Math_Analysis_HW5_Solutions.pdf - Real Analysis Homework V...

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Unformatted text preview: Real Analysis Homework V solutions October 18th, 2019 Exercise 3.1.1 Find the limit or prove that the limit does not exist. √ a. limx→c x, for c ≥ 0. b. limx→c x2 + x + 1, for any c ∈ R. c. limx→0 x2 cos(1/x) d. limx→0 sin(1/x) cos(1/x) e. limx→0 sin(x) cos(1/x) √ √ a. We’ll prove limx→c x = c . By Lemma 3.1.7, it is enough √ to prove that for any √ sequence {xn } ⊂ [0, ∞) that converges to c, limn→∞ xn = c . Consider any such sequence. By Proposition 2.2.6, q √ √ lim xn = lim xn = c. n→∞ n→∞ b. Again, we’ll use lemma 3.1.7 to prove limx→c x2 + x + 1 = c2 + c + 1 by showing that for any sequence xn → c , x2n + xn + 1 → c2 + c + 1 . Suppose xn → c . Then, by limit arithmetic, lim x2n + xn + 1 = lim x2n + lim xn + lim 1 = c2 + c + 1. n→∞ n→∞ n→∞ n→∞ c. We’ll use the squeeze theorem for limits (Corollary 3.1.11). Notice that for all x. Moreover, −x2 ≤ x2 cos(1/x) ≤ x2 lim x2 = 0, x→0 lim −x2 = − lim x2 = 0 x→0 x→0 by Example 3.1.5 and limit arithmetic (Corollary 3.1.12). Thus, limx→0 x2 cos(1/x) = 0 . 1 d. Let f (x) = sin(1/x) cos(1/x) By the contrapositive of Lemma 3.1.7, the limit will fail to exist if there does not exist an L such that for all xn → 0, f (xn ) → L . This means we can prove limx→0 f (x) does not exist by finding a sequence xn → 0 where f (xn ) does not converge. . Let xn = 4 π(1+2n) . Then, xn → 0 , but 1 1 cos x x  nπ π n π π  = sin + n cos + n 4 2 4 2 (−1)n = 2 f (xn ) = sin does not converge. e. We will use the squeeze theorem (Corollary 3.1.11). We have that −| sin(x)| ≤ sin(x) cos(1/x) ≤ | sin(x)| since −1 ≤ cos(1/x) ≤ 1 . On the other hand, | sin(x)| → 0 since sin x is continuous so sin x → sin 0 = 0 and so Corollary 3.1.10 shows that | sin x| → 0 as x → 0. Thus, limx→0 sin(x) cos(1/x) = 0 . Exercise 3.1.5 Let A ⊂ S. Show that if c is a cluster point of A, then c is a cluster point of S. Note the difference from Proposition 3.1.15. The statement in the exercise has both different hypotheses and conclusions from Proposition 3.1.15. In Proposition 3.1.15, we assume everything given in the exercise, together with the condition that for some α > 0, (A \ {c}) ∩ (c − α, c + α) = (S \ {c}) ∩ (c − α, c + α). As a result of the stronger hypotheses, Proposition 3.1.15 tells us that c is a cluster point of A if and only if c is a cluster point of S, while in the exercise we can only prove that c is a cluster point S if it is a cluster point of the subset A. Proof. We will prove that there exists a sequence of points in S converging to c . Since c is a cluster point of A, there exists a sequence of points {xn } ⊂ A converging to c . But, {xn } ⊂ S since A ⊂ S, so xn is a sequence of points in S converging to c , which shows that c is a cluster point of S. 2 Exercise 3.1.6 Let A ⊂ S. Suppose c is a cluster point of A and it is also a cluster point of S. Let f : S → R be a function. Show that if f (x) → L as x → c, then f |A (x) → L as x → c. Note the difference from Proposition 3.1.15. The statement in the exercise has both different hypotheses and conclusions from Proposition 3.1.15. In Proposition 3.1.15, we assume everything given in the exercise, together with the condition that for some α > 0, (A \ {c}) ∩ (c − α, c + α) = (S \ {c}) ∩ (c − α, c + α). As a result of the stronger hypotheses, Proposition 3.1.15 tells us that limx→c f (x) = limx→c f |A (x) (in particular, if one limit exists, so does the other), while in the exercise we can only prove that assuming limx→c