Introductory Analysis I Fall 2014 Homework #9 Due: Wednesday, November 19

1. Here is an easy one, to serve as warmup. Assume M is a compact metric space and N is a metric space. Assume that fn : M → N for each n ∈ M. Assume that for every p ∈ M there exists a neighborhood Vp of p such

that (fn|Vp ) converges uniformly on Vp. Prove: (fn) converges uniformly on M.

Proof. By compactness, there exist p1, . . . , pn ∈ M such that M = Sn V . Let > 0 be given. By hypothesis (f | ) is Cauchy in the j=1 pj n Vpj uniform norm for each j = 1, . . . , n; thus there exist N1,...,Nj such that

|fn(p) − fm(p)| < if n, m ≥ Nj, p ∈ Vpj , j = 1, . . . , n. Clearly, if n, m ≥ max(N1,...,Nn), then |fn(p) − fm(p)| < for all p ∈ M. The sequence (fn) is thus Cauchy in the uniform norm, hence converges uniformly.

2. Here is an exercise from the second set of pre-lim problems in the textbook; it is #70 on page 266. The book phrases it as follows: Let A be the set of all positive integers that do not contain the digit 9 in their decimal expansion. Prove that X 1 < ∞. a a∈A P∞ In other words, we consider the series n=1 an where the sequence (1/an) is 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, . . . At ﬁrst it is the almost the same as the sequence of positive integers, but eventually bigger and bigger gaps occur. k k+1 Solution. For k = 0, 1, 2,..., let Ak = {a ∈ A : 10 ≤ a < 10 }. These are numbers with k + 1 digits and some elementary counting tech- niques show that there are 8·9k numbers with k digits that do not contain the digit 9. Since all are ≥ 10k we see that k X 1 9 a ≤ 8 · 9k = 8 . 10k 10 a∈Ak Assume now F is a ﬁnite subset of A. There is then k ∈ N ∪ {0} such that Sk A ⊂ j=0 Aj, hence

k k j ∞ j X X X a X 9 X 9 a ≤ 8 ≤ 8 = 80 < ∞. ≤ 10 10 a∈F j=0 a∈Aj j=0 j=0 The set of all ﬁnite sums is bounded; the result follows. 2

3. Prove the following result ﬁrst proved by Euler: Let p1, p2,... be the X 1 sequence of primes. Then = ∞. In view of this result and the p j=1 j result of the previous exercise, what do you think appears more frequently in the number sequence: A prime, or an integer whose decimal expansion has no digit equal to 9? Hints: For a given integer N select n as in the solution of Exercise 10 of Homework 8; that is, n is such that {p1, . . . , pn} is the set of all primes ≤ N. Recall also formula (1) from that solutions set, which is also valid if s = 1. The terms of the summation on the right side of (1) comprise all numbers that decompose into a product of (powers of) the primes p1, . . . , pn; in particular every number from 1 to N. Thus

N n X 1 Y 1 ≤ n 1 − 1 n=1 k=1 pk

Prove that if 0 ≤ x ≤ 1/2 then 1 − x ≤ e2x. Use this to conclude that

N X 1 Pn 2 ≤ e k=1 pk . n n=1

Solution. In the hints, “0 ≤ x ≤ 1/2 then 1 − x ≤ e2x,” should have been “0 ≤ x ≤ 1/2 then (1 − x)−1 ≤ e2x.

Let N ∈ N and let p − 1, . . . , pn be all the primes ≤ N. If k ≤ N, then j1 jn k = p1 ··· pn for some (unique) choice of j1, . . . , jn ∈ N ∪ {0}. Thus 1 = p−j1 ··· p−jn k 1 n and is thus a member of the summation appearing on the right hand side of equation (1) of the solution pages of Homework 8, with s = 1. It follows, by that formula, that N n X 1 Y 1 ≤ . (1) n 1 − 1 n=1 k=1 pk Claim: If 0 ≤ x ≤ 1, then (1 − x)−1 ≤ e2x. Letting ϕ(x) = e2x(1 − x), the claim is equivalent to claiming that ϕ(x) ≥ 1 for 0 ≤ x ≤ 1/2. Now ϕ0(x) = e2x(1 − 2x) ≥ 0 for 0 ≤ x ≤ 1, thus ϕ is increasing in that interval, hence ϕ(x) ≥ ϕ(0) = 1 for all x ∈ [0, 1/2], establishing the claim. −1 It follows that (1 − p−1)−1 ≤ e2p for all primes p and (1) implies

