SOLUTION OF HW9
MINGFENG ZHAO
April 28, 2013
1. [5 Points] Define
x+ =




 x
if x ≥ 0



 0
if x < 0.
Prove that f (x) = xk+ is continuously differentiable if k is an integer greater than one.
Proof. It is easy to see that f (x) = xk+ is continuously differentiable on R\{0}. When k ≥ 2 and
k ∈ N, we know that f (x) = xk for all x > 0, so f 0 (x) = kxk−1 for all x > 0. Since f (0) = 0 and
k > 1, then
lim
x→0+
xk
f (x) − f (0)
= lim
= 0.
x→0+ x
x−0
Also we know that f (x) = 0 for all x ≤ 0, then f (x) = 0 for all x < 0 and f 0 (0−) = 0. Hence
we know that f 0 (0) exists and f 0 (0) = 0. In summary, we know that f (x) = xk+ is continuously
differentiable if k is an integer greater than one.
2. [5 Points] Given a closed set A, construct a continuously differentiable function that has A as its
set of zeros.
Proof. Since A is closed, then R\A is open, which implies that there exists countablely many ai <
T
bi ∈ R with i ≥ 1 such that (ai , bi ) (ai+1 , bi+1 ) = ∅ for all i ≥ 1 and
R\A =
∞
[
(ai , bi ).
i=1
1
2
MINGFENG ZHAO
For each i ≥ 1, let fi (x) = (x − ai )2+ (x − bi )2+ , then it is easy to see that fi (x) = 0 if and only if
x∈
/ (ai , bi ). By the result of Problem 1, we know that fi is continuously differentiable on R. Consider
the function
f (x) =
∞
X
fi (x),
∀x ∈ R.
i=1
For any x ∈ R, then either x ∈ A or x ∈ (ai , bi ) for some unique i, which implies that either
f (x) = 0 or f (x) = fi (x), and f is continuously differentiable on R. On the other hand, if f (x) = 0,
since fi ≥ 0 on R, then fi (x) = 0 for all i ≥ 1, which implies that x ∈
/ (ai , bi ) for all i ≥ 1, that is,
x ∈ A. In summary, we know that f is continuously differentiable on R and f (x) = 0 if and only if
x ∈ A.
3. [5 Points] Prove that if f is any continuous one-to-one function on an interval (a, b), then either f
is strictly increasing or strictly decreasing.
Proof. Assume that f is neither strictly increasing nor strictly decreasing, then we have two cases:
Case I: there exist some x1 , x2 , x3 ∈ (a, b) such that x1 < x2 < x3 , f (x1 ) ≤ f (x2 ) and f (x2 ) ≥ f (x3 ).
Since f is continuous, then there exists some y1 ∈ (x1 , x2 ) and y2 ∈ (x2 , x3 ) such that f (y1 ) = f (y2 ),
which contradicts with the assumption that f is one to one.
Case II: there exist some x1 , x2 , x3 ∈ (a, b) such that x1 < x2 < x3 , f (x1 ) ≥ f (x2 ) and f (x2 ) ≤
f (x3 ). Since f is continuous, then there exist some y1 ∈ (x1 , x2 ) and y2 ∈ (x2 , x3 ) such that f (y1 ) =
f (y2 ), which contradicts with the assumption that f is one to one.
In summary, we know that either f is strictly increasing or strictly decreasing.
4. [5 Points] Show that a polynomial of even order (6= 0) has either a global maximum or a global
minimum but not both.
SOLUTION OF HW9
3
Proof. Assume that f (x) = ax2n + · · · with a 6= 0. If a > 0, then we know that lim
|x|→∞
f (x) = ∞,
which implies that f has a global minimum in R, but is unbounded above. If a < 0, then we know
that lim
|x|→∞
f (x) = −∞, which implies that f has a global maximum in R, but is unbounded below.
5. [5 Points] Let fn denote the nth iterate of f , f1 = f , f2 (x) = f (f1 (x)), · · · , fn (x) = f (fn−1 (x)).
Express fn0 in terms of f 0 . Show that if 0 ≤ a ≤ |f 0 (x)| ≤ b for all x, then an ≤ |fn0 (x)| ≤ bn .
Proof. By the definition of fn , we know that
f10 (x)
=
f 0 (x)
f20 (x)
=
[f (f1 (x))]0 = f 0 (f1 (x))f10 (x) = f 0 (f1 (x))f 0 (x)
f30 (x)
=
[f (f2 (x))]0 = f 0 (f2 (x))f20 (x) = f 0 (f2 (x))f 0 (f1 (x))f 0 (x).
By induction, we know that for all n ≥ 2, we have
fn0 (x) = f 0 (x) ·
n−1
Y
f 0 (fi (x)).
i=1
Since 0 ≤ a ≤ |f 0 (x)| ≤ b for all x, then
|fn0 (x)|
n−1
Y
0
0
f (fi (x))
= f (x) ·
i=1
0
= |f (x)| ·
n−1
Y
|f 0 (fi (x))|
i=1
≥ an
|fn0 (x)|
n−1
Y
0
0
= f (x) ·
f (fi (x))
i=1
= |f 0 (x)| ·
n−1
Y
|f 0 (fi (x))|
i=1
≤ bn .
4
MINGFENG ZHAO
6. [5 Points] Suppose f is a C 2 function on an interval (a, b) and the graph of f lies above every secant
line. Prove that f 00 (x) ≤ 0 on the interval.
Proof. By the assumption, we know that for any x, y, z ∈ (a, b) with x < y < z, we have
f (y) ≥
f (z) − f (x)
(y − x) + f (x)
z−x
That is, we have
f (z) − f (x)
f (y) − f (x)
≥
,
y−x
z−x
∀x < y < z.
Now for any small h > 0, let x = y − h and z = y + h, we have
f (y) − f (y − h)
f (y + h) − f (y − h)
≥
.
h
2h
That is, we have
f (y + h) + f (y − h) − 2f (y)
≤ 0.
2h2
By taking h → 0 and L’Hospital rule, we know that f 00 (y) ≤ 0.
7. [5 Points] Suppose f is C 1 on an interval and f 0 satisfies the Holder condition of order α, |f 0 (x) −
f 0 (y)| ≤ M |x − y|α for all x and y in the interval, where α is a fixed value, 0 < α ≤ 1. Show that
|∆2h f (x)| ≤ c|h|1+α . How does the constant c relate to the constant M ?
Proof. By the definition of ∆2h f , we know that
∆2h f (x)
=
∆h f (x + h) − ∆h f (x)
= f (x + 2h) − f (x + h) − f (x + h) + f (x)
= f (x + 2h) − f (x + h) − [f (x + h) − f (x)]
Z
1
Z
1
df (x + h + th) −
=
0
Z
=
0
df (x + th)
0
1
f 0 (x + th + h)h dt −
Z
0
1
f 0 (x + th)h dt
SOLUTION OF HW9
Z
5
1
[f 0 (x + h + th) − f 0 (x + th)]h dt.
=
0
So we get
|∆2h f (x)|
Z
≤
1
|f 0 (x + h + th) − f 0 (x + th)| dt
|h|
0
Z
≤
|h|
1
M |x + h + th − x − th|α dt
0
= |h|1+α M.
So we can just take c = M .
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu