Homework 9, Advanced Calculus 1

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Homework 9, Advanced Calculus 1

A topological space X issequentially compactif every sequence has a convergent subsequence.

A compact space is sequentially compact (cf. Theorem 2.37 of Rudin, which is valid for general topological spaces). The first two problems below will prove that for metric space, the two compactness properties are actually equivalent.

1. Prove that for a sequentially compact metric space (X, d), every open cover has acountablesubcover.

Solution: The statement follows from our earlier homework exercises. First we recall Exercises 23,24 of Chapter 2 (Problem 2, 3 of Homework 4). The two problems together say that if every infinite subset of a metric space (X, d) has a limit point, then it has a countable base. A sequentially compact metric space certain satisfies the sufficient condition, since every infinite subset contains a sequence, and the limit of its convergent subsequence is a limit point. Let U = {Ui}i=1 be a countable base: every open subset is a union of sets inU.

Given a open coverX =∪αGα of X, for every α, we have Gα=∪i_αUi_α where Ui_α ∈ U. The set {Ui_α}α,i_αis a subset ofU and therefore countable, which we re-label by{Ui_j}j ⊂ {Ui}i and we have X =∪αGα=∪jUi_j.For eachj, pick ONEGα that containsUi_j, call itGj. The choice is possible due to theaxiom of choice. We then clearly haveX =∪jGj, a countable subcover.

2. Use Problem 1 to show that (X, d) is compact.

Solution: LetX=∪_αG_α. If there is a finite subcover, we are done. If not, by Problem 1, there is a countable subcoverX =∪^∞_j=1Gj. Suppose this cover has NO finite subcover. Then we may take a sequence as follows. Letx₁∈G₁, and inductively, for each k, take x_k ∈ ∪/ ^k_j=1G_j possible due to the assumption that the open cover above has no finite subcover. SinceX is sequentially compact, the sequence{xk}k has a convergent subsequence, which we still call{xk} for convenience. Now xk →xask→ ∞andx∈Gn for some n. Sincexk →x, there isK ∈Nso thatxk ∈Gn∀k > K.

IncreasingK if necessary, we may assume thatK > n. But then we have xk ∈Gn for somek > n, contradicting our construction thatxk ∈ ∪/ ^k_j=1Gj. We conclude that the countable open cover above has a finite subcover and thereforeX is compact.

3. Rudin Chapter 7 Exercise 8

Solution: Letgm(x) =Pm

n=1cnI(x−xn). We first show thatgm⇒fon [a, b]. Since|cnI(x−xn)| ≤

|cn| ∀xand P

|cn|converges, this follows easily from Theorem 7.10 (the Weirstrass M-Test). The continuity of f then follows from continuity of each g_m, which follows from continuity of each I(x−x_n) on [a, b]\{xj}.

Forx∈[a, b]\{x_j}, we havex6=x_n. There exists δ >0 so that (x−δ, x+δ)∩ {x_j}=∅. For all y∈(x−δ, x+δ), we see thatx_nis less or greater than BOTHxandy. ThereforeI₍x−x_n) =I(y−x_n) and the function is a constant on (x−δ, x+δ) which is certainly continuous.

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Solution: Sincefn ⇒f and each fn is continuous, so is f. For all >0, there exist N ∈Nand δ >0 so that

• |fn(t)−f(t)|< ₂ for alln > N andt∈E.

• |f₍t)−f(x)|<₂ if|t−x|< δ.

IncreasingN if necessary, since xn →x, we haved(xn, x)< δ fornlarge enough. We then have

|fn(x_n)−f(x)| ≤ |fn(x_n)−f(x_n)|+|f(xn)−f(x)|<

for allnsatisfying the requirement above.

The converse is false. Consider fn(x) = _x2+(1−nx)^x² ² on (0,1), which has a pointwise limit f = 0.

