Assignment-6
(Due 07/30)
1. Let sequences fn and gn converge uniformly on some set E ⊂ R to f and g respectively
(a) Construct an example such that fn gn does not converge uniformly on E.
Solution: Take fn = gn = x + 1/n and E = R. Clearly fn , gn → x uniformly on R. Now
fn gn = (x + 1/n)2 , and we claim that this does not converge uniformly to x2 . To see this, we
let hn (x) be the sequence of the squares and expand
1 2
2x x2
hn (x) = x +
= x2 +
+ 2.
n
n
n
But then we see that
1
≥ 2,
n
no matter how big of an n we choose. This contradicts uniform convergence.
|hn (n) − x2 | = 2 +
(b) Prove that fn gn does converge uniformly if f and g are bounded on E.
Solution: Let |f (t)|, g(t)| < M for all t ∈ E. Without loss of generality, we can assume that
M > 1. Given ε > 0, choose N such that for all t ∈ E and all n > N ,
|f (t) − fn (t)|, |g(t) − gn (t)| <
ε
.
3M
Note also that if ε is small enough, then for all n > N and all t ∈ E,
|fn (t)| ≤ |fn (t) − f (t)| + |f (t)| ≤
ε
+ M < M + 1.
3M
Next, note that
fn (t)gn (t) − f (t)g(t) = fn (t)gn (t) − fn (t)g(t) + fn (t)g( t) − f (t)g(t),
and so by triangle inequality, for all t ∈ E and all n > N ,
|fn (t)gn (t) − f (t)g(t)| ≤ |fn (t)gn (t) − fn (t)g(t)| + |fn (t)g( t) − f (t)g(t)|
= |fn (t)||gn (t) − g(t)| + |g(t)||fn (t) − f (t)|
ε
ε
≤ (M + 1)
+M
< ε.
3M
3M
2. Show that the sequence of functions
nx
n+1
does not converge uniformly on all of R, but does converge uniformly on bounded intervals (a, b). What
is the point-wise limit on R?
fn (x) =
1
Solution: We show that fn (x) → x point-wise on R.
|fn (x) − x| =
nx
|x|
−x =
.
n+1
n+1
At x = 0, fn (0) = 0 for all n, and so there is nothing to prove. If x 6= 0, given any ε > 0, we can
pick N large enough so that |x|/(N + 1) < ε. THen for n > N we see that
|fn (x) − x| < ε.
Of course the N depends on x and hence we have only proved point-wise. But notice that if x ∈ (a, b)
in a bounded interval, then there is an M such that |x| < M .Then we can simply choose N > M/ε,
then for any n > N and x ∈ (a, b), it is easy to see that
|fn (x) − x| < ε,
and hence the convergence is uniform on bounded intervals.
We now claim that the convergence is NOT uniform on all of R.
That is we need to show that there exists ε > 0, a subsequence nk → ∞ and points xnk ∈ R such
that
|fnk (xnk ) − xnk | > ε.
For this, let ε = 1, nk = k and xk = k + 2. Then by the calculation above
|fk (xk ) − xk | =
k+2
|xk |
=
> 1.
k+1
k+1
Note that we cannot use the method of Mn , since fn is not a bounded function on R.
3. If g : [0, 1] → R is a continuous function such that g(1) = 0, show that the sequence of functions
{g(x)xn }∞
n=1 converges uniformly on [0, 1].
Solution: We claim that the sequence converges uniformly to zero. Let ε > 0. Since g(1) = 0 and
g is continuous at 1, there exists a δ such that
|g(x)| < ε
for all x ∈ [1 − δ, 1]. So for x ∈ [1 − δ.1],
|g(x)xn | < |g(x)| < ε.
Since g is also bounded on [0, 1], letting M = supt∈[0,1] |g(t)|, for x ∈ [0, 1 − δ] we have
|g(x)xn | ≤ M |x|n ≤ M (1 − δ)n .
Choose N big enough so that (1 − δ)n < ε/M for all n > N . Then for all n > N and all x ∈ [0, 1],
|g(x)xn | < ε,
showing that
u.c
g(x)xn −−→ 0
on [0, 1].
