Advanced Calculus Final Exam

1. Let{s_n}be a sequence of real numbers, letE be the set of subsequential limits of{s_n}defined by E=

x∈R∪ {±∞} | ∃ {s_nk} ⊂ {s_n}such that lim

k→∞s_nk =x

and let the upper and lower limits of{s_n}be denoted by lim sup

n→∞

s_n=s^∗and lim inf

n→∞ s_n=s_∗respectively.

Assume that

lim sup

n→∞

s_n=s^∗ = lim

n→∞sup{s_m|m≥n}=inf{sup{s_m|m≥n} |n∈N}, lim inf

n→∞ s_n=s_∗ = lim

n→∞inf{s_m|m≥n}=sup{inf{s_m|m≥n} |n∈N}.

(a) FindE,and the upper and lower limits of{s_n}if i.

4% {s_n}=Q,i.e.{s_n}is a sequence containing all rationals.

Solution: Since ¯Q=R =⇒ E=R,s^∗=∞ands_∗=−∞.

ii.

4% s_n= (−1)ⁿ

1+ (1/n). Solution: Since lim

n→∞s_2n=1 and lim

n→∞s_2n−1=−1 =⇒ E={±1},s^∗=1 ands_∗=−1.

(b)

8% Prove that

n→∞lims_n=s∈R ⇐⇒ lim sup

n→∞

s_n=lim inf

n→∞ s_n=s∈R.

Solution:

(=⇒)Givenε>0,since lim

n→∞s_n=s∈R,there existsN∈Nsuch that ifn≥N, thens−ε<s_n<s+ε

=⇒ s−ε≤inf{s_n|n≥N} ≤sup{s_n|n≥N} ≤s+ε

=⇒ lim sup

n→∞

s_n=lim inf

n→∞ s_n=s∈R. (⇐=)Givenε>0,since lim sup

n→∞

s_n=lim inf

n→∞ s_n=s∈R,there existsN∈Nsuch that ifn≥N,thens−ε ≤inf{s_n|n≥N} ≤sup{s_n|n≥N} ≤s+ε

=⇒ s−ε≤inf{s_n|n≥N} ≤s_n≤sup{s_n|n≥N} ≤s+ε

=⇒ lim

n→∞s_n=s∈R.

2.

8% Lets₁=√ 2,and

s_n+1= q

2+√

s_n forn=1,2,3, . . . . Prove that lim

n→∞s_nexists.

Solution: Forn≥m≥2,since

|s_n−s_m| = | q

2+√ s_n−1−

q 2+√

s_m−1|

= |2+√

s_n−1−2−√ s_m−1| p2+√

s_n−1+p 2+√

s_m−1

= |√

s_n−1−√ s_m−1| s_n−1+s_m−1

= |s_n−1−s_m−1| s_n−1+s_m−1 √

s_n−1+√ s_m−1

< |s_n−1−s_m−1| 2√

2·2·2^1/4 sinces_n≥√

2∀n∈N

≤ |s_n−1−s_m−1| 4 Thus forn≥m≥N≥2,we have

|s_n−s_m| ≤ |s_n−1−s_m−1| 4

≤ |s_n−(N−1)−s_m−(N−1)| 4^N−1

≤ 2−√ 2 4^N−1 . This implies that{s_n}is a Cauchy sequence inRand lim

n→∞s_nexists.

3.

8% Find the radius of convergence of the power series

∞

∑

n=1

n³ 3ⁿzⁿ.

Solution: For eachz6=0∈C,since

n→∞limsup

(n+1)³zⁿ⁺¹ 3ⁿ⁺¹ · 3ⁿ

n³zⁿ

= lim

n→∞sup(n+1)³|z|

3n³ = |z|

3 , the power series

∞

∑

n=1

n³

3ⁿzⁿconverges if|z|<3 and the radius of convergence isR=3.

4.

8% Let{s_n|n=0,1,2, . . .}be a complex sequence and letσ_ndenote its arithmetic means defined by σ_n= s₀+s₁+· · ·+s_n

n+1 forn=0,1,2, . . . . If lim

n→∞s_n=s,prove that lim

n→∞σn=s.

Solution: For eachε>0,since lim

n→∞s_n=s,there existsN₁∈Nsuch that ifn≥N₁then|s_n−s|<ε.

Also sinces_n−sconverges to 0,s_n−sis bounded and there existsM≥0 such that

|s_n−s| ≤M ∀n∈N.

This implies that there existsN₂∈Nsuch that ifn≥N₂then

N1

k=0

∑

|s_k−s|

n+1 ≤ (N₁+1)M n+1 <ε.

Thus ifN=max{N₁,N₂}and ifn≥N, then|σ_n−s|=

n

∑

k=0

s_k−s n+1

≤

N1

∑

k=0

|s_k−s|

n+1 +

n

∑

k=N₁+1

|s_k−s|

n+1 <ε+ε·n−N₁ n <2ε.

Hence, lim

n→∞σ_n=s.

5.

