x^ + Let e^ = {1,0,..,,0) € IR . We shall prove that for every p » 0 with p = 1, u e IR, a e IRt and x e {- m) x {R^ \ if x = h(p,u), then x + ae^ = h(p, u + a). Note first that u(x + ae^) £ u + a, that is, x + ae^

satisfies the constraint of the EMP for (p, u + cch Let ye!R^ ana u(y) * u +

ol. Then uiy - ae.) ^ u, Hence p-(y - ae.) ^ p*x. Thus p-y ^ p*(x + aej,

Therefore, for every £ e {2,...,L>, u € IR, and u1 e iR, hjp,u) = ft^ip.u1). That is, the Hicksian demand functions for goods 2,...,L are independent of utility levels. Thus, if we define h(p) = h(pt0), then h(p,u) = hip) + ue^.

Since Mp, u + a) = Mp,u) + ae^t we have e(p, u + a) = e{p,u) + a.

Thus, if we define e{p) = e(p,0), then e{p,u) = e(p) + u.

3.E.8 We use the utility function u(x) ~ To prove (3.E.1),

/ , «.x -a., .a-1 a l-a, <x,t A-a -a a-1 , e(p,v(p,w)) = a (1 - a) p^p2 (a {1 - a) pi p2 w) - w,

v{p,e(p,u)) - a (1 - a) p p2 (a (1 - ex) p1p2 u) = u.

-a x(p,e(p,u)) = (a "(1 - a)a lpC^p^~CCu)(oi/pli (1 - oij/p^)

ocp.

3.E.9 First, we shall prove that Proposition 3.D.3 implies Proposition 3.E.2 via (3.E.1). Let p » 0, p' » 0, u € {R, u' € R, and a 2= 0.

(i) Homogeneity of degree one in p: Let a > 0. Define w = e(p,u), then u = v{p,w) by the second relation of (3.E.1). Hence e(aptu) = e(ap,v(p?w)) = e(ap,v(ap,aw)) = aw « ae(p,u), where the second equality follows from the homogeneity of and the third from the first relation of (3.E.1),

(ii) Monotonicity: Let ur > u. Define w = e(p,u) and w7 « e(ptu')t then u = v(p,w) and uJ = v(p,w'L By the monotonicity of v{ •, 0 in w, we must have w' > w, that is, e(p\u) > e(pfu).

Next let p' £ p. Define w = e(p,u) and w' = e(p\u), then, by the second relation of (3.E.1), u - v(p,w) = v(p',w'). By the monotonicity of v(*,-), we must have w* wt that is, e(p',u) e(p,u).

(iii) Concavity: Let a € [0,1). Define w = e(p,u) and w* = e(p\u), then u = v(pyw) = v{p\w), Define p" = ap + (1 ~ a)p" and w" = aw + (1 - a)w\ Then, by the cuasiconvexity of v{-,-), v(p",w") ^ u. Hence, by the monotonicity of v(*,') in w and the second relation of (3.EJ), w" ^ e(p",u). that is, e{ap + (I - a)p\ u) ae(p,u) + (1 - a)e(p\u).

(iv) Continuity: It is sufficient to prove the following statement: For any sequence with (pn»un) (p,u) and any w, if e(pntun) s w for every n, then e(p,u) ^ w; if e(pntun} w for every n, then e(p,u) ^ w. Suppose that e(pD,un) ^ w for every n. Then, by the monotonicity of v( -, • 5 in w, and the second relation of (3.E.1), we have u° ^ v(pn,w) for every n. By the continuity of us v(p,w). By the second relation of (3.E.1) and the monotonicity of in w, we must have e(p,u) ^ w. Tne same argument can be applied for the case with e(pn,un) £ w for every n.

Let's next prove that Proposition 3.E.2 implies Proposition 3-D.3 via (3.E.1). Let p » 0, p' » 0, w € (R, w' e IR, and a 2: 0.

(i) Homogeneity: Let a > 0. Define u = v(p,w). Then, by the first relation of (3.E.1), e[p,u) = w. Hence v(ap,aw) = v(aptae(p,w}) = v(ap,e(ccp,u)) = u = v(p,w), where the second equality follows from the homogeneity of ei-,0 and the third from the second relation of (3.E.1).

(ii) Monotonicity: Let w' > w. Define u = v(p,w) and u' = v(p,w')f then e(p,u) = w and e(p,ur) = w\ By the monotonicity of e(*,0 and w* > w, we

Next, assume that p' > p. Define u = v(p,w) and u' ~ v{then e(p,u} = e(p\u') = w. By the monotonicity of e( ■, *) and p' 2: p, we must have u* ^ u, that is, v(p,w) > v(p\w).

(iii) Quasiconvexity: Let a € [0,1]. Define u = v(p,w) and u' - v(p\w'). Then e(p,u) = w and e(p,u') - w\ Without loss of generality, assume that u' £ u. Define p,r ~ ap + (1 - a)pr and w = aw + (1 - a)w\ Then e{p'\u') 2: ae(ptuy) + (1 - a)e(p\u')

£ ae(p,u) + (I - a)e(p\u') = aw + (1 - a)wr - w", where the first inequality follows from the concavity of et-.u), the second from the monotonicity of in u and u' 2: u. We must thus have v(p",wM

(iv) Continuity: It is sufficient to prove the following statement. For any sequence with Cpn,wn) (p,w) and any u, if vtp^w11) s u for every n, then Wp,w) £ u; if v(pn,wn) i u for every n, then v(p,w) £ u.