2
4. (a) Define a sequence of functions by
(
fn (x) =
1, x = 1, 21 , · · · , n1
0, otherwise.
Calculate the pointwise limit function f . Is each fn continuous at zero? Does fn → f uniformly on
R. Is f continuous at 0?
Solution: The limit function is
(
1, x = 1, 12 , 13 · · ·
f (x) =
0, otherwise.
The function fn is continuous everywhere except at x = 1, 1/2, · · · , 1/n, whereas the limit
function is discontinuous at the reciprocals of all natural numbers. The theorem on uniform
convergence and continuity does not automatically imply that the convergence is not uniform,
but we nevertheless claim that fn does not converge uniformly. Suppose the convergence is
uniform, then there exists an N such that for all x ∈ R,
|fN (x) − f (x)| ≤
But then if x =
1
N +1 ,
1
.
2
we have that
fN
1 1 = 1,
−f
N +1
N +1
a contradiction.
(b) Repeat the exercise with the functions
(
gn (x) =
and
x, x = 1, 21 , · · · , n1
0, otherwise,

1
1

x, x = 1, 2 , · · · , n−1
hn (x) = 1, x = n1


0, otherwise.
Solution:
• The sequence gn (x). The pointwise limit is clearly
(
x, x = 1, 12 , · · ·
g(x) =
0, otherwise
Again, we note that gn is continuous everywhere except at x = 1, 1/2, · · · , 1/n while g(x)
is continuous everywhere except the reciprocals of natural numbers.
u.c
Claim. gn −−→ g on R.
Proof. Let us consider Mn = supR |gn (x) − g(x)|. Now, |gn (x) − g(x)| = 0 everywhere,
except when x = 1/m for m > n, in which case the difference is 1/m. So
Mn =
3
1
n→∞
−−−−→ 0,
n+1
u.c
and hence gn −−→ g on R.
• The sequence hn (x). The pointwise limit in this case is
(
x, x = 1, 12 , · · ·
h(x) =
0, otherwise,
exactly as above. Again hn is continuous everywhere except at x = 1, 1/2, · · · , 1/n while
h(x) is continuous everywhere except the reciprocals of natural numbers. But in this case
we claim.
Claim. hn does not converge uniformly to h on R.
Proof. To see this, we again consider Mn = supR |hn (x) − h(x)|. Now, |gn (x) − g(x)| = 0
everywhere, except when x = 1/m for m ≥ n. Moreover, we have that
(
1
1
1
, m>n
−h
|= m 1
hn
m
m
1 − n , m = n,
and so Mn = 1 − 1/n for n > 2 and hence
lim Mn = 1 6= 0.
n→∞
.
5. Let {fn } be a sequence of real-valued continuous functions fn : E → R for some subset E ⊂ R. Suppose
u.c
fn −−→ f on E. Show that
fn (xn ) → f (x)
for every sequence of points xn → x in E. Is the conclusion true if fn → f only pointwise? Either
provide a proof, or a counter example. Is the converse of the above statement true?
Solution: Given ε > 0, there exists N1 such that for any n > N1 and any y ∈ E,
|f (y) − fn (y)| <
ε
.
2
Since f is continuous, there also exists N2 such that for all n > N2 ,
|f (xn ) − f (x)| <
ε
.
2
Now let N = max(N1 , N2 ). If n > N we then have
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| <
ε ε
+ = ε.
2 2
The conclusion does not hold assuming only pointwise convergence. Consider for instant fn (x) = xn
on [0, 1]. Then fn → f , pointwise, where
(
1, x = ‘
f (x) =
, 0, otherwise.
Now let xn = 1 − 1/n → 1. Then
fn (xn ) → e−1 6= f (1) = 1.
4
The converse is also not true. Consider the same function as above, fn (x) = xn , but on (0, 1). Then
fn → f = 0 pointwise but not uniformly. For any x ∈ (0, 1) and any sequence xn → x, we claim
that fn (xn ) → 0 = f (x). To see this, let x < δ < 1. Then since xn → x, for N large enough, xn < δ.
n→∞
but then fn (xn ) < δ n −−−−→ 0. This proves the claim. But since fn does not converge uniformly to
f , we see that the converse does not hold.