N n X 1 Y 1 Pn 1 ≤ e pk = e k=1 pk . n n=1 k=1 3

Taking logarithms, n N X 1 X 1 ≥ . pk n k=1 n=1 Letting N → ∞, we also have n → ∞ and the right hand side of the last displayed inequality diverges to ∞. This implies that the left hand side also has to diverge to ∞ as n → ∞.

The answer to the last question is that the frequency at which primes appear (i.e., on the average, the number of primes in any interval of integers) is higher than that of integers with a decimal expansion with all digits diﬀerent from 9. 4. More on monotone functions. In the previous homework you learned (if you did not know it already) that a monotone function can only have a countable set of discontinuities. The surprising fact is that every countable subset of an interval [a, b] can be the set of discontinuities of a monotone function. For example, there are monotone functions that are discontinuous at all rational points, continuous elsewhere. Here you are required to ﬁll in the details of the constructions of such a function.

Let −∞ < a < b < ∞ and let (xn) be a sequence of distinct points of the interval (a, b). Deﬁne f :[a, b] → R as follows. Select cn for n ∈ N so that P∞ cn > 0 and cn < ∞. For x ∈ [a, b] let S(x) = {n ∈ N : xn < x}. Pn=1 Deﬁne f(x) = n∈S(x) cn. Prove: (a) f is increasing in [a, b].

(b) f has a jump at xn for each n ∈ N; in fact f(xn+) − f(xn−) = cn for each n ∈ N, (c) f is continuous at all x 6= xn, n ∈ N.

Comments and hints. The points xn could be all over the place so it is quite possible (as it would be if (xn) is an ordering of the rational numbers in the interval) that S(x) is an infinite set for all x ∈ (a, b). Of P course S(a) = ∅ and one defines cn = 0. If S(x) is a finite set there P n∈∅ is no problem in how n∈S(x) cn is defined. If it is infinite it is countably infinite and one can either define X X cn = sup{ cn : F is a finite subset of S(x)} n∈S(x) n∈F

or, equivalently, one can order S(x) into a sequence {n1, n2,...} and deﬁne P P∞ n∈S(x) cn = k=1 cnk . The way one does this ordering is irrelevant; the end result is always the same. Proving that f is increasing is easy, I think. For the rest, a direct proof can be sort of messy. Here is a more elegant way of proceeding. Let H be the Heaviside function; that is the function 4

defined by 0, if x < 0, H(x) = 1, if x ≥ 0. P∞ Prove that f(x) = n=1 cnH(x − xn); in fact, prove the series P∞ n=1 cnH(· − xn) converges uniformly to f. Use this to complete the proof. Solution. The hints were a bit muddled. That the function f defined in the statement of the exercise is the same as the function defined by f(x) = P∞ P∞ n=1 cnH(x − xn) is, I think, obvious. I will write f = n=1 gn, where gn(x) = cnH(x − xn) for n ∈ N, x ∈ (a, b). The series converges uniformly because |gn(x)| ≤ cn for all n ∈ N, x ∈ (a, b), so uniform convergence is P guaranteed by the Weierstrass M-test, since cn < ∞. By a result seen in class (and in the textbook), the limit of a uniformly convergent sequence (or series) of functions from a metric space to another is continuous at each point where all the terms of the sequence (or series) are continuous. Now x 7→ H(x − xn) is continuous at all points x 6= xn. Thus, if x 6= xn for n ∈ N, then gn is continuous at x for all n, and f is continuous at all points x 6= xn, n ∈ N. Let n ∈ N. Then gm is continuous at xn for P m 6= m. Defining G(x) = m6=n gm(x), we have f = gn + G and since G is continuous at x we see that f(xn+) − f(xn−) = gn(x+) − gn(x−) = cn. At this point we proved

(a) f has a jump at xn for each n ∈ N; in fact f(xn+) − f(xn−) = cn for each n ∈ N, (b) f is continuous at all x 6= xn, n ∈ N. It remains to be seen that f is increasing. This is obvious from the ﬁrst deﬁnition of f.