8% LetX,Y be metric spaces and f :X →Y be a map fromX intoY. Prove that f is continuous onX in the sense ofε−δ definition is equivalent to that f⁻¹(V) ={x∈X | f(x)∈V}is open inX for every open setV ⊂Y,i.e. prove that

∀ε >0, ∀p∈X, ∃δ >0 such that f(B_δ(p))⊂B_ε(f(p))

⇐⇒ f⁻¹(V) ={x∈X| f(x)∈V}is open inXfor every open setV ⊂Y, whereB_δ(p) ={x∈X |d_X(x,p)<δ}andB_ε(f(p)) ={y∈Y |d_Y(y,f(p))<ε}.

Solution:

(=⇒)LetV be an open subset ofY and letpbe a point in f⁻¹(V).Since f(p)∈V andV is an open subset ofY,there exists anε>0 such that

B_ε(f(p))⊂V.

With this givenε,the hypothesis implies that there exists aδ >0 such that

f(B_δ(p))⊂B_ε(f(p))⊂V =⇒ B_δ(p)⊂ f⁻¹(V) =⇒ pis an interior point of f⁻¹(V).

Since pis an arbitrary point in f⁻¹(V),this implies that f⁻¹(V)is open inX.

(⇐=) For eachε >0 and for each p∈X, sinceB_ε(f(p))is an open subset ofY, f⁻¹(B_ε(f(p))) is open inX by the hypothesis. Next since p∈ f⁻¹(B_ε(f(p))), pis an interior point of f⁻¹(B_ε(f(p))) and there exists anδ >0 such that

B_δ(p)⊂ f⁻¹(B_ε(f(p))) =⇒ f(B_δ(p))⊂B_ε(f(p))

6. LetX be a compact metric space,Y be metric spaces.

(a)

8% If f :X →Y is a continuous mapping, prove that f(X)is compact.

Solution: Let{V_α |α∈I}be an open cover of f(X) ={f(p)| p∈X}.

Since f(X)⊂ ^[

α∈I

V_α =⇒ X ⊂ f⁻¹(^[

α∈I

V_α) = ^[

α∈I

f⁻¹(V_α) =⇒ {f⁻¹(V_α)|α∈I}is a cover ofX, and since f is continuous and V_α is an open subset of Y, f⁻¹(V_α) is an open subset of X and {f⁻¹(V_α)|α ∈I} is an open cover of X. The compactness of X implies that there exist α1,· · ·,αm∈I such that

X ⊂

m

[

i=1

f⁻¹(V_αi) = f⁻¹(

m

[

i=1

V_αi) =⇒ f(X)⊂

m

[

i=1

V_αi =⇒ f(X)is a compact subset ofY.

(b)

8% If f :X →Y is a continuous mapping, prove that f is uniformly continuous onX.

Solution: For eachε>0 and for eachp∈X,since f is continuous onX,there exists aδ(p)>0 such that

f(B_δ(p)(p))⊂B_ε/2(f(p)) ∀p∈X (∗).

SinceX is compact andX⊆ ^[

p∈X

B_δ(p)/2(p),there existp₁,· · ·,p_m∈X such that

X ⊆

m

[

i=1

B_δ(pi)/2(p_i).

Letδ = min

1≤i≤m

δ(p_i)

2 and letx,ybe any points inX such thatd(x,y)<δ.Since x∈X ⊆

m

[

i=1

B_δ(pi)/2(p_i) =⇒ x∈B_δ(pk)/2(p_k) for some 1≤k≤m =⇒ d(x,p_k)<δ(p_k) 2 and

d(y,p_k)≤d(y,x) +d(x,p_k)<δ+δ(p_k)

2 ≤δ(p_k) =⇒ x,y∈B_δ(p

k)(p_k).

By(∗),this implies that

f(x), f(y)∈B_ε/2(f(p_k)) =⇒ d(f(x), f(y))≤d(f(x), f(p_k)) +d(f(p_k), f(y))< ε 2+ε

2 =ε.

Hence f is uniformly continuous onX. 7.

6% LetX be a metric space and let

C(X) ={f | f :X →Cis a complex-valued continuous, bounded function defined onX}.

For each f ∈C(X),let the functionk k_X :C(X)→Rbe defined by kfk_X =sup

x∈X

|f(x)|=sup{|f(x)| |x∈X}.

Prove thatk k_X is a norm function onC(X),i.e. prove that

i. kfk_X ≥0 ∀f ∈C(X)andkfk_X =0 if and only if f(x) =0 ∀x∈X.

Solution: For each f ∈C(X),since f is bounded,kfk_X =sup

x∈X

|f(x)| ≥0 exists. Furthermore, sincekfk_X ≥ |f(x)| ≥0∀x∈X,kfk_X =0 if and only if f(x) =0 ∀x∈X.

ii. kc fk_X =|c| kfk_X ∀f ∈C(X), ∀c∈C.