6. For n = 1, 2, · · · and x ∈ R, define
x
.
1 + nx2
Show that {fn } converges uniformly to a differentiable function f on R, and that the equation
fn (x) =
f 0 (x) = lim fn0 (x)
n→∞
is correct for all x 6= 0 but false at x = 0. Why does this not contradict the theorem on uniform
convergence and differentiation?
Solution: It√
is clear that the pointwise limit is 0. Now by completing squares it is easy to see that
1 + nx2 > 2x n, and so
1
x
< √ .
|fn (x)| =
1 + nx2
2 n
Given ε > 0 choosing N > 2ε2 , we see that the sequence converges to zero uniformly on R. Using
quotient rule,
(1 + nx2 ) − 2nx2
1 − nx2
fn0 (x) =
=
.
(1 + nx2 )2
(1 + nx2 )2
It is easy to see (by pulling out an n2 from the denominator and n from the numerator, or by using
L’Hospital’s rule) that if x 6= 0 then fn0 (x) → 0 = f 0 (x). If x = 0, then f 0 (0) = 0 but fn0 (0) = 1 for
all n, and so
lim fn0 (0) 6= f 0 (0).
n→∞
7. Let fn : [0, 1] → R be a sequence of bounded functions. That is for each n, there exists an Mn such that
|fn (t)| < Mn
u.c
for all t ∈ [0, 1]. Suppose in addition that fn −−→ f .
(a) Show that f is bounded on [0, 1]. Hint. Apply the definition of uniform convergence with ε = 1.
Solution: There exists an N such that
|f (t) − fN (t)| < 1
for all t ∈ [0, 1]. But then by triangle inequality,
|f (t)| ≤ |f (t) − fN (t)| + |fN (t)| < 1 + MN ,
and so f is bounded on [0, 1].
(b) Show that the sequence of functions is uniformly bounded. That is, show that there is an M such
that
|fn (t)| < M
5
for all t ∈ [0, 1] and all n. Hint. Show that for large n, fn can be bounded essentially by the bound
for f .
Solution: Let M0 = supt∈[0,1] |f (t)|. By the first part, we know that M0 < ∞. There exists
an N such that for all n ≥ N and all t ∈ [0, 1],
|fn (t) − f (t)| < 1.
BUt then for n ≥ N and t ∈ [0, 1],
|fn (t)| ≤ |fn (t) − f (t)| + |f (t)| < 1 + M0 .
Now let M = maxM0 + 1, M1 ., · · · , MN −1 . Then clearly for any fn and any t ∈ [0, 1],
|fn (t)| < M.
(c) Suppose additionally fn ∈ R[0, 1] for all n. Prove or disprove Z 1−1/n
Z 1
lim
fn (t) dt =
f (t) dt.
n→∞
Hint.
0
0
1−1/n
Z
1
Z
1/n
Z
fn (t) dt −
fn (t) dt =
0
fn (t) dt
0
0
Solution: As in the hint,
1−1/n
Z
1
Z
Z
1/n
fn (t) dt −
fn (t) dt =
0
0
fn (t) dt.
0
u.c
Since fn −−→ f ,
1−1/n
Z
lim
n→∞
1
Z
f (t) dt − lim
fn (t) dt =
0
Z
n→∞
0
1/n
fn (t) dt.
0
To compute the second term we estimate
Z
1/n
Z
fn (t) dt ≤
0
1/n
|fn (t)| dt ≤
0
M
→0
n
as n → ∞. This completes the proof.
8. Find the radius and interval of convergence of the following series.
P∞ (2x+1)n
1.
n=1
n
Solution: We can rewrite this as
∞
X
2n 1 n
x+
.
n
2
n=1
Written this way, the center is clearly a = −1/2. To find the radius of convergence we compute
2n 1/n
lim sup
= 2,
n
n→∞
6
and so R = 1/2. So the series definitely converges on (−1, 0). Next, we check boundary points.