3. Textbook, Chapter 4, Exercise 8, p. 252. Solution.

The problem was to prove or disprove that the sequence of functions fn : R → R deﬁned by 1 2 n fn(x) = cos(n + x) + log 1 + √ sin (n x) n + 2 is equicontinuous. The answer is yes; to prove it we ﬁrst establish some preliminary results.

Lemma 1 Let I be an interval in R and let fn : I → R for n ∈ N. Assume that f is diﬀerentiable at all interior points of I (if I is open, 0 that’s all points of I). If there exists M ≥ 0 such that |fn(x)| ≤ M for all 0 x ∈ I , n ∈ N, then (fn) is equicontinuous. 5

Proof. Let > 0. Take δ = /(M + 1) (I use M + 1 to avoid saying: We may assume M > 0.) If x, y ∈ I, x 6= y. By the mean value theorem there is c between x and y such that f(x) − f(y) = f 0(c)(x − y), thus |f(x) − f(y)| ≤ M|x − y| for all x, y ∈ I (trivially if x = y). Thus |x − y| < δ implies |f(x) − f(y)| ≤ M|x − y| < Mδ = M/(M + 1) < .

Lemma 2 Let (fn), (gn) be two equicontinuous sequences of functions from a metric space M to R. Then the sequences (fn + gn), (cfn) (where c ∈ R) are also equicontinuous.

Proof. Left as exercise.

We are ready for the exercise. Let gn, hn : R → R be defined by gn(x) = 1 2 n cos(n+x), hn(x) = log 1 + √ sin (n x) for x ∈ R. Both functions n + 2 are differentiable on R. 0 |gn(x)| = | sin(n + x)| ≤ 1 for all x ∈ R, n ∈ N. By Lemma 1, (gn) is equicontinuous. Since fn = gn+hn, by Lemma ]reflem 2 it suffices to prove that (hn) is equicontinuous. For this purpose, let > 0 be given. Now limx→0 log(1 + x) = log1 = 0; there is thus η > 0 such√ that if |x| < η then | log(1 + x)| < /2. Taking now N ∈ N such that 1/ N + 2 < η we see that if n ≥ N, x ∈ R, then √ 1 2 n √ sin (n x) ≤ 1 n + 2 < η n + 2 1 2 n and it follows that log 1 + √ sin (n x) < /2; i.e, n + 2

|fn(x)| < /2 ∀ n ≥ N, x ∈ R. (2) Assume now n ≤ N. In this case

2N N 2N N |fn(x) − fn(y)| ≤ √ |x − y| < √ δ = . 3 − 1 3 − 1 6

If n > N, then by (2), |f (x) − f (y)| ≤ |f (x)| + |f (y)| < + = . n n n n 2 2

We proved that |fn(x) − fn(y)| < for all n ∈ N whenever |x − y| < δ. We are done. 4. Textbook, Chapter 4, Exercise 13, p. 252.

Solution. To decide: Assume fn : R → R for n ∈ N and (fn|K ) is pointwise equicontinuous and pointwise bounded for each compact subset K of R.