f (x) exists, then limx→c f |A (x) exists and is equal to limx→c f (x). Proof. We will use the  – δ definition. Let  > 0. Since f (x) → L as x → c, there exists a δ > 0 such that for all x ∈ S \ {c} where |x − c| < δ , |f (x) − L| <  Choose δ. Then, for x ∈ A \ {c} with |x − c| < δ, we have that x ∈ S \ {c} since A ⊂ S, so |f |A (x) − L| = |f (x) − L| <  by the definition of δ. Exercise 3.1.7 Find an example of a function f : [−1, 1] → f |A (x) → 0 as x → 0, but the limit of f (x) as apply Proposition 3.1.15. Define f (x) by ( −1 f (x) := 1 R such that for A := [0, 1], the restriction x → 0 does not exist. Note why you cannot −1 ≤ x < 0 0≤x<1 We claim that (1) limx→0 f |A (x) = 1 but (2) limx→0 f (x) does not exist. (1) limx→0 f |A (x) = 1: Let xn be an arbitrary sequence in A = [0, 1] converging to 0. Then, lim f (xn ) = lim 1 = 1 n→∞ n→∞ so, by Lemma 3.1.7, limx→0 f |A (x) = 1. (2) limx→0 f (x) does not exist: Let xn = (−1) . Then, xn → 0 is a sequence of terms in n n [−1, 1], but f (xn ) = (−1) does not converge . Thus, by Lemma 3.1.7, the limit does not exist. Note that Prop 3.1.15 does not apply because for any α > 0, (A \ {0} ∩ (0 − α, 0 + α) = (0, min{1, α}) is never equal to (S \ {0} ∩ (0 − α, 0 + α) = (max{−1, −α}, min{1, α}) \ {0} . n 3 Exercise 3.1.8 Find example function f and g such that neither the limit of f (x) nor g(x) exists as x → 0, but such that the limit of f (x) + g(x) exists as x → 0. Let f (x) be given by ( −1 1− ≤ x < 0 f (x) := 1 0≤x<1 and let g(x) = −f (x). We proved in Exercise 3.1.7 that limx→0 f (x) does not exist limx→0 g(x) does not exist either. If it did, then limx→0 f (x) = limx→0 −g(x) = − limx→0 g(x) would exist as well, contradicting Exercise 3.1.7. However, limx→0 f (x) + g(x) = limx→0 0 = 0. Exercise 3.1.10 Let c be a cluster point of A ⊂ R and f : A → R be a function. Suppose for every sequence {xn } in A such that lim xn = c, the sequence {f (xn )}∞ n=1 is Cauchy. Prove that limx→c f (x) exists. For this exercise and Exercise 3.1.11, we will use the following interleaving lemma: Lemma Interleaving lemma. Suppose xn → c and yn → c. Then, if zn is given by ( xn n is even zn = yn n is odd then zn → c. Proof. We will use the definition of convergence of sequences. Let  > 0. Since xn → c, there is an M1 such that for n ≥ M1 , |xn − c| < , and similarly for yn there is an M2 such that for n ≥ M2 , |yn − c| < . Let M = max{M1 , M2 }, and consider n ≥ M . There are two cases: either n is even, or it is odd. If n is even, then zn = xn and |zn − c| = |xn − c| <  because n ≥ M ≥ M1 . Similarly, if n is odd, then zn = yn , and because n ≥ M ≥ M2 , |zn − c| = |yn − c| < . Since in either case |zn − c| < , zn → c. With the lemma in hand, we now turn to the proof for the exercise: Proof. By lemma 3.1.7, it is enough to prove that there exists an L such that for any subsequence {yn } ⊂ A \ {c} with yn → c, {f (xn )} → L. Our first task is to find L. Since c is a cluster point, by Proposition 3.1.2 there is a sequence xn is A \ {c} converging to c. By hypothesis, the sequence {f (xn )} is Cauchy, so by Theorem 2.4.5 it is convergent, and so it has some limit. Call this limit L. 4 We now must prove that for any {yn } ⊂ A \ {c} with yn → c, f (yn ) → L. Consider such a sequence yn . By hypothesis, f (yn ) is Cauchy, so the sequence of function values converrges to some limit L0 , and we must prove L = L0 . Now, let zn be the sequence defined by zn = xn if n is even and zn = yn is n is odd. Again, {f (zn )} converges to some limit, which we will call L00 . But, f (z2n ) = f (x2n ) is a subsequence of both {f (zn )} and {f (xn )}, so it converges to both L and L00 . By the uniqueness of limits, L = L00 . Similar arguments with f (z2n−1 ) = f (y2n−1 ) show that L0 = L00 , so L0 = L and lim f (yn ) = L. This shows that limx→c f (x) = L. Exercise 3.1.11 Prove the following stronger version of one direction of Lemma 3.1.7: Let S ⊂ R, c be a cluster point of S, and f : S → R be a function. Suppose that for every sequence {xn } in S \ {c} such that lim xn = c the sequence {f (xn )} is convergent. Then show f (x) → L as x → c for some L ∈ R. This follows from Exercise 3.1.10. Proof. Let A = S \ {c}. Then, c is also a cluster point of A (by Proposition 3.1.15(i) with α = 1), and the hypotheses given in this exercise tell us that for every sequence {xn } in A which converges to c, the sequence {f (xn )} is Cauchy (since by Theorem 2.4.5, convergent sequences are Cauchy). Thus, by Exercise 3.1.10, there is an L ∈ R such that lim f |A (x) = L. x→c But, again, S \ {c} ∩ (c − 1, c + 1) = A \ {c} ∩ (c − 1, c + 1), so by Proposition 3.1.15 lim f (x) = lim f |A (x) = L. x→c x→c Exercise 3.2.3 Let f : R → R be defined by ( x f (x) := x2 if x is rational, if x is irrational. Using the definition of continuity directly prove that f is continuous at 1 and discontinuous at 2. First, the proof that f is continuous at 1. Proof. Let  > 0, and take δ = min{1, /3} . Consider x with |x − 1| < δ. There are two cases: either x is rational or it is irrational. If x is rational, then |f (x) − f (1)| = |x − 1| < δ <  5 If x is irrational, then |f (x) − f (1)| = |x2 − 1| = |x − 1||x + 1| since δ ≤ 1 , |x + 1| = |x − 1 + 2| < |x − 1| + 2 ≤ 3 so |f (x) − f (1)| = |x − 1||x + 1| < 3|x − 1| < 3δ ≤  In either case, |f (x) − f (1)| < , so f is continuous at 1. Now, the proof that f is discontinuous at 2. Proof. We must show that there exists an  > 0 such that for all δ > 0, there exists an x with |x − 2| < δ such that |f (x) − f (2)| ≥  . Let  = 2 , and consider δ > 0, arbitrary. By the Archimedean property, there is an n such √ √ that n2 < δ. Let x = 2 + 2/n . Then, √ 2 |x − 2| = < δ. n Since x is irrational , |f (x) − f (2)| = |x2 − 2| = (2 + √ 2n)2 − 2 ≥ 2 =  Exercise 3.2.7 Prove the following statement. Let S ⊂ R and A ⊂ S. Let f : S → R be a continuous function. Then the restriction f |A is continuous. Proof. We must show that f |A is continuous at every point a ∈ A. Consider a ∈ A. If a is not a cluster point of A, then Proposition 3.2.2 (i) says that f |A is continuous at a and we are done. It only remains to consider the case when a is a cluster point of A. By Proposition 3.2.2 (ii), it suffices to show that lim f |A (x) = f (a). x→a By Exercise 3.1.5, since A ⊂ S , c is a cluster point of S, and by Exercise 3.1.6 lim f |A (x) = lim f (x) = f (a) = f |A (a) x→a x→a since f is continuous . 6 Exercise 3.2.9 Give an example of functions f : R → R and g : R → R such that the function h defined by h(x) := f (x) + g(x) is continuous, but f and g are not continuous. Can you find f and g that are nowhere continuous, but h is a continuous function? Let f be the function ( 1 x is rational, f (x) = 0 x is irrational. In example 3.2.11, it was given that f is discontinuous at every point. If g(x) = −f (x), then g is also discontinuous at every point (since if g were continuous at c, then f (x) = −g(x) would be continuous at c Proposition 3.2.5(iii) and the fact that h(x) = −1 is continuous . However, h(x) = 0 = f (x) + g(x) is continuous at every point (since constant functions are continuous). 7 ...
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