• x = −1. The series is then
∞
X
(−1)n
n
n=1
which converges by the alternating series test.
• x = 0. The series is then the harmonic series which diverges.
So I = [−1, 0).
2.
2n n
n=1 n2 x .
P∞
Solution: Apply root test to conclude that radius of convergence is 1/2 and interval of convergence is [−1/2, 1/2].
3.
2n n
n=0 n! x .
P∞
Solution: Apply ratio test to conclude that radius of convergence is infinity and hence I = R.
9. Decide whether each proposition is true or false, providing a complete proof, or a counter example, as
appropriate.
P
(a) If
fn converges uniformly, then fn converges uniformly to zero.
Solution: True. Apply uniform Cauchy criteria for series with n = m + 1.
(b) If 0 ≤ fn (x) ≤ gn (x) and
P
gn converges uniformly, then
P
fn also converges uniformly.
Solution: True. Again uniform Cauchy criteria for series.
P
(c) If
fn converges
uniformly on E, then there exists constants Mn such that |fn (x)| < Mn for all
P
x ∈ E and
Mn converges.
Solution: Not necessarily true. Let fn : R → R be defined by
(
1
, x ∈ [n, n + 1)
fn (x) = n
0, otherwise
Then for any n > m and for any x ∈ R,
n
X
fk (x) <
k=m
1
,
m
and so the series satisfies
P −1uniform Cauchy criteria, and hence is uniformly convergent. But
supR |fn | = n−1 and
n diverges.
u.c
(d) If each fn is uniformly continuous on E and fn −−→ f on E, then f is also uniformly continuous on
E.
7
Solution: True. Given any ε > 0, let N ∈ N such that
sup |fN (x) − f (x)| <
E
ε
.
3
Since fN is uniformly continuous, there exists δ > 0 such that for any x, y ∈ E,
|x − y| < δ =⇒ |fN (x) − fN (y)| <
ε
.
3
So if x, y ∈ E with |x − y| < δ, then
|f (x) − f (y)| ≤ |f (x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − f (y)|
ε ε ε
< + + = ε.
3 3 3
So f is also uniformly continuous.
u.c
(e) If fn has a finite number of discontinuities on E and fn −−→ f , then f has a finite number of
discontinuities on E.
Solution: False. See example in 4(a).
(f) If fn has at most M number of discontinuities on E (where M is fixed and independent of n) and
u.c
fn −−→ f , then f has at most M number of discontinuities on E.
Solution: This is true. To prove, this suppose f has at least M +1 discontinuities p1 , · · · , pM +1 .
We first have the following observation, which follows from a simple counting argument.
Claim. There exists a k ∈ {1, 2, · · · , M + 1} such that for all N ∈ N, there exists n ≥ N such
that fn is continuous at pk .
Proof If not, then for all k ∈ {1, 2, · · · , M +1}, there exists an Nk ∈ N such that for all n ≥ Nk ,
fn is discontinuous at pk . Let N = max(N1 , · · · , NM +1 ). Then fN is discontinuous at each
p1 , p2 , · · · , pM +1 contradicting the hypothesis that fN can have at most M discontinuities.
Without loss of generality, let p1 be such that for all N ∈ N, there exists n ≥ N such that fn
is continuous at p1 . Then there exists a subsequence {fnj } such that each fnj is continuous at
p1 To see this, simply apply the claim to pick an n1 ≥ 1 such that fn1 is continuous at p1 , and
then having chosen n1 < n2 < · · · < nj−1 , apply claim with N = nj−1 , to obtain nj > nj−1
u.c
u.c
such that fnj is continuous at p1 . But then since fn −−→ f , we also must have fnj −−→ f . This
contradicts the theorem on uniform convergence and continuity, since each fnj is continuous at
p1 but f is discontinuous at p1 .
10. Define
g(x) =
∞
X
n=0
x2n
.
+1
x2n
Find the values of x where the series converges, and show that we get a continuous function on this set.
