COMMENTS ON STYLE AND SAMPLE SOLUTIONS MARK H. KIM Comments on Style I will assess the quality of exposition as well as the correctness of the arguments. Mathematical proofs are bona fide pieces of writing, and I request that you treat them as such. Please make a serious effort to communicate your ideas efficiently and elegantly. Now, I understand that English is not the first language for some of you, so it does not make sense to take points off for grammar or any other English-specific issues. There is no excuse for sloppiness, however, and points will be deducted for poor explanation of correct ideas. Here are some pointers: (1) As a general rule, you should not submit the first draft. Do the problems on scratch paper and think about how you should organize your proofs before writing them down. In particular, if you wrote down a correct proof in an incorrect order, then your proof would cease to be correct. (2) Please keep in mind that your proof is wrong by default if I, the grader, cannot understand it. If you’re a top expert in your field, then people will try their best to read your works regardless of the quality of presentation. But otherwise, most of us, including myself, have very little motivation to read someone else’s incomprehensible proofs. (3) One way to improve your proof is to employ short sentences and paragraphs. Each paragraph should only contain one main idea; each sentence should only say one thing. Math is difficult enough without having to parse overly complex writing. (4) To achieve (3), you must, of course, use actual sentences. Please write in complete sentences without shorthand notations like ∀, ∃, ⇐, ⇔, s.t., WLOG, and so on. No, using logical quantifiers does not make an argument more mathematical. And yes, using them does make your proof substantially less readable. (5) “Clearly” implies that there is a gap in your reasoning, and that you expect the reader to be able to fill it in with no trouble. This means “clearly” is not the right word to use if you are merely restating a definition, a theorem, or a lemma—in this case, there is no gap in your reasoning. It also means that “clearly” is never the right word to use in a homework write-up. If the gap has to do with a fact that everyone in the class is expected to know, then you can simply use it without comment. If not, then you really shouldn’t leave that gap unfilled. (6) Being wordy is not good, either! Include only the necessary details. The above list is inspired by James Munkres’s Comments on Style, which you can find easily on the internet. Another quick but useful resource is a little booklet called How to Write Mathematics, published by the AMS in 1973. Nicholas J. Higham’s Handbook of Writing for the Mathematical Sciences is also useful. 1 2 MARK H. KIM Sample Solutions (for Problem Set 1) Notations: Given a partition P = {x0 , . . . , xN } of [a, b] and a function f : [a, b] → R, we let Mn (f ) = max f (x) and mn (f ) = x∈[xn−1 ,xn ] min f (x). x∈[xn−1 ,xn ] Problem 1. (a) Fix a partition P of [a, b]. Since kP k ≥ |xn − xn−1 | for all 1 ≤ n ≤ N , we see that ωf (kP k) ≥ Mn − mn for all 1 ≤ n ≤ N . Therefore, Uα (f, P ) − Lα (f, P ) = N X (Mn (f ) − mn (f ))(α(xn ) − α(xn−1 )) n=1 ≤ N X ωf (kP k)(α(xn ) − α(xn−1 )) n=1 = ωf (kP k)(α(b) − α(a)), as was to be shown. Remark 1.1. In what sense does this “quantify” the result C ([a, b]) ⊆ Rα ([a, b])? Recall that a function f : [a, b] → R is Lipschitz continuous with constant K if |f (x) − f (y)| ≤ K|x − y| for all x, y ∈ [a, b]. This is equivalent to the statement that ωf (δ) ≤ Kδ for all δ ∈ [0, ∞)—please check this for yourselves! Similarly, given a fixed 0 < α ≤ 1, a function f : [a, b] → R is α-H¨ older continuous with constant K if |f (x) − f (y)| ≤ K|x − y|α for all x, y ∈ [a, b]. Again, this is equivalent to the statement that ωf (δ) ≤ Kδ α for all δ ∈ [0, ∞). From what we have just shown above, we see that the upper and lower sums of Lipschitz functions converge faster than, say, those of (1/2)-H¨older functions. Please convince yourselves why this might be true. (b) Here α is continuous instead of f , so Uα (f, P ) − Lα (f, P ) ≤ N X (Mn − mn )ωα (kP k) n=1 for each partition P of [a, b]. Monotonicity of f implies that N X (Mn (f ) − mn (f )) = MN (f ) − m1 (f ) ≤ |f (b) − f (a)|, n=1 and so N X (Mn (f ) − mn (f ))ωα (kP k) = (MN (f ) − m1 (f ))ωα (kP k) ≤ |f (b) − f (a)|ωα (kP k). n=1 The desired result now follows. Problem 2. Recall that (a) a linear combination of Riemann-Stieltjes-integrable functions is linear, and (b) the absolute value of a Riemann-Stieltjes-integrable function is Riemann integrable. COMMENTS ON STYLE AND SAMPLE SOLUTIONS 3 Since f + g + |f − g| f + g − |f − g| and min{f, g} = , 2 2 the desired result follows immediately from (a) and (b). max{f, g} = Remark 2.1. We could, of course, take two partitions—one for f , another for g—and consider their common refinement. Problem 3. Consider, for example, α(x) = x and ( 1 if x ∈ Q ∩ [a, b] f (x) = −1 if x ∈ [a, b] r Q. Note that |f | ≡ 1. Since constant functions are Riemann-integrable, |f | ∈ Rα ([a, b]). On the other hand, we claim that f ∈ / Rα ([a, b]). To see this, we fix an arbitrary partition P . Since the rationals and the irrationals are dense in R, respectively, we see that U (f, P ) = 1 and L(f, P ) = −1. The claim now follows from Riemann’s condition. Problem 4. Fix a partition P = {x0 , . . . , xN } and let In = [xn−1 , xn ] for each 1 ≤ n ≤ N . We examine three cases: (1) If f ≥ 0 on In , then f ≡ |f | on In , and so Mn (f ) = Mn (|f |) and mn (f ) = mn (|f |). (2) If f ≤ 0 on In , then f ≡ −|f | on In , and so Mn (f ) = −mn (|f |) and mn (f ) = −Mn (|f |). (3) If f changes sign on In , then Mn (f ) = Mn (|f |) and mn (f ) ≤ 0 ≤ mn (|f |). Therefore, Mn (f ) − mn (f ) ≥ Mn (|f |) − mn (|f |). In all cases, we see that Mn (f ) − mn (f ) ≥ Mn (|f |) − mn (|f |). Since n was arbitrary, the desired result now follows from Riemann’s condition. Problem 5. A pre-proof remark: I am pretty sure [a, b] is supposed to be [0, 1] for this problem. The proof below takes this assumption for granted. Let p1 , . . . , pN be the accumulation points of Ω. Fix ε > 0 and let In = [pn − ε/4N, pn + ε/4N ] ∩ [0, 1] for each 1 ≤ n ≤ N . By the definition of accumulation SN points, Ω0 = Ω r I n=1 n is a finite set. We assume without loss of generality that Ω0 is nonempty and write {q1 , . . . , qM } to represent this set. We define a partition P of [0, 1] as follows: (1) pick subintervals J1 , . . . , JN of [0, 1] such that each Jn is of length at most ε/2N , pn ∈ Jn , and N N [ [ Jn = In ; n=1 n=1 0 0 (2) pick subintervals J10 , . . . , JM such that each Jm is of length at most ε/2M 0 and qm ∈ Jm ; 0 (3) let J = J1 ∪ · · · ∪ JN ∪ J10 ∪ · · · ∪ JM and divide [0, 1] r J into finitely many subintervals of arbitrary length; 4 MARK H. KIM consolidating the endpoints of the subintervals of type (1), (2), and (3), we obtain a partition P of [0, 1]. Observe