We observe that 0≤f(x)≤ _(1−nx)^x² 2. For every x∈(0,1) and x_n → x, there exists >0 so that xn∈(x−, x+)⊂(0,1) for allnlarge enough. On that neighborhood, we have

0≤fn(x)≤ x² [1−n(x+)]², and the right hand side approaches 0 asn→ ∞.

However, the convergence is not uniform since for every n, f_n(_n¹) = 1 and therefore f_n does not converge uniformly to 0.

Solution: We first prove that f is discontinuous precisely on Q, which is countable and dense.

Clearly, every function (nx) is discontinuous at _n^p for p ∈ N with left limit = 1 and right limit

= 0. Therefore, each ^p_q ∈Q, is a discontinuity for functions{^(jqx)_j2q²}_j∈N in the series. All the other terms are continuous. Therefore lim_x→p

q

−f(x)−lim_x→p q

+f(x) = P∞ j=1

1

j²q² > 0. On the other hand, each ^(nx)_n2 is continuous on Q^c and so is the partial sum Pm

n=1 (nx)

n² . Since|^(nx)_n2 | ≤ _n¹2 and P

n 1

n² converge,Pm n=1

(nx)

n² ⇒f and thereforef is continuous onQ^c. Therefore,f is discontinuous preciselyonQ.

For the Riemann integrability, note that each (nx) is discontinuous at {_n^p}p∈N, and each bounded interval [a, b] can only contain finitely many points of them, and so is the partial sum Pm

n=1 (nx)

n² . Having finitely many discontinuities, it follows thatPm

n=1 (nx)

n² is Riemann integrable on [a, b], and so is its uniform limitf asm→ ∞.

Solution: We claim thatf satisfying these conditions is a constant function.

If not, there are pointsx < yso that =|f(x)−f(y)|>0. Since{f_n(t)} is equicontinuous, there exists δ > 0 so that |s−t| < δ ⇒ |fn(s)−fn(t)| < for all n. Pick n larger enough so that (_n^x,^y_n) ⊂(s, t). Then |fn(^x_n)−fn(_n^y)| < . But since fn(t) = f(nt), we have |fn(_n^x)−fn(_n^y)| =

|f(x)−f(y)|=, a contradiction.

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Solution: Sincefn(x) is pointwise convergent, it is pointwise bounded. Therefore{fn}satisfies the condition for Arzela-Azcoli Theorem, and therefore it has a uniformly convergent subsequence. But sincefn has a pointwise limit, the subsequence is the sequence itself.

Solution: Repeat the proof of Theorem 3.42 and conclude that P

fngn is uniformly Cauchy and therefore uniformly convergent.

Solution:

The solution is summarized from Group 4 - greatly appreciated!

First note that since |fn| ≤ g and fn ⇒f, |f| ≤ g as well. Since R∞

0 g dx <∞, R∞

0 fn dx and R∞

0 f_ndx exist and are all finite.

Second,R∞

0 g dx <∞means that the sequencesm:=Rm

1 m

g dxconverges tos=R∞

0 g dx. That is, s−sm→0 asm→ ∞. Precisely, for all >0, there isM∈Nso that

Z

[_m¹,m]^c

g dx= lim

a→0

Z _m¹

a

g dx+ lim

b→∞

Z b

m

g dx <

3 ∀m > M. (1) Now we estimate

Z ∞

0

fndx− Z ∞

0

f dx

≤ Z

[_m¹,m]^c

|fn|dx+ Z m

1 m

|fn−f|dx+ Z

[_m¹,m]^c

|f|dx.

Since|fn|,|f| ≤g, we have

Z ∞

0

f_ndx− Z ∞

0

f dx

≤2 Z

[_m¹,m]^c

g dx+ Z m

1 m

|fn(x)−f(x)|dx. (2)

Sincef_n⇒f on every compact interval, there existsN so that sup_[1

m,m]|f_n(x)−f(x)|<₃ for all n > N. Then, for alln > N, pickm > M as above, then the right hand side of (2) is less than and we are done.

Solution: In this problem, we use the fact that a monotonic function can have at most countably many discontinuities.