2n
x
Solution: Let fn (x) = 1+x
2n . If |x| ≥ 1, clearly limn→∞ fn (x) 6= 0. On the other hand, for any
x ∈ [−r, r] with r < 1, since 1 + x2n > 1, we have
|fn (x)| ≤ r2n .
8
P 2n
And since
r converges, by the Weierstrass M -test, the series converges uniformly on any subset
[−r, r] ⊂ (−1, 1). Since each function fn is continuous on (−1, 1), by the theorem on uniform
convergence and continuity, g(x) is defined and continuous on (−1, 1).
11. Let
h(x) =
∞
X
n=1
x2
1
.
+ n2
(a) Show that h is continuous on all of R.
Solution: For any x ∈ R,
x2
1
1
≤ 2,
2
+n
n
P −2
and since
n converges, by the Weierstrass M -test, the given series converges uniformly
to h(x) on all of R. Since each term is continuous on R and the convergence is uniform, the
infinite sum h(x) is also continuous on R.
(b) Is h differentiable? If so, is the derivative h0 continuous? Give complete proofs.
Solution: Let
sn =
n
X
k=1
x2
1
.
+ k2
We need to verify the two hypothesis in the theorem on uniform convergence and differentiation.
• We have already seen that sn (x) → h(x) on all of R.
• We need to check that s0n (x) converges uniformly on R. We calculate
s0n (x) = −
n
X
k=1
2x
.
(x2 + n2 )2
For any x ∈ R, it is easy to see that 2n|x| ≤ x2 + n2 , and so
2x
1
1
≤
≤ 3.
x2 + n2
n(x2 + n2 )
n
Since
P
n−3 is convergent, again by Weierstrass M -test, s0n (x) converges uniformly.
Hence by applying the theorem on uniform convergence of series and differentiation (Theorem
6.10 in Week-6 notes), we conclude that h is differentiable on R and
h0 (x) = −
∞
X
2x
.
2 + n2 )2
(x
n=1
Also, we have already seen that the series for h0 (x) converges uniformly on R, and so again by
the theorem on uniform convergence and continuity, h0 (x) is also continuous on R.
12. We saw in class that a function represented by a power series is automatically smooth, that is, it has
derivatives of all orders. The aim of this question to show that the converse might not be true. That is,
there exist smooth functions that cannot be represented by a power series.
9
(a) If P (x) is a polynomial, show that
2
lim P (x)e−x = 0.
x→∞
Hint. Use L’Hospital’s rule and an induction on the degree of the polynomial.
Solution: It is enough (Why?) to show that for any integer n ≥ 0,
lim tn e−t = 0.
t→∞
To prove this we use L’Hospital’s rule and induction. The base case n = 0 is trivial. Suppose we
have prove that the limit is zero for 1, 2, · · · n−1. To prove it for n, we notice that tn e−t = tn /et .
Since both numerator and denominator converge to infinity, we can use L’Hospital’s theorem.
So
ntn−1
tn−1
tn
= n lim
.
lim t = lim
t
t→∞
t→∞ et
t→∞ e
e
The final limit is zero because of our inductive hypothesis.
(b) Now define
(
2
e−1/t , t 6= 0
0, t = 0.
f (t) =
Show that f has derivatives of all orders at x = 0, and that f (n) (0) = 0 for all n = 1, 2, · · · . Can f
be represented by a power series in a neighborhood of t = 0?
Solution: Even though we have to find derivatives at t = 0, since they are higher order
derivatives, we are forced to compute f (n) (t) for t 6= 0. We have the following claim.
Claim. For any n, there is a polynomial Pn (x) such that for any t 6= 0,
f (n) (t) = Pn
1
t
2
e−1/t .
Proof. We proceed by induction. For n = 0, simply choose P0 (x) = 1. Having constructed,
P0 , P1 , · · · , Pn−1 , we construct Pn . By the inductive hypothesis, for t 6= 0,
f (n−1) (t) = Pn−1
1
t
2
e−1/t .