(a) Does it follow that there is a subsequence of (fn) that converges pointwise to a continuous g : R → R? (b) What about uniform convergence? The answer to the first question is yes. To the second question, not nec- essarily. We prove first there is a subsequence converging pointwise to a continuous function on R. But before we can do this we have to deal with a certain nuisance; the fact that the assumption is not that the restrictions of the sequence to compact sets are equicontinuous and (uniformly) bounded, but only pointwise equicontinuous (which I think is a fairly useless concept) and pointwise bounded. We show (and a full solution of this exercise would have had to show) that one can get rid of the pesky “pointwise” clause. Partially this is done in our textbook; in Theorem 37 (page 249). I did not cover this in class because, as I mentioned above parenthetically, I consider it a fairly useless concept. Moreover, the proof is in a later starred section. Pugh’s proof is by contradiction but it is also easy to give a better, direct proof, which is what I will do. That is, I’ll show that if K is a compact metric space and gn : K → R for n ∈ N is such that the sequence (gn) is pointwise equicontinuous and pointwise bounded, then it is equicontinuous and bounded, hence Arzela-Ascoli applies. I’ll begin with equicontinuity. Let > 0 be given. By pointwise equicontinuity, for each p ∈ K there exists δp > 0 such that |gn(q) − gn(p)| < /2 for all n ∈ N if d(q, p) < δp. The open balls B(p; δp/2) constitute an open covering of K; by compactness there is a finite subcovering

{B(p1; δp1 /2),...,B(pm; δpm /2)}. Let δ = min1≤i≤m δpi /2. Then δ > 0. Assume q1, q2 ∈ K and d(q1, q2) < δ. There is i, 1 ≤ i ≤ m such that

q1 ∈ B(pi, δpi /2). Then d(q1, pi) < δpi /2 < δpi and |gn(q1) − gn(pi)| < /2 for all n ∈ N. In addition,

d(q2, pi) ≤ d(q2, q1) + d(q1, p) < δ + δpi /2 ≤ δpi /2 + δpi /2 = dpi ,

so that also |gn(q2) − gn(pi)| < /2 for all n ∈ N. Thus |g (q ) − g (q )| ≤ |g (q ) − g (p )| + |g (p ) − g (q )| < + = n 1 n 2 n 1 n i n i n 2 2 2 7 for all n ∈ N. This proves equicontinuity. Now that we have equicontinuity, we see that pointwise boundedness implies boundedness. In fact, we may assume now (gn) is an equicontinuous sequence from K to R and assume (gn(p)) is a bounded sequence of real numbers for every p ∈ K. You might recall that this was all that was needed to prove the Arzela- Ascoli theorem, so a quick argument is to say that by the same proof of the Arzela-Ascoli theorem, the sequence has a uniformly convergent subsequence. More to the point, every subsequence has a uniformly convergent subsequence, and boundedness follows easily from this. But I’ll give a direct proof. By equicontinuity (with = 1), there is δ > 0 such that if d(p, q) < δ, then |gn(p) − gn(q)| < 1 for all n ∈ N. By compactness, since {B(p; 1)}p∈K is an open covering of K, there exist p1, . . . , pr ∈ K Sr such that K = j=1 B(pj; 1). By the pointwise boundedness, there exist C1,C2,...,Cr such that |gn(pj)| ≤ Cj for all n ∈ N. Let q ∈ K. Then there is j ∈ {1, . . . , r} such that q ∈ B(pj, 1); thus

|gn(q)| ≤ |gn(q) − gn(pj)| + |gn(pj)| < 1 + Cj ≤ max Cj + 1. 1≤j≤r

The sequence is uniformly bounded by max1≤j≤r Cj + 1. Concerning our sequence, it follows it is equicontinuous and uniformly bounded on compact sets, thus Arzela-Ascoli applies to all restrictions of the sequence to a compact set.

We turn to the proof as such. It is done by a diagonal process similar to the one used to prove Arzela-Ascoli. We construct an ∞×∞ integer matrix ν ∞ such that for i = 2, 3,..., the i-th row, namely (ν(i, j))j=1 is a subsequence of the (i − 1)-st row as follows. By Arzela-Ascoli, the sequence (fn) restricted to the compact set [−1, 1] has a uniformly convergent subsequence

