Math_Analysis_HW7_Solutions.pdf - Real Analysis Homework...

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This preview shows page 1 out of 6 pages. Unformatted text preview: Real Analysis Homework VII solutions November 8th, 2019 Exercise 3.4.3 Show that f : (c, ∞) → R for some c > 0 and defined by f (x) := 1 x is Lipschitz continuous. Proof. By Definition 3.4.7, we must show that there exists an L such that for all x, y > c, |f (x) − f (y)| < L|x − y|. We compute that for x, y ∈ (c, ∞), 1 1 y − x |x − y| 1 = |f (x) − f (y)| = − = ≤ 2 |x − y| . x y xy xy c Thus, f is Lipschitz continuous with Lipschitz constant c−2 . Exercise 3.4.7 Let f : (0, 1) → R be a bounded continuous function. Show that the function g(x) := x(1 − x)f (x) is uniformly continuous. Proof. By Proposition 3.4.6, g is uniformly continuous if and only if limx→0 g(x) and limx→1 g(x) exist . Thus, we only need to show the limits exist . We begin with limx→0 g(x). Since f is bounded, we have that |f (x)| < B for all x ∈ [0, 1]. Hence, −Bx ≤ x(1 − x)f (x) ≤ Bx. By the Squeeze theorem (Lemma 2.2.1), limx→0 g(x) = 0 . A similar argument shows that limx→1 g(x) = 0 . In this case, −B(1 − x) ≤ x(1 − x)f (x) ≤ B(1 − x) so we can again use the Squeeze theorem to conclude. 1 Exercise 3.4.10 a. Find a continuous f : (0, 1) → R and a sequence {xn } in (0, 1) that is Cauchy, but such that {f (xn )} is not Cauchy. b. Prove that if f : R → R is continuous, and {xn } is Cauchy, then {f (xn )} is Cauchy. a. Let f (x) = x1 , xn = Cauchy, but 1 n for each n. Then, f is a continuous function on (0, 1) and {xn } is f (xn ) = n so {f (xn )} is not Cauchy. b. Here are two proofs: Proof. By Theorem 2.4.5, it is enough to show {f (xn )} is convergent . Since {xn } is Cauchy, it converges to some x. By the Bridge Theorem (Prop 3.2.2), f (xn ) → f (x) , so {f (xn )} is convergent and hence Cauchy. Proof. Suppose {xn } is Cauchy. Then, it is bounded, so {xn : n ∈ N} ⊂ [−R, R] for some positive R. The restriction of f to [−R, R], f |[−R,R] , is a continuous function on a bounded set, so by Theorem 3.4.4 it is uniformly continuous . Lemma 3.4.5 then assures us that {f (xn )} = {f |[−R,R] (xn )} is Cauchy. Exercise 3.4.16 Suppose f : S → R and g : [0, ∞) → [0, ∞) are functions, g is continuous at 0, g(0) = 0, and whenever x and y are in S we have |f (x) − f (y)| ≤ g(|x − y|). Prove that f is uniformly continuous. We remark that the function g is called the modulus of continuity of f . The case g(x) = Lx corresponds the f being Lipschitz with Lipschitz constant L. Proof. Since S is an arbitrary set, we cannot use the theorems about uniform continuity of functions defined on intervals. Thus, we argue by the definition (3.4.1). Consider  > 0, arbitrary. We must show that there exists a δ such that for any x, y ∈ S with |x − y| < δ, |f (x) − f (y)| <  . Since g is continuous at 0 and g(0) = 0, there exists a δ > 0 such that for every z with 0 ≤ z < δ , |g(z)| <  Then, for any x, y ∈ S with |x − y| < δ, we have that |f (x) − f (y)| ≤ g(|x − y|) <  as was to be shown. 2 Exercise 3.5.1 Show that the limit at ∞ or −∞ is unique if it exists. The argument follows that same line of reasoning as Proposition 2.1.6 and Proposition 3.1.4. Proof. Suppose L1 and L2 are both limits of f (x) as x → ∞. Then, for any  > 0 there exists an N1 such that for x ≥ N1 , |f (x) − L1 | < 2 , and similarly there exists an N2 such that for x ≥ N2 , |f (x) − L2 | < 2 . Choose x ≥ max{N1 , N2 }. Then, by the Triangle inequality, |L1 − L2 | ≤ |f (x) − L1 | + |f (x) − L2 | <  Since this holds for all  > 0, it follows that L1 = L2 . This show the limit as x → ∞ is unique. If we define g(x) = f (−x), then lim f (x) = lim g(x), x→−∞ x→∞ so we have also shown that the limit as x → −∞ is unique. Exercise 4.1.5 Define f : R → R by ( x2 f (x) = 0 if x ∈ Q else Prove that f is differentiable at x = 0, but discontinuous at all points except 0. Proof. By Definition 4.1.1, the derivative at 0 is f (x) − f (0) x→0 x−0 f (x) = lim x→0 x f 0 (0) = lim Observe that −|x| ≤ f (x) ≤ |x|. x So by the squeeze theorem, f 0 (0) = 0 . To see that f is not continuous away from 0, let x 6= 0 be arbitrary. Either x is rational, or it is irrational. If x is rational, let {xn } be a sequence of irrational numbers converging to x . Then, lim f (xn ) = 0 6= f (x) = x2 n→∞ so, by the (contrapositive of) the bridge theorem, f is discontinuous at x. Similarly, if x is irrational, we let {xn } be a sequence of rational numbers converging to x . Then,   lim f (xn ) = lim x2n = n→∞ n→∞ lim xn n→∞ so f is discontinuous at x by the bridge theorem. 