Solution: For each f ∈C(X)and for eachc∈C,sincekc fk_X =sup

x∈X

|c f(x)|=|c|sup

x∈X

|f(x)|=

|c|kfk_X =⇒ kc fk_X =|c| kfk_X.

iii. kf+gk_X ≤ kfk_X+kgk_X ∀f,g∈C(X).

Solution: For any f,g∈C(X),sincekf+gk_X =sup

x∈X

|f(x) +g(x)| ≤sup

x∈X

|f(x)|+sup

x∈X

|g(x)|= kfk_X+kgk_X =⇒ kf+gk_X ≤ kfk_X+kgk_X.

8. (a)

8% Let {f_n} and f be bounded, continuous complex-valued function defined on E. Prove that f_n converges uniformly on E to f if and only if lim

n→∞kf_n− fk_E = lim

n→∞sup

x∈E

|f_n(x)− f(x)| =0, i.e.

prove that

∀ε>0, ∃N=N(ε)∈Nsuch that ifn≥N,then|f_n(x)−f(x)| ≤εfor allx∈E

⇐⇒ lim

n→∞kf_n−fk_E = lim

n→∞sup

x∈E

|f_n(x)−f(x)|=0.

Solution:

(=⇒)Suppose

∀ε>0, ∃N=N(ε)∈Nsuch that ifn≥N,then|f_n(x)−f(x)| ≤εfor allx∈E

=⇒ ∀ε>0, ∃N=N(ε)∈Nsuch that ifn≥N,thenkf_n− fk_E =sup

x∈E

|f_n(x)−f(x)| ≤ε

=⇒ lim

n→∞kf_n−fk_E = lim

n→∞sup

x∈E

|f_n(x)−f(x)|=0.

(⇐=)For eachε>0,since lim

n→∞kf_n−fk_E= lim

n→∞sup

x∈E

|f_n(x)−f(x)|=0,there existsN=N(ε)∈ Nsuch that

ifn≥N, thenkf_n−fk_E ≤ε.

Since|f_n(x)− f(x)| ≤sup

z∈E

|f_n(z)−f(z)|=kf_n−fk_E for allx∈X,this implies that ifn≥N, then|f_n(x)−f(x)| ≤ε ∀x∈X.

(b)

8% LetE be a subset of a metric space X and let f_n:E→Cbe a sequence of continuous complex- valued functions defined on E. If f_n converges uniformly to f :E →C on E, prove that f is continuous onE.

Solution: For eachε>0,since f_nconverges uniformly onEto f,there is an integerNsuch that ifn≥Nand ifx∈E then|f_n(x)−f(x)|< ε

3.

For each p∈E,since f_N is continuous onE,there exists aδ >0 such that ifq∈Eandd_X(p,q)<δ then|f_N(p)−f_N(q)|<ε

3. Hence for eachp∈E and for eachq∈E satisfyingd_X(p,q)<δ,we have

|f(p)−f(q)| ≤ |f(p)−f_N(p)|+|f_N(p)−f_N(q)|+|f_N(q)−f(q)|<ε.

This implies that f is continuous onE.

9. (a) LetF ={f_n(x) =xⁿ|n∈N, x∈[0,1]}.

i.

2% Show thatF is bounded on[0,1].

Solution: For eachn∈N,sincekf_nk_[0,1]= sup

x∈[0,1]

|x|ⁿ≤1,F is bounded on[0,1].

ii.

4% Show thatF is not uniformly equicontinuous on[0,1].

Solution: For eachn∈N,letx_n=e−ln2/n.Thenx_nis a sequence of points in(0,1)such that lim

n→∞x_n= 1 while f_n(x_n) = 1

2 6=1= f_n(1) for all n∈N. Hence F is not uniformly equicontinuous on[0,1].

(b) LetF ={f_n(x) = 1

1+ (x−n)² |n∈N, x∈[0,∞)}and let f(x) = lim

n→∞f_n(x)forx∈[0,∞).

i.

2% Show thatF is bounded on[0,∞).

Solution: For eachn∈N,sincekf_nk_[0,∞)= sup

x∈[0,∞)

1

1+ (x−n)²≤1,F is bounded on[0,∞).

ii.

4% Show thatF is uniformly equicontinuous on[0,∞).

Solution: For eachn∈Nand for eachx∈[0,∞),since

|f_n⁰(x)|=

−2(x−n) 1+ (x−n)²2

≤ 1

1+ (x−n)² ≤1.

This implies that for eachn∈Nand for anyx,y∈[0,∞),by the Mean Value Theorem, there existsz_nlying betweenxandysuch that

|f_n(x)−f_n(y)|=|f_n⁰(z_n)| |x−y| ≤ |x−y|

HenceF is uniformly equicontinuous on[0,∞).

iii.

2% Show that lim sup

n→∞

kf_n−fk_[0,∞)6=0.

Solution: For each x∈[0,∞), since lim

n→∞f_n(x) =0 =⇒ f(x) =0 for all x∈[0,∞). This implies that lim sup

n→∞

kf_n−fk_[0,∞)=lim sup

n→∞

kf_nk_[0,∞)≥lim sup

n→∞

|f_n(n)|=16=0.