (VI 5.6) also A, F(q)dq - fA f(p)A(p)dp| < rl(6). +) + 0 as -+ 0, we have proved that fA F(q)dq = fA f(p)A(p)dp and (VI 5.2) is thereby proved. (b) Now let A be any measurable subset of G, and f(x,y), (x,y) E A, any nonnegative measurable function on A. Then there are compact sets C, n = 1,2,..., such that C c A, C c Cn, IC < + n = 1,2,..., and n' ' n ' n n+1 n A-A I 0 where A= U C If T(C) A' = T(A ), then A'-A' = 0 0 0 n n n n 0o o where A' = U C'. Now let us define f (p), p A, by taking f (p) = 0 in o n n n n A-C; f (P) = f(p) if f(p) < n and p c C; and f (p) = n if f(p) > n and n f) n n - p c C. Then f (p) -+ f(p) as n - oo for all p e A, and if F (q) = f (T1 n n o n n q 21

q c A', then F (q) - F(q) as n -+ for all q c A. Finally, both sequences n o [f (p) p e A, n=l,2,..], [F (q), q e: A', n=l,2,...] are nonnegative monon n tone nondecreasing sequences. By part (a) above we have /C F (q)dq = fn(p)A(p)dp, n=l,2,..., ' n ' n n n and also A' F(q)dq = f (p)A(p)dp, n = 1,2,.... Al n An By Lebesgue monotone convergence theorem, F is L-integrable if and only if jA' F (q)dq as a finite limit as n + o. Analogously, f A is L-integrable n if and only if A f (p)A(p)dp has a finite limit as n +. Therefore, A n:A'F(q)dq = f f(P)A(p)dp whenever one of these integrals exists, and then the other one also exists, and equality holds. (c) Now, let f(x,y), (x,y) c G, be any measurable function. Let G, G be the subsets of G where f > 0, of f < 0 respectively. Then, if U = T(G ), U = T(G ), by force of part (b) above we have + F(q)dq = / + f(p)A(p)dp, f F(q)dq = J f(p)A(p)dp. U G U G (VI 5.7) The first relation in (VI 5.7) holds if we know that either F is L-inte+ + grable in U, or that fA is L-integrable in G, and an analogous statement 22

holds for the second relation in (VI 5.7). By combining the two relations in (VI 5.7) we have f F(q)dq = f(p)A(p)dp, (VI 5.8) U r and this relation certainly holds if we know that F is L-integrable in U, or that fA is L-integrable in G. Finally, by (VI 5.8) and statement (VI 3.iii), we deduce statement (VI 5.i). Statement (VI 5.i) can be extended in the following way: (VI 5.ii) Let T be a one to one transformation from G to U which preserves measurable sets. Assume that F is integrable on U and there are countably many sets U, n=l,2,..., such that U c U; U U = U-B with BI = 0; and n n n+l n n T is of class K on each U. Then relation (VII 5.1) holds. n This theorem follows from (VI 5.i) by application of Lebesgue monotone convergence theorem. Statement (VI 5.ii) applies for instance in the transformation of cartesian into polar coordinates. Bibliographical notes. The property of differentiability almost everywhere of Lipschitz functions was proved by H. Rademaker [94]. The concept of transformation of class K (or class K, I > 1) was studied by C. B. Morrey [82b]. For the concept of topological degree mentioned in the proof of (VI 4.iii) the reader may see P. Alexandroff and H. Hopf [3], or T. Rado and P. Reichelderfer [96]. Another proof of (VI 4.iii) is given in thelast mentioned work 6, p. 132]. 23

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