a n+2 a n+1 M n a 2 a 1. (2)

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1 Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside S, so you cnnot use Theorem 3., p. 60). Proof. Let ε > 0. Since f is uniformly continuous on S, there exists δ > 0 such tht x y < δ implies f(x) f(y) < ε for ll x, y S. (1) For this δ > 0 there exists N Z + such tht for n, m > N, x n x m < δ. By (1) it follows tht for n, m > N, f(x n ) f(x m ) < ε. Hence {f(x n )} is Cuchy.. Let M > 0 nd let f : D R, D R, stisfy the condition f(x) f(y) M x y for ll x, y D. Show tht f is uniformly continuous. Proof. Let ε > 0. Put δ = Hence f is uniformly continuous. ε. Then if x y < δ, M f(x) f(y) M x y Mδ = M ε M = ε < ε. 3. Suppose tht for some constnt M with 0 < M < 1, n+ n+1 M n+1 n, n = 1,, 3,.... Prove tht the sequence { n } is Cuchy. Proof. We will first show tht n+ n+1 M n 1. () This is true for n = 1 s ssumed. Suppose it is true for n = k. Then Hence () is true for ll n. k+3 k+ M k+ k+1 M k 1.

2 For fixed N nd n = N + r, we hve n N N+r N+r 1 + N+r 1 N+r + + N+1 N M N+r M N 1 1 = 1 M N 1 [1 + M + + M r 1 ] 1 M N 1 1 M. (3) Given ε > 0, let 1 M N 1 1 M < ε. Then M N 1 < (1 M)ε. Tke logrithms. 1 (N 1) ln M < ln N 1 > or N > (1 ε)ε 1 (1 M)ε ln 1 since ln M < 0 ln M (1 M)ε ln 1 ln M Suppose tht f nd g re continuous functions on the closed interl [, b] such tht f(r) = g(r) for every rtionl number r [, b]. Prove tht f(x) = g(x) for ll x [, b]. Proof. Suppose tht f(z) g(z) for some irrtionl number z in [, b]. Let f(z) g(z) = α. For ε = α, there exists δ > 0 such tht f(z) f(x) < ε nd g(z) g(x) < ε whenever z x < δ (δ = min(δ 1, δ )). Let r be rtionl number with z r < δ. Then f(r) = g(r). Moreover, α = f(z) g(z) f(z) f(r) + g(z) g(r) α 4 + α 4 = α, contrdiction. Hence f(z) = g(z) for ll z [, b]. 5. Let u n+1 = u n + 1, u 1 = 1. () Show tht {u n } is bounded nd monotone. Proof. Let f(x) = x + 1. Then f 1 (x) = > 0 for x > 1. Hence f is x + 1 incresing. Consider the point x = Then f(x ) = x. For u 1 = 1 < x,

3 u = f(u 1 ) < f(x ) = x. And by induction, we hve u n < x. Hence {u n } is bounded bove by x. Since u = = > u 1, it follows by the sme resoning tht {u n } is monotoniclly incresing. Hence by the Bolzno-Weierstrn Theorem, {u n } must converge. (b) Find lim n u n. Proof. Let x + 1 = x. Then x = since it is negtive. is the limit point s 1 5 is discrded 6. Let S be the spce of ll rtionl numbers, with d(p, q) = p q, nd E is the set of ll rtionl numbers p such tht < p < 3. Prove tht (i) E is closed nd bounded. (ii) E is not compct. Proof. E = {p : < p < 3} S [ = ( 3, ) (, ] 3) S = (I 1 I ) S () E is clerly bounded by 3 nd 3. Let x E, then there is sequence {p n } in E with d(p n, x) 0 s n. Now {p n } is either in I 1 or I, sy I n. Hence it either converges in I, nd hence x I 1 or it converges to either 3 or which re not in our spce. Hence E is closed. (b) Consider the open cover F = {V n : i = 1,, 3,... } where V n = {p S : 3 p < 3}. This cover hs not finite subcover. Hence it is not compct. Another solution: Tke the sequence {p n }, p n = 3 1. Then {p n n} hs no convergent subsequences. n < n+1 7. Let E be nonempty subset of metric spce (S, d). Define the distnce from x S to the set E by ρ(x) = glb d(x, y). () Prove tht ρ(x) = 0 if nd only if x E. (b) Prove tht ρ : S R is uniformly continuous on S.

4 Proof. () Let p(x) = 0. Then glb d(x, y) = 0. Hence there is sequence {y n } in E with d(x, y n ) 0 s n. Thus x E. For the converse, let x E. If x E, then s d(x, x) = 0, p(x) = 0. If x E\E, then there exists y n E, y n x s n or d(y n, x) 0 s n. Hence p(x) = 0. (b) p : S R. For x 1, x S p(x 1 ) p(x ) = glb d(x 1, y) glb d(x, y) glb d(x 1, y) d(x, y) (4) But d(x 1, y) d(x 1, x ) + d(x, y) or d(x 1, y) d(x, y) d(x 1, x ). Similrly d(x, y) d(x 1, y) d(x 1, x ). Thus d(x 1, y) d(x, y) d(x 1, x ). Hence p(x 1 ) p(x ) d(x 1, x ). Given ε > 0, let δ = ε. If d(x 1, x ) < δ, then p(x 1 ) p(x ) d(x 1, x ) < δ = ε. 8. Suppose tht f is continuous on n open intervl I contining x 0, suppose tht f is defined on I except possibly t x 0, nd suppose tht lim f (x) = L. Prove tht f (x 0 ) = L. Proof. f f(x) f(x 0 ) (x 0 ) = lim x x 0 f(x 0 ). Now if the limits exists. Since f is continuous, lim f(x) = L f(x) f(x 0 ) lim = 0 x x 0 x x 0 0 = lim f (x) 1 = L. 9. Let f nd g be continuous functions on [, b], g is positive nd monotoniclly decresing nd g (x) exists on [, b]. Prove tht there exists point ξ [, b] such tht ξ f(x)g(x) d(x) = g() f(x) dx. Proof. Let h(x) x f(t) dt. Since g is positive, either 0 f(x)g(x) dx g() f(t) dt

5 or g() f(t) dt f(x)g(x) dx 0. f(x)g(x) dx In either cse, is between h() = 0 nd h(b). By the Intermedite Vlue g() Theorem, there exists ξ between nd b such tht f(x)g(x) dx g() = h(ξ) = ξ f(t) dt. 10. Suppose tht f is continuous t x = such tht f() < 1. Prove tht there exists n open intervl I = ( δ, + δ), δ > 0, such tht for ll x I, f(x) M < 1, for some fixed constnt M. Proof. Let L = f() < 1. If such n M does not exist for ny δ, there exists sequence {x n } tht converges to with f(x n ) > L + 1. Since f is continuous, f(x n ) f() s n. But this is not possible s f(x n ) f() > 1 L for ll n.

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