Differentiating by using chain and product rules,
1 −1 1 2 2
0
−1/t2
·
f (n) (t) = Pn−1
e
+
P
e−1/t
n−1
2
3
t
t
t
t
h
1 2 1 1 i
0
−1/t2
= Pn−1
− Pn−1
· 2 e
.
t
t3
t
t
So if we let
0
Pn (x) = 2Pn−1 (x)x3 − Pn−1
(x)x2 ,
then Pn (x) is a polynomial such that for all t 6= 0,
f (n) (t) = Pn
1
t
2
e−1/t .
Now, coming back to the problem, we compute f (n) (0) by induction. The base case n = 0 is
trivial since by definition f (0) = 0. Suppose now we have shown that f (n−1 (0) exists an is zero.
10
Then,
f (n−1) (h) − f ((n−1) (0)
h→0
h
1
2
1
e−1/h
(by claim)
= lim Pn−1
h→0 h
h
f (n) (0) = lim
= lim kPn−1 (k)e−k
2
k→∞
=0
where the final equality follows from part(a) and the observation that kPn−1 (k) is again a
polynomial in k. By definition then the Taylor of f centered at t = 0 is trivial, that is
Tf (t; 0) = 0
for all t. Since f (t) is never zero when t 6= 0, it follows that f (t) 6= Tf (t; 0) for all t 6= 0,
or in other words, even though f ∈ C ∞ (R), f is not represented by a power series in any
neighborhood of t = 0.
11
Periodicity of sine and cosine.
This exercise is unimportant from the point of view of doing well in the course, but highly recommended
for those with a wish to explore the non-trivial origins of one of the most trivial of high-school facts.
Recall that sin x, cos x : R → R are defined by the power series
sin x =
∞
X
(−1)n 2n+1
x
(2n + 1)!
n=0
∞
X
(−1)n 2n
cos x =
x .
(2n)!
n=0
In class, we saw that as a consequence, we have the following basic properties:
d cos x
d sin x
= cos x,
= sin x
dx
dx
sin(−x) = sin(x), cos(−x) = cos(x)
sin(x + y) = sin x cos y + cos x sin y
cos(x + y) = cos x cos y − sin x sin y
sin2 x + cos2 x = 1.
Our aim is to prove the following theorem.
Theorem. There exists a real number π > 0 such that sin(x + 2π) = sin(x) for all x ∈ R. Moreover, if
β ∈ R such that sin(x + β) = sin(x) for all x, then β = 2nπ for some n ∈ Z.
Proof.
(i) Show that if 0 ≤ x ≤
√
2, then for all n = 0, 1, 2, · · · ,
x4n+2
x4n
−
≥ 0.
(4n)! (4n + 2)!
√
Hence
√ show that cos x > 0 if x ∈ [0, 2]. In particular, this shows that there is no root of cos x in
[0, 2].
√
(ii) Show that cos 2 < − 13 , and hence show that there is at least one root of cos x in [ 2, 2].
(iii) Next, show that sin x ≥
x
3
when x ∈ [0, 2]. Use this to show that | sin x| ≥
|x|
3
for all x ∈ [−2, 2].
√
(iv) √
Use this and the addition formulas to conclude that cos x has a unique root in [ 2, 2]. Hint. If
2 ≤ x1 < x2 ≤ 2 were two roots, then show that sin(x2 − x1 ) would have to be zero. But this
should contradict the above inequality.
(v) Let this unique root be ζ and define π = 2ζ. Show that cos nπ = (−1)n for all integers n. In
particular, show that cos 2nπ = 1.
(vi) Hence, show that sin(x + 2π) = sin(x) and cos(x + 2π) = cos(x). It now remains to show that any
other period has to be an integer multiple of 2π.
(vii) Show that if sin(x + β) = sin x for all x, then sin(β/2) = 0 and hence cos β = 1.
(viii) To finish off the proof of the theorem, show that if cos β = 1, then β = 2nπ for some integer n. Hint.
Without loss of generality, let β > 0. There is a natural number n such that −π ≤ β − 2nπ < π.
Show that sin(β/2 − nπ) = 0 but that this contradicts the inequality in (iii) unless β = 2nπ.
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