Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Homework #5
Solutions
1. Let α and c be real numbers, c > 0, and f is defined on [−1, 1] by
α
x sin(xc ) if x 6= 0
f (x) =
0
if x = 0.
Find all the values of the parameters α and c such that:
(a) f is continuous.
(b) f 0 (0) exists.
(c) f 0 is continuous.
(d) f 00 (0) exists.
Solution: First, notice that the only “problematic” point is 0. Everywhere else, the function
is differentiable infinitely many times. Thus:
(a) Since whenever limx→0 f (x) exists we have
sin(xc ) = lim xα+c ,
lim f (x) = lim xα+c ·
x→0
x→0
x→0
xc
then limx→0 f (x) = 0 if and only if α > −c.
(b) We have:
f (x)
sin(xc ) = lim xα−1+c ·
= lim xα−1+c ,
x→0 x
x→0
x→0
xc
0
and hence f (0) exists if and only if α ≥ 1 − c. In fact,
0 if α > 1 − c
0
f (0) =
1 if α = 1 − c.
f 0 (0) = lim
(1)
(c) Away from zero we have:
0
f (x) = αx
α−1
c
sin(x ) + cx
α+c−1
c
cos(x ) = x
α−1+c
sin(xc )
c
α·
+ c · cos(x ) ,
xc
(2)
and hence
0
lim f (x) =
x→0
0
if α > 1 − c
α + c = 1 if α = 1 − c and c ≥ 0.
It then follows from (1) that f 0 is continuous if and only if either α > 1 − c or α = 1 − c
and c ≥ 0.
1
(d) It follows from (1) and (2) that when α > 1 − c,
00
f (x) = x
α−2+c
sin(xc )
c
α·
+ c · cos(x )
xc
exists if and only if α ≥ 2 − c. When α = 1 − c and c ≥ 0,
i
1 h sin(xc )
c
α·
−
1
+
c
cos(x
)
−
1)
x→0 x
xc
α c
2c−1
= (use the L’hospital rule)
lim x
− −
x→0
6 2
f 00 (x) = lim
exists if and only if c ≥ 1/2 (note that
α
6
+
c
2
=
1−c
6
+
c
2
=
1+2c
6
6= 0).
2. Let a ∈ R be a given real number. Suppose that a function f : R → R satisfies
f 0 (x) < 0 < f 00 (x)
if x < a,
f 0 (x) > 0 > f 00 (x)
if x > a.
and
Prove that f 0 (a) does not exists.
Hint: To see informally the reason why f 0 (a) doesn’t exist, draw the graph of, for example,
(x − a)2
if x < a,
f (x) =
1
1 − 1+x−a if x ≥ a.
The monotonicity of f (x) on (a, ∞) and (−∞, a) dictates f 0 (a) = 0 if the derivative exists.
But f 0 (a) = 0 means that a is a local maximum of f, and hence f 00 (0) < 0. This is impossible
since f 00 is decreasing and positive on x < a. To make a formal proof from this argument,
(a)
use the mean value theorem f (x)−f
= f 0 (yx ) and the fact that the derivative is monotone
x−a
on each side of a, to deduce that if f 0 (a) exists then
f 0 (a) = lim f 0 (yx ) = sup f 0 (x) = inf f 0 (x) = 0.
x→a
x>a
x<a
Use then the mean value theorem for the derivative
assertion f 0 (a) = 0 contradicts the assumptions.
f 0 (x)−f 0 (a)
x−a
Solution: Since
0 < f 00 (x)
if x < a,
f 0 (x) is increasing for x < a. Since
0 > f 00 (x)
if x > a,
2
= f 00 (zx ) to check that the
f 0 (x) is decreasing for x > a.
To get a contradiction, assume that f 0 (a) exists and is equal to b. Then
f (x) − f (a)
.
x→∞
x−a
b = lim
By the mean value theorem, for any x 6= 0, we have
f (x) − f (a)
= f 0 (yx ),
x−a
for some yx within the interval containing a and x as its endpoints. Since f 0 (x) is monotone
and bounded on the both half-lines x > a as well as x < a, this implies that
f 0 (a) = 0 = sup f 0 (x) = inf f 0 (x).
x>a
x<a
But then, the mean value theorem implies that for any x > a there exists zx ∈ (a, x) such
that
f 00 (zx ) =
f 0 (x) − f 0 (a)
f 0 (x)
=
> 0,
x−a
x−a
in contradiction to the assumption that f 00 (z) < 0 for any z > a. The calculation formally
confirms the heuristic suggested by the picture (see the Hint) that the problem with the
existence of the derivative is caused by the assumption that f is concave for x > a.