Suppose that v(p ,w ) < u for every n. Then, by the monotonicity of e(-,0 u and the first relation of (3.E.1), we have wn £ e(pn,u) for every n. By th continuity of e{*,*)> w < e(p,u), We must thus have v{p,w) s u. The same argument can be applied for the case with v(pn,wn) £ u for every n.

An alternative, simpler way to show the equivalence on the concavity/ cuasiconvexity and the continuity uses what is sometimes called the epigraph.

For the concavity/quasiconvexity, the concavity of e(-,u) is equivalent to the convexity of the set Up,w): e(pTu) 2: w) and the quasi-convexity of vf - j is equivalent to the convexity of the set {(p,w): v{p,w) * u) for every u. But (3.E.1) and the monotonicity imply that v(p,w) £ u if and only if e(p,u) ^ w\ Hence the two sets coincide and the quasiconvexity of v(-) is equivalent to the concavity of e{-,u).

As for the continuity, the function e(-) is continuous if and only if both {(p,w,u): e(p,u} ^ w) and {(p,w,u); e(p,u) 2: w} are closed sets. The function v(-) is continuous if and only if both {(p,w,u): v(p,w) £ u} and {(p,w,u); v(p,w) ^ u) are closed sets. But, again by (3,E.l) and the monotonicity,

<(p,w,u)f e{p,u} * w) = {(p,w,u): v(p,w) > u}; <(ptwfu): e(p,u) 2: w} = {(p,wTu): v(p,w] ^ u}. Hence the continuity of e(0 is equivalent to that of vM.

3.E.10 [First printing errata: Proposition 3.E.4 should be Proposition 3.E.3J

Let's first prove that Proposition 3.D.2 implies Proposition 3.E.3 via the relations of (3.E.1) and (3.E.4). Let p e R^ and u e R.

(i) Homogeneity: Let a > 0. Define w - e(p,u), then u = v(p,w) by the second relation of (3.E.1), Hence h(ap,u) = x(ap,eiap,u)) = jc(ap,ae(ptu)) ® x{p,e{p,u}) = h(p,u), where the first equality follows from by the first relation of (3.E.4), the second from the homogeneity of e(-,u), the third from the homogeneity of *:(*,-)> and the last from by the first relation of (3.E.4),

(ii) No excess utility: Let (p,u) be given and x e h(ptu). Then x € x(p,e(p,u)) by the first relation of (3.E.4). Tnus u(x) ~ v(p,e{p,u)) = u by the second relation of (3.E.I).

\iii) Convexity/Uniqueness: Obvious.

Let's first prove that Proposition 3.E.3 implies Proposition 3.D.2. via the relations of (3.E.1) and (3.E.4). Let p € \RL and w e IR.

(i) Homogeneity: Let a > 0 and define w 2= e{p,u), then v{p,w) - u. Hence x(ap,aw) = hiap,v(ap,ctw)) = h(ap,v(p,w)) = h(ptv(p,w)) - xip,w), where the first equality follows from the second relation of (3.E.4), the second from the homogeneity of v(-), the third from the homogeneity of h( * ) in pt and the last from the first relation of (3.E.4).

(ii) Walras* law: Let (p(w) be given and x e x(p,w). Then x e h(p,v(p,w)) by the second relation of (3.ET.4). Thus p-x = e(p,v(p,w}) ~ w by the definition of the Hicksian demand and the first relation of (3.E.1). (iii) Convexity /Uniqueness: Obvious,

3.F.I Denote by A the intersection of the half spaces that includes K> then clearly A r> K. To show the inverse inclusion, let x e K, then, since K is a closed convex set, the separating hyperplane theorem (Theorem M.G.2) implies that there exists a p * 0 and c, such that p*x < c < p-x for every x € K. Then {z e rS p'z i c> is a half space that includes K but does not contain x Hence x g A. Thus K d A.

3.F.2 If K is not a convex set, then there exists x e K and y e K such that (l/2)x + {l/2)y £ K, as depicted in the figure below. The intersection of all the half-spaces containing K (which also means containing x and y) will contain the point (l/2)x + (l/2)y, since half-spaces are convex ana the intersection of convex sets is convex. Therefore, the point (l/2)x + (l/2)v cannot be separated from K.

As for the second statement, if K is not convex, then there exist x <= K, y e K, and a e [0,1] such that ax + (1 - a)y * K. Since every half space that includes K also contains ax + (1 - a)y, it cannot be separated from K.

3.G.1 Since the identity v{p,e{p,u)) - u holds for all p, differentiation with respect to p yields

By Roy's identity,

(3v(p,e(p,u))/dw}(- >r{p,e(p,u)) + V^e(p,u)} = 0.

By ov(p,e(p,u))/3w > 0 and h(p,u) = xrip^ip^)}, we obtain h(p,u] = 7 e(pTu)

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