that there are at most N intervals of type (1), and each yields the following estimate ε max χΩ (x) − min χΩ (x) ≤ 1 · length(Jn ) ≤ . x∈Jn x∈Jn 2N Similarly, there are at most M intervals of type (2), and each yields the following estimate: ε 0 max0 χΩ (x) − min0 χΩ (x) ≤ 1 · length(Jm . )≤ x∈Jm x∈Jm 2M For (3), we note that χΩ is uniformly zero on [0, 1] r J, whence max χΩ (x) − min χΩ (x) = 0 x∈I x∈I whenever I is a subinterval of type (3). Combining the above observations, we conclude that ε ε ε ε ·N + · M = + = ε. U (χΩ , P ) − L(χΩ , P ) ≤ 2N 2M 2 2 Since ε was arbitrary, we conclude from Riemann’s condition that χΩ is integrable. Sample Solutions (for Problem Set 2) Problem 1. By compactness, we can find x0 ∈ [a, b] such that f (x0 ) = M . Fix ε > 0 and invoke the continuity of f to find δ > 0 such that f (x) ≥ M − ε for all x ∈ [x0 − δ, x0 + δ] ∩ [a, b]. Find c, d ∈ R such that [c, d] = [x0 − δ, x0 + δ] ∩ [a, b]. Positivity of f on [a, b] now implies that "Z #1/n "Z #1/n b d f (x)n dx f (x)n dx ≥ a c "Z #1/n d (M − ε)n dx > c (M − ε)(d − c)1/n , = whence sending n → ∞ yields the lower bound "Z #1/n b f (x)n dx lim n→∞ Since ε was arbitrary, we conclude that "Z #1/n b lim n→∞ ≥ M − ε. a n f (x) dx ≥ M. a We now claim that the lower bound is optimal, i.e., "Z #1/n b f (x)n dx lim n→∞ a = M. COMMENTS ON STYLE AND SAMPLE SOLUTIONS 5 To establish the upper bound, we invoke the monotonicity of the Riemann integrabl to conclude that "Z #1/n "Z #1/n b b f (x)n dx ≤ M n dx M (b − a)1/n . a a Sending n → ∞, we see that "Z #1/n b n ≤ M, f (x) dx lim n→∞ a as was to be shown. Problem 2. For each n ∈ N, we define a partition Pn = {x0 , . . . , xn } of [0, 1] and a set of points T = {x1 , . . . , xn } by setting xk = k/n for each 0 ≤ k ≤ n. Observe that kPn k = 1/n, and that n X k 1 (1) S(f, Pn , Tn ) = f . n n k=1 Since kPn k → 0 as n → ∞, the desired result now follows from Corollary 52.6 in the textbook. Problem 3. Since f is Lipschitz with constant C, ωf (δ) ≤ Cδ for all δ ∈ [0, ∞), where ωf is the modulus of continuity of f : see Remark 1.1 from Homework 1. Homework 1, Problem 1(a) implies that U (f, P ) − L(f, P ) ≤ CkP k (2) for each partition P of [0, 1]. We now let Pn and Tn denote the sets defined in Problem 2. Fix n ∈ N. Since L(f, Pn ) ≤ S(f, Pn , Tn ) ≤ U (f, Pn ), we see that Z 1 L(f, Pn ) − Z f dx ≤ S(f, Pn , Tn ) − 0 1 Z f dx ≤ U (f, Pn ) − 0 1 f dx, 0 and so Z 1 S(f, Pn , Tn ) − (3) f dx ≤ U (f, Pn ) − L(f, Pn ). 0 Similarly, Z L(f, Pn ) ≤ 1 f dx ≤ U (f, Pn ), 0 and so Z L(f, Pn ) − S(f, Pn , Tn ) ≤ 1 f dx − S(f, Pn , Tn ) ≤ U (f, Pn ) − S(f, Pn , Tn ), 0 whence Z 1 f dx − S(f, Pn , Tn ) ≤ U (f, Pn ) − L(f, Pn ). (4) 0 6 MARK H. KIM Combining (3) and (4), we obtain the estimate Z 1 f dx − S(f, P , T ) (5) n n ≤ U (f, Pn ) − L(f, Pn ). 0 Now, (2) implies that C , n as kPn k = 1/n. It now follows from that (1) and (5) Z n 1 X k 1 f dx − f ≤ U (f, Pn ) − L(f, Pn ), 0 n n (6) U (f, Pn ) − L(f, Pn ) ≤ k=1 as was to be shown. Problem 4. Similarly as in Problem 3, we have the estimate ωf (δ) ≤ Cδ. For a partition P of [0, L], Homework 1, Problem 1(a) implies that (7) U (f, P ) − L(f, P ) ≤ CkP kL. Fix 1/2 < α < 1. For each n ∈ N, we define a partition Pn0 = {x0 , . . . , xn } of [0, n1−α ] and a set of points Tn0 = {x1 , . . . , xn } by setting xk = k/nα . Then n k 1 X f (8) S(f, Pn0 , Tn0 ) = α . n nα k=1 Now, an argument analogous to the one given in Problem 3 yields the estimate Z 1−α n f dx − S(f, Pn0 , Tn0 ) ≤ U (f, P ) − L(f, P ), 0 whence (7) and (8) imply that Z 1−α n n 1 X k 1 f dx − α f ≤ C · α · n1−α = C · n1−2α . α 0 n n n k=1 Since 1/2 < α < 1, we see that −1 < 1 − 2α < 0, whence Z 1−α n n 1 X k lim f dx − α f ≤ lim Cn1−2α = 0. n→∞ 0 n nα n→∞ k=1 As 1/2 < α < 1 also implies that n1−α → ∞ as n → ∞, the desired result now follows. Problem 5. A pre-proof remark: as per your request, the proof below no longer depends on integration by parts. p Let f (x) = cos(x2 ) and, for each n ∈ N, xn = (n − 1/2)π. For each a > 0, the function f is continuous on the interval [0, a], whence f ∈ R([0, a]). This, in particular, implies that Z xn+1 bn = cos(x2 ) dx xn COMMENTS ON STYLE AND SAMPLE SOLUTIONS 7 is a real number for each n ∈ N. Note by periodicity of the cosine function that Z xn+1 bn = | cos(x2 )| dx. xn Observe also that Z xn+1 cos(x2 ) dx = (−1)n bn (9) xn for all n ∈ N, whence by the linearity of the integral we have the identity Z √π/2 Z a ∞ X cos(x2 ) dx = (10) lim cos(x2 ) dx + (−1)n bn . a→∞ 0 0 n=1 We claim that (10) is finite, which is the desired result. To this end, we note first that cos(x2 ) ∈ R([0, x1 ]), and so Z x1 cos(x2 ) dx < ∞. 0 It therefore suffices to show that ∞ X (11) (−1)n bn n=1 is finite. To this end, we recall the following result: Lemma 5.1 (Leibniz alternating-series test). Let (cn )∞ n=1 be a sequence of real numbers. If (cn )∞ n=1 is a decreasing sequence of nonnegative numbers converging to zero, then ∞ X (−1)n cn n=1 converges to a finite number. Proof. Summation by parts. (You must have learned this in Analysis I, yes?) (bn )∞ n=1 . (bn )∞ n=1 It thus suffices to check the hypothesisR for By definition, is a q sequence of nonnegative numbers. Since p g dx ≤ (maxa≤x≤b |f |) (q − p) for each interval [p, q] and every g ∈ R([p, q]), we see that Z xn+1 lim bn = lim | cos(x2 )| dx n→∞ n→∞ x n ! r r √ 1 1 n+ − n− ≤ lim 1 · π n→∞ 2 2 = 0. It remains to show that (bn )∞ n=1 is a decreasing sequence. To this end, we shall make use of the following lemma: Lemma 5.2 (Integration by substitution). Suppose that g ∈ R([p, q]), ϕ : [p0 , q 0 ] → [p, q] is an invertible differentiable map such that ϕ(p0 ) = p and ϕ(q 0 ) = q, and (g ◦ ϕ)ϕ0 ∈ R([p0 , q 0 ]). Then the formula Z q Z q0 g(ϕ(x))ϕ0 (x) dx g(x) dx = p p0 8 MARK H. KIM holds. Rt Proof of lemma. Let F (t) = p g(x) dx. By the fundamental theorem of calculus, Z q g(x) dx = F (q) − F (p). p Now, (F ◦ ϕ)0 = (F 0 ◦ ϕ)ϕ0 by the chain rule. It then follows from the fundamental theorem of calculus that Z q0 g(ϕ(x))ϕ0 (x) dx = (F ◦ ϕ)(q 0 ) − (F ◦ ϕ)(p0 ). p0 Since (F ◦ ϕ)(q 0 ) − (F ◦ ϕ)(p0 ) = F (ϕ(q 0 )) − F (ϕ(p0 )) = F (q) − F (p), the desired result follows. Fix n ∈ N. Applying integration by substitution with the map ϕ1 (t) = see that Z √(n+1/2)π bn = | cos(x2 )| dx √ √ t, we (n−1/2)π Z (n+1/2)π = (n−1/2)π | cos t| √ dt. t We invoke integration by substitution once more with the map ϕ2 (s) = s + (n − 1)π to obtain the identity Z (n+1/2)π | cos t| √ dt. bn = t (n−1/2)π Z 3π/2 | cos(s + (n − 1)π)| p = dt. s + (n − 1)π π/2 By periodicity of cosine, we see that Z 3π/2 | cos s| p dt. bn = s + (n − 1)π π/2 Since | cos s| | cos s| ≥√ s + nπ s + (n − 1)π for all s ∈ [π/2, 3π/2], it follows that bn ≥ bn+1 . We have thus shown that (bn )∞ n=1 is a decreasing sequence of nonnegative numbers converging to zero. The Leibniz alternating-series test implies that (11) converges to a finite number. It now follows that cos(x2 ) is improper Riemannintegrable on [0, ∞), as was to be shown. p