(fnj ) to some continuous function g1 :[−1, 1] → R; define ν(1, j) = nj. Thus (fν(1,j)|[−1,1]) converges uniformly to g1. Assume ν(i, j) defined for ∞ some i ∈ N and all j ∈ N so that (ν(i, j))j=1 is a strictly increasing sequence of integers and (fν(i,j)|[−i,i]) converges uniformly (as j → ∞) to a continuous function gi :[−i, i] → R. We now consider the sequence (fν(i,j)) restricted to [−i − 1, i + 1]. By Arzela-Ascoli it has a subsequence converging uniformly to some continuous gi+1 :[−i − 1, i + 1] → R.A subsequence of a sequence is defined by a subsequence of the indices of the sequence; that is, there exists a strictly increasing sequence of positive integers (jk) such that (fν(i,jk)|[−i−1,i+1]) converges uniformly to gi+1 as k → ∞. We define ν(i + 1, k) = ν(i, jk) for k = 1, 2,.... This concludes the definition of the matrix ν.

It should be clear that the limit functions gi satisfy gi+1|[−i,i] = gi for i = 1, 2, 3 ...; that is simply because the sequence (fν(i+1,j)|[−i,i]) is a subsequence of (fν(i,j)|[−i,i]), hence their limits have to coincide. We can thus deﬁne a function g : R → R by g(x) = gi(x) if |x| ≤ i and this 8

function will be well deﬁned. It will also be continuous; if x ∈ R select i ∈ N so |x| < i; then g = gi in a neighborhood of x and the continuity of gi implies that of g. Returning to our original sequence of functions, the diagonal process is completed by selecting the diagonal sequence (fν(k,k)); ∞ this sequence is a subsequence of (fν(i,j))j=1 for k ≥ i; more precisely, for ∞ ∞ i = 1, 2,...,(fν(k,k))k=i is a subsequence of (fν(i,j))j=1, thus it, and hence ∞ also the full sequence (fν(k,k))k=1, converges to gi = g uniformly on [−i, i]. In particular the subsequence (fν(k,k)) of (fn) converges pointwise to g on R. This takes care of the ﬁrst question.

Concerning the second question, the answer is no. There are many coun- terexamples, the following rather simple one shows that even if we strengthen “pointwise bounded” to “bounded,” the conclusion might still be false. Deﬁne ﬁrst f : R → R by sin x, 0 ≤ x ≤ π, f(x) = 0 , x < 0 or x > π.

(Any continuous function that’s zero outside of a compact set will do.) Deﬁne fn : R → R by fn(x) = f(x − n). This is trivially equicontinuous (given > 0, the δ > 0 that works for f works for all fn) and uniformly bounded (by 1). It is also clear that it converges pointwise to 0. The convergence is uniform on compact sets. But the convergence is clearly not uniform on all of R; if ∈ (0, 1) there is no N such that |fn(x)| < for all n ≥ N. The same goes for any subsequence of this sequence. 5. Textbook, Chapter 4, Exercise 22, p. 254 Solution. In this exercise one assumes E is an equicontinuous and bounded subset of C0. The mission is: (a) Prove x 7→ sup{f(x): f ∈ E} is continuous. (b) Show (a) fails without equicontinuity. (c) Show that this continuous sup-property does not imply equicontinuity. (d) Assume that the continuous sup-property is true for each subset F ⊂ E. Is E equicontinuous? Prove or disprove. Proof of (a) It is mentioned at the beginning of the exercise set that C0 stands for C([a, b]); I will be a bit more general because with the same eﬀort one can prove the result for C(K), K a compact metric space. So in this part we are given a bounded, equicontinuous subset of C(K) a compact metric space and we want to prove F : x 7→ sup{f(x): f ∈ E} is continuous. Notice that because E is bounded, we have F (p) ≤ M for all p ∈ K, where M is a bound for E. Assuming E 6= ∅, we’ll also have F (p) ≥ −M for all p ∈ K. 9

Let > 0 be given. By equicontinuity there is δ > 0 such that |f(p) − f(q)| < for all p, q ∈ K with d(p, q) < δ, all g ∈ E. Assume now p, q ∈ K, d(p, q) < δ. For each g ∈ E, f(q) < f(p) + ≤ F (p) + , thus F (p) + is an upper bound of {f(q): f ∈ E}. Thus F (q) ≤ F (p) + . Reversing the roles of p, q we get F (p) < F (q) + . We proved |F (p) − F (q)| < if d(p, q) < δ; i.e., F is uniformly continuous, hence continuous. Proof of (b) I am writing down this proof (or disproof) before seeing any of your answers. One of you may have come up (or found) a simpler proof. For n ∈ N deﬁne fn : [0, 2] → R by  0, if 0 ≤ x ≤ 1,  1 fn(x) = nx − n, if 1 < x ≤ 1 + n ,  1 1, if 1 + n < x ≤ 2.

