Download MA 504 1. Suppose f is a positive continuous function

MIDTERM 2 PRACTICE PROBLEMS - MA 504
PAULINHO TCHATCHATCHA
1. Suppose f is a positive continuous function on Rn such that lim|x|→∞ f (x) = 0 [i.e.,
for all > 0, there is an N so that |f (x)| < form all x with |x| > N ]. Show that f is
uniformly continuous on Rn .
Solution.
First we see that by Theorem 4.19, f is uniformly continuous on BN = {x ∈ Rn : |x| ≤ N },
for all N > 0. So let > 0 be given and, since lim|x|→∞ f (x) = 0, there exists N > 0 such
that |f (x)| < if |x| ≥ N . Since f is uniformly continuous on BN , there exists δ > 0 such
that
|f (x) − f (y)| < if |x − y| < δ and x, y ∈ BN .
We see that if |x| > N , then either Bδ (x) ∩ BN = ∅ or there exists y ∈ Bδ (x) with |y| < N,
where Bδ (x) = {z ∈ Rn : |x − z| < δ}. In the first case, if z ∈ Bδ (x), then |z| > N, so
|f (x) − f (z)| ≤ |f (x)| + |f (z)| < + = 2.
In the second case, then let y ∈ Bδ (x) with |y| < N, and let x0 ∈ Bδ(x) with |x0 | = N. We
have then
|f (x) − f (y)| ≤ |f (x) − f (x0 )| + |f (x0 ) − f (y)| ≤ |f (x)| + |f (x0 )| + |f (x0 ) − f (y)| < 3,
note that |x0 − y| < δ.
Therefore
|f (x) − f (y)| < 3,
whenever |x − y| < δ. Hence f is uniformly continuous.
2. Let f be a continuous function on R to R which does not take on any of its values
twice. It it true that f must either be strictly increasing (in the sense that if x0 < x00 then
f (x0 ) < f (x00 )) or strictly decreasing?
Solution.
Answer: SIM! (YES!).
Let x ∈ R. Since f does not take on any of its values twice, then if y > x either f (x) < f (y) or
f (x) > f (y). Assume without loss of generality that f (x) < f (y). Then if z ∈ R, x < z < y,
we have that f (z) > f (x), because if f (z) < f (x), then since f (y) > f (x) > f (z), by the
intermidate value theorem there exists w, z < w < y such that f (w) = f (x), a contradiction
with the fact that f does not take on any of its values twice. Similarly f (z) < f (y), because
of f (z) > f (y), then f (x) < f (y) < f (z) so there would be a w, x < w < z such that
f (w) = f (y), a contradiction. Hence f is strictly increasing.
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Note that if f is not continuous, then the result is no longer valid. For instance, consider
f (x) = x if x 6= 1, 2 and f (1) = 2, f (2) = 1. Then f is neither strictly increasing, nor
strictly decreasing.
∞
3. Suppose that {sn }∞
n=1 and {tn }n=1 are sequences of positive real numbers such that
sn
sn
lim inf
> 0 and lim sup
< ∞.
n→∞ tn
tn
n→∞
Prove that there is a constant M > 0 such that
1
tn ≤ sn ≤ M tn for all n = 1, 2, 3, ...
M
Solution.
Since
sn
lim sup
< ∞,
tn
n→∞
there exists M0 > 0 such that
sn
lim sup
< M0 .
tn
n→∞
sn
By theorem 3.17, there exists N0 such that n ≥ N0 implies
< M0 . Let
tn
sN 0
s1
, ...,
M1 = max
, M0 .
t1
tN0
Then
sn
< M0 ,
tn
for all n = 1, 2, 3, ....
Similarly, since
lim inf
n→∞
sn
>0
tn
there exists M2 > 0 such that
sn
1
>
,
tn
M2
for all n = 1, 2, 3, ....
Now let M = max{M1 , M2 }. We have
sn
1
1
1
M>
>
≥
⇒
tn ≤ sn ≤ M tn for all n = 1, 2, 3, ...
tn
M2
M
M
4. What is the radius of convergence of the power series
∞
∞
X
X
(3n)!n! n
(a)
(2 + (−1)n )n z n (b)
x
(4n)!
n=0
n=0
(c)
∞
X
n=1
x
nn
(d)
∞
X
n=0
n2 xn
MIDTERM 2 PRACTICE PROBLEMS - MA 504
3
Solution.
(a) We have
p
n
(2 + (−1)n )n = lim sup 2 + (−1)n = 3,
1
so by theorem 3.39, the radius of convergence is R = .
3
(b) We have
lim sup
(3n+3)!(n+1)!
(4n+4)!
(3n)!n!
(4n)!
=
(3n + 3)! (n + 1)! (4n)!
(3n + 3)(3n + 2)(3n + 1)(n + 1)
=
,
(3n)!
n! (4n + 4)!
(4n + 4)(4n + 3)(4n + 2)(4n + 1)
so
(3n + 3)(3n + 2)(3n + 1)(n + 1)
33
= 4,
n→∞ (4n + 4)(4n + 3)(4n + 2)(4n + 1)
4
4
4
and the radius of convergence is R = 3 .
3
(c) We have
√
n
n−1
n
xnn = xn /n = xn
lim
so by the root test, we see that the series converges if |x| < 1 and diverges if |x| ≥ 1, ie, the
radius of convergence is R = 1.
(d) We have
(n + 1)2
= 1,
lim
n→∞
n2
hence the radius of convergence is R = 1.
5. Let
∞
X
an be an absolutely convergent series of numbers and {bn }∞
n=1 a bounded se-
n=1
quence. Show that
∞
X
∞
X
an bn is also convergent. Can we still have the same conclusion if
n=1
an is convergent, but not absolutely convergent?
n=1
Solution.