3 2 = x2 6= f (x) = 0 Exercise 4.1.8 Let f : I → R be differentiable. Given n ∈ Z, define f n by f n (x) = (f (x))n . (If n < 0, assume f (x) 6= 0). Prove that (f n )0 (x) = n[f (x)]n−1 f 0 (x). Proof. This is a result of the chain (Prop. 4.1.10) and product (Prop. 4.1.8) rules. We recall and prove the following result from Exercise 4.1.3: Proposition . For n ∈ Z, (xn )0 = nxn−1 (if n < 0, we require that x 6= 0). Proof. For n ≥ 0, we proceed by induction. If n = 0, then the proposition reads 10 = 0, which is certainly true. Now, assume that for some n, (xn )0 = nxn−1 . Then, by the product rule, 0 xn+1 =x0 (xn ) + x (xn )0  =1 (xn ) + x nxn−1 =(n + 1)xn which proves the result for n ≥ 0. For n < 0, the quotient rule states that  0 1 −(−n)x−n−1 n 0 (x ) = = = nxn−1 −n −2n x x which is the desired result. We now return to the problem. Let g(x) = xn . Then, f n (x) = (g ◦ f )(x), so (f n )0 (x) = (g ◦ f )0 (x) = g 0 (f (x))f 0 (x) = n[f (x)]n−1 f 0 (x) as desired. Exercise 4.1.10 Let I1 , I2 be intervals, f : I1 → I2 be a bijective function, and g : I2 → I1 be its inverse. Suppose both that f is differentiable at c and g is differentiable at f (c). Use the chain rule to find a formula for g 0 (f (c)). On one hand, by the chain rule, (g ◦ f )0 (c) = g 0 (f (c))f 0 (c) On the other hand, g ◦ f is the identity function g ◦ f (x) = x for all x ∈ I1 , so (g ◦ f )0 (c) = 1. Equating these two expressions and simplifying, we find g 0 (f (c)) = 4 1 . f 0 (c) Exercise 4.1.11 Suppose f : I → R is bounded and g : I → R is differentiable at c ∈ I with g(c) = g 0 (c) = 0. Show that h(x) = f (x)g(x) is differentiable at c. Proof. As stated in the hint, we cannot apply the product rule since f is not known to be differentiable. However, we can attempt the same manipulations from the proof of the product rule. Consider the difference quotient h(x) − h(c) x−c f (x)g(x) − f (0)g(0) = lim x→c x−c f (x)g(x) − f (x)g(c) + f (x)g(c) − f (c)g(c) = lim x→c x−c h0 (c) = lim x→c Now, we observe that g(c) = 0, so we can drop the last two terms in the numerator. This leaves us with h0 (c) = lim f (x) x→c g(x) − g(c) x−c Now, since f is bounded, |f (x)| ≤ M for all x ∈ I. Thus, g(x) − g(c) g(x) − g(c) ≤M 0 ≤ f (x) x−c x−c But, since g 0 (c) = 0, the limit on the right goes to 0 as x → c. By the squeeze theorem, it follows that f (x) g(x)−g(c) → 0 as x → c, and hence that x−c h0 (c) = lim f (x) x→c g(x) − g(c) = 0. x−c Since the limit of the difference quotient exists, h is differentiable at 0. Exercise 4.1.15 Prove the following simple version of L’Hôpital’s rule. Suppose f : (a, b) → R and g : (a, b) → R are differentiable functions whose derivatives f 0 and g 0 are continuous functions. Suppose that at c ∈ (a, b), f (c) = 0, g(c) = 0, and g 0 (x) 6= 0 for all x ∈ (a, b), and suppose the limit as x → c 0 (x) of fg0 (x) exists. Prove that f (x) f 0 (x) lim = lim 0 . x→c g(x) x→c g (x) 5 Proof. Notice that since f 0 and g 0 are continuous with g 0 (c) 6= 0, the limit on the right is just f 0 (c) . Thus, it suffices to prove that g 0 (c) f (x) f 0 (c) = 0 . x→c g(x) g (c) lim We proceed as follows: f (x) f (x) − f (c) = lim x→c g(x) x→c g(x) − g(c)   lim = lim  x→c  = = f (x)−f (c) x−c f (x)−f (c) x−c  (c) limx→c f (x)−f x−c  limx→c  f (x)−f (c) x−c f 0 (c) g 0 (c) where we used the arithmetic rules for limits to change the limit of the quotient to a quotient of limits in the third line. 6 ...
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