One sees without excessive diﬃculties that the set E = {fn : n ∈ N} is a bounded subset of C([0, 2]) but

0, if 0 ≤ x ≤ 1, sup f (x) = n 1, if 1 < x ≤ 2, n∈N is not continuous. There is no need to check that E is not equicontinuous; it can’t be by part (a). Proof of (c) A trivial counterexample is to take any bounded non equicontinuous set of functions and then add a nice continuous function larger than all the functions in the set. For example, let E be the set from part (b) and consider E0 = E ∪ {h} where h(x) = 1 for all x ∈ [0, 2]. Then sE0 is a non equicontinuous, bounded subset of C([0, 2]) and sup{f : f ∈ E0} = h is continuous. Answer to the question in (d) The answer is no.

For n ∈ N deﬁne fn : [0, 1] → R by  1, if 0 ≤ x < 1 ,  n+1    −2n(n + 1)x + 2n + 1, if 1 ≤ x < 1 ( 1 + 1 ),  n+1 2 n n+1 fn(x) =  2n(n + 1)x − 2n − 1, if 1 ( 1 + 1 ) ≤ x < 1 ,  2 n n+1 n    1 1, if n ≤ x ≤ 1.

The following graphs shows f1, f2 and f5: 10

f1 f2 f3

Consider the set E = {fn : n ∈ N}. It is fairly obvious that this set is not equicontinuous; perhaps the most immediate way to see this is to see that no subsequence can converge uniformly. If F is any finite subset of E, then sup{f(x): f ∈ F} = max{f(x): f ∈ F} and the maximum of a finite set of continuous functions is continuous. If F is infinite, then for every x > 1, there will be fn ∈ F with 1/n ≤ x; then fn(x) = 1 implying that sup{f(x): f ∈ F} = 1. If x = 0, then fn(x) = 1 for all n ∈ N. The conclusion is that if F is infinite, then sup{f(x): f ∈ F} = 1 for all x ∈ [0, 1]. Thus the continuous sup-property holds for all subsets of E, but E is not equicontinuous.

Some of you came up with an even easier counterexample, namely the n set {fn : n ∈ N} where fn : [0, 1]toR is deﬁned by fn(x) = x . This is not an equicontinuous family; it converges pointwise to the discontinuous function f satisfying f(1) = 1, but f(x) = 0 for all x ∈ [0, 1). Because it is a decreasing sequence (f1 ≥ f2 ≥ · · · ) one sees that the sup of any subset of these functions is the element of this set with the smallest index, hence in the set, hence continuous. The following exercises, from Berkeley pre-lims, are optional but I think you should try to do some or all of them. They are nice exercises and could appear in qualiﬁers. Please, if you can’t do them, don’t do them. Don’t just go and copy the solution from somewhere. As usual, hints are provided upon request. 6. Problem 3 of the More Pre-lim Problems of the textbook, page 258. Solution. Since this is the same as Exercise 22, which I placed into Homework 10, I won’t provide a solution yet. 7. Problem 5 of the More Pre-lim Problems of the textbook, page 259. 11

Solution. One had to prove that if (Pn) is a sequence of real polynomials of degree ≤ 10 with limn→∞ Pn(x) = 0 for all x ∈ [0, 1],then (Pn) converges uniformly to 0.

