MATH 243 Analysis 2
Midterm Exam Solutions Page 1 of 6 March 14, 2022
1. (10 marks) Let I be an interval, let f : I → R be differentiable and let f′ be bounded on I. Prove that f is Lipschitz continuous on I i.e. prove that there exists a K > 0 such that for all x,u∈I itholdsthat|f(x)−f(u)|≤K|x−u|.
Sincef′ isboundedonIthereexistsaK>0suchthat|f′(x)|≤Kforallx∈I. Wewill prove that for this K it holds that |f(x)−f(u)| ≤ K|x−u| for all x,u ∈ I. If x = u this is obvious. Now assume that x ̸= u; due to symmetry we may assume w.l.o.g. that x < u. We will now apply the mean value theorem to the function f on the interval [x, u]; the conditions of this theorem are satisfied since f is differentiable (and thus also especially continuous) on all of I ⊇ [x,u]. Thus there exists a c ∈]x,u[ (and thus in I) such that
f′(c) = f(x) − f(u) x−u
Then f(x) − f(u) = f′(c)(x − u) and thus |f(x) − f(u)| = |f′(c)| · |x − u| ≤ K|x − u|. In any case we thus have |f(x) − f(u)| ≤ K|x − u| which is what we had to prove.
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MATH 243 Analysis 2
Midterm Exam Solutions Page 2 of 6 March 14, 2022
1 ≤ k ≤ n be such that 0 ∈]xk−1,xk]. Prove that U(f,P) = 2, and that 2 + xk−1 − xk if 0 ∈ ]xk−1, xk[
L(f,P)= 2+xk−1−xk+1 if0=xk
(b) ProvethatifthemeshofP islessthanδ,then2−2δ
MATH 243 Analysis 2
Midterm Exam Solutions Page 3 of 6 March 14, 2022
n k−1 n L(f,P)=mi(xi −xi−1)=(xi −xi−1)+ (xi −xi−1)
i=1 i=1 i=k+2 Telesc. sum
= (xk−1 −x0)+(xn −xk+1)=(xk−1 +1)+(1−xk+1)=2+xk−1 −xk+1 This proves the formula for L(f, P ) in all cases.
(b) Let the mesh of P be less than δ. Then
2+xk−1 −xk =2−(xk −xk−1)>2−δ>2−2δ
L(f,P)= 2+xk−1 −xk+1 =2−(xk+1 −xk)−(xk −xk−1)>2−δ−δ=2−2δ We thus have 2 − δ < L(f, P ) in all cases.
(c) It follows from parts (a) and (b) that for any partition P of [−1, 1] of mesh less than δ it holds that
2 − 2δ < L(f,P) ≤ L(f) ≤ U(f) ≤ U(f,P) = 2
Consequently, 2−2δ < L(f) ≤ U(f) ≤ 2 for all δ > 0. Hence 2 = L(f) = U(f), which
means that f is Riemann integrable on [−1, 1], and that
MATH 243 Analysis 2
Midterm Exam Solutions Page 4 of 6 March 14, 2022
3. (10 marks) Let S := {n1 : n ∈ N}. Consider the function f : [0,1] → R,
sin (1/x) if x ∈ S 0 otherwise
1. Solution
Let ε > 0. Note that −1 ≤ f (x) ≤ 1 for all x ∈ [0, 1]; this motivates the following definitions
Use the squeeze theorem to prove that f is Riemann integrable on [0,1] and compute Solution:
for α, ω : [0, 1] → R:
−1 if x ∈ [0, 4ε ] α(x):= f(x) ifx∈]4ε,1]
1 if x ∈ [0, 4ε ] ω(x):= f(x) ifx∈]4ε,1]
Then α and ω are constant on [0, 4ε ], and are thus Riemann integrable on this interval.
Now note that ]4ε,1] ∩ S is finite since n1 > 4ε ⇔ n < 4ε which is only satisfied by finitely many n ∈ N. This means that α and ω are constantly zero on [4ε,1] except at finitely many points. It now follows from assignment 4 that both α and ω are Riemann integrable on [4ε,1]. Finally, it follows from additivity that α and ω are Riemann integrable on [0, 1] and that
(ω − α) = 2 < ε 0
1 4ε α =
εε α = (−1)4 = −4
εε ω= ω=1·4=4
α and ω thus satisfy all conditions of the squeeze theorem and we conclude that f is Riemann integrable on [0,1]. Finally, it follows from α(x) ≤ f(x) ≤ ω(x) for all x ∈ [0,1] that
1ε11ε α = −4 ≤ f ≤ ω = 4
000 ε1ε 1
i.e. that −4 ≤ f ≤ 4 for all ε > 0. But this means that f = 0. 00
MATH 243 Analysis 2
Midterm Exam Solutions Page 5 of 6 March 14, 2022
2. Solution
Let ε > 0. Consider the following functions α, ω : [0, 1] → R:
−1 ifx∈S 0 otherwise
Then α(x) ≤ f(x) ≤ ω(x) for all x ∈ [0,1] since the sine function only attains values between
1 ifx∈S 0 otherwise
−1 and 1. Note that ω is Riemann integrable on [0, 1] with
5. Note furthermore that α = −ω which implies that α is Riemann integrable on [0, 1] as well.
This implies that
(ω − α) = 2ω = 2 ω = 0 < ε
All conditions of the squeeze theorem are thus satisfied, and we conclude that f is Riemann integrable on [0,1]. Finally, it follows from α(x) ≤ f(x) ≤ ω(x) on [0,1] that
0=α≤f≤ω=0 000
ω = 0 as shown on assignment
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MATH 243 Analysis 2
Midterm Exam Solutions Page 6 of 6
4. (10marks)Letfn : R→R,fn(x):= 1 foralln∈N. 1+nx2
(a) Prove that (fn) converges pointwise on R and find the limit function f. (b) Prove that (fn) converges uniformly on [a, ∞[ for any a > 0.
(c) Prove that (fn) does not converge uniformly on [0,∞[. (d) Does (fn) converge uniformly on ]0, ∞[? Justify!
(a) Let x ̸= 0. Then
lim(fn(x))=lim 1+nx2 =lim 1 +x2 =lim1 +x2=x2 =0
forallx̸=0. Forx=0weget: fn(0)=1foralln∈N. Thuslim(fn(0))=1. Summarizing:
1 ifx=0 fn →f onRwheref(x)= 0 ifx̸=0
March 14, 2022
1 1 lim1 0 nn
(c) 1. Solution: fn is continuous on R and thus on [0, ∞[ for all n ∈ N. However, f is discon- tinuous at x = 0. Since uniform convergence preserves continuity it follows immediately that fn ̸⇒ f on [0, ∞[.
2. Solution: We use the sequential criterion for non-uniform convergence. Let xn := √1 n
1 1 1 1 |fn(x) − f(x)| = − 0 = < ≤
1+nx2 1+nx2 nx2 na2 for all x ∈ [a, ∞[ and for all n ∈ N.
Nowletε>0andletN> 1 .Then|f(x)−f(x)|≤ 1 ≤ 1 <εforallx∈[a,∞[ a2ε n na2Na2
andalln∈N. Hencefn ⇒f on[a,∞[.
and let ε := 12. Then
foralln∈N. Thusfn ̸⇒f on[0,∞[.
1 1 |fn(xn)−f(xn)|= −0= ≥ε
(d) Since xn ∈ ]0, ∞[ for all n ∈ N, it follows from the 2. solution to part (c) above that fn ̸⇒ f on ]0, ∞[.
1+n·n1 2