Since{fn(x)}are uniformly bounded, it is pointwise bounded and therefore{fn(x)}has a convergent subsequence at everyx∈Q∩[0,1]. Since this set is countable, there is a subsequence{f_n_i}_iof{f_n} that converges at everyr∈Q∩[0,1] to, say f(r). We extend the domain off to the entire [0,1] by

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f(x) = sup

r≤x,rational

f(r).

We check thatf is monotonically increasing. It is an increasing function onQ∩[0,1]. Indeed, take two rationals r < s. f_n_i(r) ≤ f_n_i(s) for all i and so are their limits as i → ∞. For any x < y, take a rational number r∈ (x, y). Then f(r) ≥f(r⁰) for all rational numbersr⁰ ≤x. Therefore, f(y)≥f(r)≥sup_r0∈Q≤xf(r⁰) =f(x).

Next we prove that fn_i(x)→ f(x) as i→ ∞ for all x at whichf is continuous. Let >0, there existsδ >0 so that|t−x|< δ⇒ |f(t)−f(x)|< ₂. Take two rational numbersr, s∈(x−δ, x+δ) so thatr < x < s. Monotonicity ofgi implies that

fn_i(r)≤fn_i(x)≤fn_i(s).

Sincefn_i(r), fn_i(s)→f(r), f(s) respectively, there isI∈Nso thatfn_i(s)≤f(s) +₂ andfn_i(r)≤ f(r) + ₂ for all i > I. Furthermore, since r, s ∈ (x−δ, x+δ), continuity of f implies that f(s)≤f(x) +₂ andf(r)≥f(x)−₂. Combining all the estimates, we have

f(x)−≤f_n_i(x)≤f(x) + for alli > I. Therefore,f_n_i(x)→f(x).

Finally, since f(x) is monotonic, there are at most countably many points of discontinuities. The functions fn_i converge at each of those point, and therefore we may take a further subsequence {f_n⁰

i}, which converges at every point of discontinuity. The subsequence of course still converge at point of continuity off, and the proof is completed.

11. Prove Theorem 7.17, with additional assumption thatf_n⁰ is continuous for alln.

Solution: Since each f_n⁰(x) is continuous, it is integrable and by the Fundamental Theorem of Calculus, we have

f_n(x) =f_n(x₀) + Z x

x₀

f_n⁰(t)dt.

Then for allx∈[0,1], we have

|fn(x)−fm(x)| ≤ |fn(x0)−fm(x0)|+ Z x

x0

|f_n⁰(t)−f_m⁰ (t)|dt.

Take > 0, since fn converge uniformly, it is uniformly Cauchy and there is N1 ∈ N so that

|f_n⁰(t)−f_m⁰ (t)|< ₂ for all t∈ [0,1] and n, m > N1. Since{fn(x0)} converges, there isN2 ∈N so that|fn(x0)−fm(x0)|< ₂ for all n, m > N2. TakeN=max(N1, N2), then|fn(x)−fm(x)|< for alln, m > N,x∈[0,1] and therefore{fn} is uniformly Cauchy and convergent.

Consider, fort6=x,

φn(t) = fn(t)−fn(x)

t−x =

Rt

xf_n⁰(s)ds t−x . Sincef_n⁰ uniformly converge to, say,g, we have

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φn(t)→ Rt

xg(s)ds t−x

asn→ ∞. On the other hand, sincefn⇒f, we haveφn(t)→ ^f(t)−f(x)_t−x asn→ ∞and therefore f(t)−f(x)

t−x = Rt

xg(s)ds t−x . LetG(t) =Rt

xg(s)ds. ThenG(x) = 0 and we have, by the Fundamental Theorem of Calculus,

t→xlim Rt

xg(s)ds t−x = lim

t→x

G(t)−G(x)

t−x =G⁰(x) =g(x), and therefore we have

t→xlim

f(t)−f(x)

t−x =g(x), The left hand side is precisely the definition forf⁰(x).