. Then Φ is a diffeomorphism from (0, ) onto (0, 1)

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1 : uv i Set F s, t s α 1 t β 1 e s+t and Φu, v. Then Φ is a u1 v diffeomorphism from 0, 0, 1 onto 0,, and JΦu, v u. Hence, ΓαΓβ F s, t dsdt F Φu, vjφu, v dudv 0, 0, 0,1 u α+β 1 e u v α 1 1 v β 1 dudv Γα + βbα, β. 0, 0,1 ii Clearl, b Theorem 5.1.8, 1 R 1 + λ dx ω 1 0, r r λ dr. ext, take Φr r 1+r. Then Φ is a diffeomorphism from 0, onto 0, 1 and JΦr r 1+r. Thus 0, r r λ dr 1 Φr 1 1 Φr λ 1 JΦr dr 0, 1 t 1 1 t λ 1 dt 1 B, λ. 0,1 iii First note that 1 ξ 1 dξ 1 ξ 1 dξ. 1,1 0,1 ext, check that Φξ ξ is a diffeomorphism from 0, 1 onto itself and that JΦξ ξ. Thus, 1 + ξ 0,1 1 dξ 1 1 0,1 0,1 Φξ 1 1 Φξ 1 JΦξ dξ t 1 1 t 1 dt 1 B 1, 1.

2 5..1 i B Fubini s Theorem and Theorem 5.1.8, 0, r 1,1 S 1 F r Ξρ, ω µ dρ dω 1 ρ 1 r F 1 ρ 1 rω, rρ dr λs 1dω dρ 0, S 1 1 ρ 1 F 1 ρ 1 x, ρ λr dx dρ R 1 F Φx, λr +1dx d. 1,1 1,1 R 1,1 ii Clearl Φ is one-to-one and onto. In fact, if u, v R R \ {0, 0}, then Φ 1 u, v u + v 1 u u. v To compute JΦ, first note that the Jacobian matrix for Φ at x, can be expressed as 1 1 I R 1 1 x, x where x is thought of as a row vector and x is the corresponding column vector. Using column operations, one can eliminate the first entries from the + 1st column to see that 1 det 1 I R 1 1 x x 1 1 I R 0 det x 1 1 dr 1 1. B combining this with i, we know that r r Ξρ, ω µ dρ dω dr 0, 1,1 S 1 F R +1 \{0} F Φ JΦ dλ R +1. ow appl Theorems 5.. and to arrive at r r Ξρ, ω µ dρ dω dr 0, 1,1 S 1 F 0, r S F rω λ S dω dr. R +1 F dλ R+1

3 iii Using F s of the form suggested, it becomes clear from ii that ψω λ S dω ψ Ξρ, ω µ dr dω, S 1,1 S 1 which shows that λ S Ξ µ. iv First use orthogonal orthogonal invariance to see that f θ, η R +1 λs dη S is the same for all θ S. Thus, it is suffices to treat the case when θ is the unit vector in the direction of the + 1st coordinate, in which case the preceding shows that f θ, η R +1 λs dη fη +1 λ S dη S S fρ µ dρ dω ω 1 fρ1 ρ 1 dρ. 1,1 1,1 S : To see that M is a hpersurface, define F : U R R b F u, fu, note that M {u, U R : F u, 0}, and check that F u, u1 fu,..., u 1 fu, 1. Hence, F never vanishes. To show that Ψ, U is a global coordinate chart, first note the Ψ is obviousl one-to-one from U onto M. Second, because ui Ψu e i + ui fue, where e 1,..., e is the standard, orthonormal basis for R, it is clear that { ui Ψu,..., u 1 Ψu} is a linearl independent subset of R, and that ui Ψu, F u, 0 0 for each 1 i 1. R Hence, Ψ, U is a global coordinate chart for M. Further, from the preceding it is eas to see that ui Ψ, uj Ψ I R R 1 + f f, 1 i,j 1 and following the hint one sees that JΨ 1 + f. 5..6: i For each x M choose r x > 0 so the Bx, r x G and x F never vanishes on Bx, r x. Then, because M is connected, H x M Bx, r x is a connected, open neighborhood of M, H G, and x F never vanishes on H. Thus, without loss in generalit, I will assume that x F > 0 on H. ext, if x, H and Φx Φ, then x 1,..., x 1 1,..., 1 and 0 F x 1,..., x 1, F x 1,..., x 1, x x F x 1,..., x 1, ξ dξ, [x, ] which is possible onl if x. Hence, since JΦ x F, Φ is a diffeomorphism on H, and so, b the Inverse Function Theorem, f exists and has the asserted properties. 3

4 4 ii Since we know that 0 ui F u, fu xi F u, fu + x F u, fu ui fu, 1 + fu i1 x i F u, fu x F u, fu Thus, the desired result follows from Exercise : Given the hint, there is nothing more to do. F u, fu x F u, fu : The reduction to the case when 0 G and ζ 0 is a simple matter fz of translation. To see that z z zfz z for z 0, appl the chain rule and the fact that 1 z is analtic in C \ {0} and therefore, b the Cauch Riemann 1 equation, satisfies z z 0. Checking that η C b R ; [0, 1] is elementar calculus, and the rest follows b appling Exercise to f r in the regions G and B0, r : The onl part that does not follow immediatel from the outline is the derivation of Minkowski s Inequalit for p from Schwarz s Inequalit. But, b Schwarz s Inequalit, f 1 + f dµ f1 dµ + f1 dµ + f 1 dµ + f 1 f dµ + f dµ 1 1 f1 dµ f dµ + f dµ, from which Minkowski s Inequalit for p is clear. f dµ 6.1.9: Obviousl, if there exists an e q S 1 such that e q, q x R > 0 for all x C, then q / C. ow suppose that q / C. Choose p C so that q p min{ q x : x C}, and set e q q p q p. Then, for an x C, 0 d q θx 1 θp θ0 q p, p x, dθ R and so e q, p x R 0 for all x C. Hence eq, q x R e q, q p R + e q, p x R q p > 0 for all x C. ext, let F be given, and define p accordingl. If p / C, then there would exist an e S 1 such that e, p F > 0. On the other hand, R e, p F R dµ e, p F dµ R 0,

5 and so p must be an element of C. Turning to the proof of Jensen s Inequalit, it is eas to check that g 1 g is a continuous, concave function on C if g 1 and g are. Thus, if one knows Jensen s Inequalit when g is bounded, then g F dµ lim g n F dµ lim g n n n F dµ g 5 F dµ. ow, assume that g is bounded. Then it is clear that Ĉ is a closed, convex subset of R +1 and that ˆF is a µ-integrable, Ĉ-valued function. Hence F dµ g F dµ Ĉ, which is equivalent to saing that Jensen s Inequalit holds.