At first glance one would assume that this obviously has to be an applica- tion of Arzela-Ascoli. A big question is whether one has to prove uniform convergence to 0 in [0, 1] or in R. Since R is not compact, it probably means in [0, 1], but we can keep this open to the end. To apply Arzela-Ascoli, we need to prove that the sequence of polynomials is bounded and equicontinuous. Both will follow if we prove the coefficients of these polynomials remain bounded. Because of the degree restriction, we only have 11 coefficients to worry about. If we can show the coefficients remain bounded then the polynomials and their derivatives remain bounded on bounded intervals, bounded derivatives imply equicontinuity and Arzela-Ascoli applies. Of course, Arzela-Ascoli only gives us a subsequence but, since every converging subsequence must have 0 as a limit, the whole sequence converges uniformly to 0. But in trying to do this we might realize that we can bypass Arzela-Ascoli completely. But we need a bit of notation. Write

10 X k Pn(x) = ankx , n = 1, 2, 3 .... k=0

∞ Claim: The 11 sequences (ank)n=1 converge to 0 as n → ∞. In fact, select 11 distinct points x1, . . . , x11 in (0, 1]. The matrix

 2 10  1 x1 x1 ··· x1 2 10  1 x2 x2 ··· x2  X =    . . . . .   . . . . .  2 10 1 x11 x11 ··· x11 is invertible (it is a Vandermonde matrix) and we have     Pn(x1) an0  Pn(x2)   an1    = X   .  .   .   .   .  Pn(x11) an,10

Thus     an0 Pn(x1)  an1   Pn(x2)    = X−1   .  .   .   .   .  an,10 Pn(x11) 12

−1 Writing X = (bkj)0≤k,j≤10 (it is convenient to index from 0 to 10 rather than 1 to 11), then

10 X an,k = bkjPn(xj+1), 0 ≤ k ≤ 10, n ∈ N. (4) j=0

But (Pn) converges pointwise to 0 in [0, 1] thus, given > 0 there are N1,...,N11 inN such that if n ≥ Nj, then |Pn(xj)| < /B, where

10 X B = max |bkj|, 0≤k≤10 j=0

for 1 ≤ j ≤ 11. It follows from (4) that if n ≥ max(N1,...,N11), then

10 10 X X |a | ≤ |b ||P (x )| ≤ |b | ≤ B = n,k kj n j+1 kj B B j=0 j=0

for k = 0,..., 10. The claim is established.

It is now immediate that (Pn) converges uniformly to 0 not only on [0, 1] but also on every bounded interval of R. In such an interval we have |x| ≤ M for some M, and then for x in that interval

10 ! X k |Pn(x)| ≤ M max |ank| → 0 as n → ∞. 0≤k≤10 k=0 Th convergence is uniform because the right hand side of the inequality above does not depend on x (as long as |x| ≤ M).

However, uniform convergence on all of R is out. A simple counterexample 1 is taking Pn(x) = n x. The sequence (Pn) satisﬁes the conditions of the exercise, yet does NOT converge uniformly to 0 on R. 8. Problem 6 of the More Pre-lim Problems of the textbook, page 259.

Solution. . To prove: Let (an) be a sequence of non-zero real numbers. The sequence of functions 1 fn(x) = sin(anx) + cos(x + an) an has a subsequence converging to a continuous function. Solution. The function x 7→ sin x/x, deﬁned for x 6= 0, extends to a continuous function on all of R deﬁning its value at0 to be 1. Since limx→∞ sin x/x = 0, the function assumes a maximum and a minimum 13 value; in particular it is bounded. There is M ≥ 0 such that | sin x/x| ≤ M for all x. (A bit of calculus shows that M = 1, but that is not important). The cosine function is, of course, bounded by 1. Thus

1 1 |fn(x)| ≤ sin(anx) +| cos(x+an)| = |x| sin(anx) +| cos(x+an)| ≤ M|x|+1 an xan for all n ∈ N, x ∈ R. It follows that the sequence (fn) is bounded on compact (closed and bounded) intervals. Moreover,

0 fn(x) = cos(anx) − sin(anx), thus 0 |fn(x)| ≤ | cos(anx)| + | sin(anx)| ≤ 2 for all n ∈ N, x ∈ R. By the mean value theorem, the sequence (fn) is equicontinuous. The result now follows from exercise 4, part a.