∞
X
Since
an is an absolutely convergent series of numbers,
n=1
∞
X
|an | < ∞.
n=1
Since {bn }∞
n=1 is a bounded sequence, there exists a M > 0 such that |bn | < M for every n.
Hence
∞
∞
∞
X
X
X
an b n ≤
|an ||bn | ≤ M
|an | < ∞,
n=1
n=1
n=1
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PAULINHO TCHATCHATCHA
so it converges.
Now consider
∞
X
(−1)n
n=1
n
.
It is easy to see that the series above converges, but it is not absolutely convergent. Let
bn = (−1)n . Then
∞
∞
X
X
1
(−1)n
(−1)n =
does not converges.
n
n
n=1
n=1
6. Assume f has a finite derivative in (a, b), and is continuous on [a, b] with f (a) = f (b) = 0.
prove that for every λ ∈ R there exists c ∈ (a, b) such that f 0 (c) = λf (c).
Solution.
By the mean value theorem there exists s ∈ (a, b) such that
0 = f (b) − f (a) = (b − a)f 0 (s) ⇒ f 0 (s) = 0.
So if f (s) = 0, f 0 (s) = 0 = λf (s) for every λ ∈ R. Otherwise f (s) 6= 0. Assume without loss
of generality that f (s) > 0, if not, consider −f instead of f .
Let λ ∈ R be given.
Since f is continuous at s, there exits a neighborhood of s, say (s − δ, s + δ) ⊂ (a, b), δ > 0,
such that f is positive on it, ie, f (x) > 0 for every x ∈ (s − δ, s + δ).
We see that there exist a1 , b1 ∈ (s − δ, s + δ) such that a1 < b1 and f (a1 ) = f (b1 ). Indeed,
otherwise by the previous problem 2, f would be strictly increasing or strictly decreasing,
but then it is not hard to show that f 0 (x) > 0 for all x ∈ (s − δ, s + δ) or f 0 (x) < 0 for all
x ∈ (s − δ, s + δ). This is a contradiction with f 0 (s) = 0.
Then we can consider the function h defined by
h(x) = log f (x) + λx.
By the generalized Mean Value Theorem (Theorem 5.9) with g(x) = λx, there exists c ∈
(a1 , b1 ) ⊂ (a, b) such that
[h(b1 ) − h(a1 )]g 0 (c) = [g(b1 ) − g(a1 )]h0 (c) ⇒ λ(b1 − a1 )λ = λ(b1 − a1 )
f 0 (c)
f (c)
⇒ f 0 (c) = λf (c).
7. Let f : R → R be a uniformly continuous function. Show that there exist constants A
and B such that
|f (x)| ≤ A + B|x|, for all x ∈ R.
Solution.
Since f is uniformly continuous, there exists δ > 0 such that
|f (x) − f (y)| < 1,
MIDTERM 2 PRACTICE PROBLEMS - MA 504
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whenever |x − y| < δ.
Let x ∈ R. Assume x 6= 0. Let n ∈ Z be a integer such that nδ ≤ x < (n + 1)δ. We have
f (x) − f (0) = f (x) − f (nδ) +
n
X
(f (jδ) − f ((j − 1)δ))
j=1
⇒ |f (x)| ≤ |f (0)| + |f (x) − f (nδ)| +
n
X
|f (jδ) − f ((j − 1)δ)|
j=1
⇒ |f (x)| ≤ |f (0)| + δ + |n|δ
|f (0)|
δ(|n| + 1)
δ(|n| + 1)
1
|f (x)|
⇒
≤
+
≤ |f (0)| +
≤ |f (0)| + 1 +
.
1 + |x|
1 + |x|
1 + |x|
|n|δ
|n|
|f (x)|
Then
≤ |f (0)| + 2, so
1 + |x|
|f (x)| ≤ |f (0)| + 2 + (|f (0)| + 2)|x|.
8. Show that a monotone function f : [a, b] → R is continuous if and only if its image
f ([a, b]) is an interval.
Solution.
Suppose that f is continuous. Then by theorem 4.22 f ([a, b]) is connected, since [a, b] is
connected. By Theorem 2.47, f ([a, b]) has the property that if x ∈ f ([a, b]), y ∈ f ([a, b]),
and x < z < y, then z ∈ f ([a, b]), ie, f ([a, b]) is a line segment. Since f ([a, b]) is compact,
because [a, b] is compact and f is continuous, f ([a, b]) is an interval.
Conversely assume that f ([a, b]) is an interval. Assume that f is not continuous. Then there
exists x ∈ [a, b] such that f is not continuous at x. Since f is monotone, the discontinuities
are of first type, ie, both limits
lim f (t),
t→x−
lim f (t)
t→x+
exist, but they are different. If x = a, then we only consider the right limit, and if x = b we
only consider the left limit.
Assume x ∈ (a, b) and assume without loss of generality that f is monotone increasing,
otherwise consider −f instead of f . Then
lim f (t) < lim+ f (t).
t→x−
t→x
Let > 0 be such that
lim f (t) < lim+ f (t) − .
t→x−
t→x
Since f is monotone increasing
f (y) ≤ lim− f (t),
for y < x,
f (y) ≥ lim+ f (t),
for y ≥ x.
t→x
and
t→x
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PAULINHO TCHATCHATCHA
But there exits no z ∈ [a, b] such that
f (a) ≤ lim− f (t) < f (z) = lim+ f (t) − < lim+ f (t) ≤ f (b),
t→x
t→x
a contradiction with f ([a, b]) being a interval.
One can argue in a similar way if x = a, or x = b.
Therefore f is continuous.
t→x