3.
(a) Let g(y) = arcsin y 3 . Compute g 0 (y) for y ∈ (0, 1).
(b) Suppose that f : R → R is differentiable and that there exists n ∈ N such that
f (tx) = tn f (x)
for any t > 0 and x ∈ R. Show that xf 0 (x) = nf (x) for all x ∈ R.
Solution:
(a) g(y) = arcsin y 3 and hence by the definition of the arcsin function, g(y) ∈ − π2 , π2 . In
calculations below we occassioanllly use the fact that this imples cos g(y) ≥ 0.
First method: Use a table of derivatives or the inverse function theorem to verify that
(arcsin x)0 = √
1
.
1 − x2
The chain rule therefore implies:
3 0
1
3 0
3y 2
(arcsin y ) = (y ) · p
=p
.
1 − (y 3 )2
1 − y6
3
Second method: Observe that sin g(y) = y 3 . Differentiate this identity to obtain
g 0 (y) cos g(y) = 3y 2 .
Conclude that
g 0 (y) =
3y 2
3y 2
3y 2
p
=p
=
.
cos g(y)
1 − y6.
1 − sin2 g(y)
Third method: First, observe that sin g(y) = y 3 and hence
that
p
3
sin g(y) = y. This implies
g(y) = f −1 (y),
where f (x) =
√
3
sin x. Next, by the chain rule,
p
1
1
f 0 (x) = (sin x)−2/3 cos x = (sin x)−2/3 1 − sin2 x.
3
3
√
Thus, if f (x) = y = 3 sin x, then
f 0 (x) =
p
p
1
1
sin x)−2/3 1 − sin2 x = y −2 1 − y 6 .
3
3
It follows from the inverse function theorem, that
0
g 0 (y) = f −1 (y) =
1
3y 2
p
=
.
f 0 (x)
1 − y6
(b) Differentiate both sides of the identity f (tx) = tn f (x) with respect to t. This yields:
xf 0 (tx) = ntn−1 f 0 (x).
Now plug in t = 1 into the last identity.
4. Solve Exercise 80 in Chapter 2 of the textbook.
Solution:
(a) Suppose that the graph Gf = {(p, y) ∈ M × R : y = f p} is a union of two clopen
subsets of M × R. Say,
H = {(p, y) ∈ A × R : y = f p}
and
F = {(p, y) ∈ B × R : y = f p},
4
where A and B are two disjoint subsets of M whose union is entire M. We will next
show that if f is continuous, then A and B are clopen in M, and hence M is a
disconnected set in contrary to the assumption. To this end, observe that if pn ∈ A
and limn→∞ pn = p ∈ M, then, since f is continuous, limn→∞ f (pn ) = f (p). Since
(pn , f pn ) ∈ H and H is a closed set, (p, f p) ∈ H. Hence p ∈ A. Thus A is a closed set.
Similarly, B is closed. In particular, no point of A is a cluster point of B. Thus all
points of A are its interior points, and hence A is open.
(b) The result is actually true if M ⊂ R. Therefore, to construct a counterexample, one
has to use higher-dimensional domains. Consider, for instance, f : R2 → R defined as
follows:
y if x ≥ 0 and y ≥ 0,
f (x, y) =
0 otherwise.
The graph of this function is connected, but the function is discontinuous, for example,
at (0, 1). Indeed, f (0, 1) = 1 but f (−ε, 1) = 0 for any ε > 0. Notice that two parts of
the graph (corresponding, respectively, to the arguments in the first quadrant and the
rest of R2 ) meet at the origin, and hence the graph is connected.
We remark in passing that it is not hard to show that if f : M → R is continuous and
Gf is connected, then M must be connected.
(c) Suppose that M is path-connected. Take any two points p, q ∈ M. Let h : [a, b] → M
be a path connecting p ∈ M and q ∈ M. That is, h is continuous, p = h(a), and
q = f (b). Define g : [a, b] → M by setting g(x) = f h(x) and F : [a, b] → M × R by
setting
F (x) = h(x), g(x) .
Then F : [a, b] → M × R is a path which connects (p, f p) and (q, f q) in the graph of
the functions f. Since (p, f p) and (q, f q) are arbitrary points in the graph, the graph
is path-connected.
(d) Recall that a function F (x) = F1 (x), F2 (x) : [a, b] → M × R is continuous if and only
if both its components F1 : [a, b] → M and F2 : [a, b] → R are continuous. Thus, if F is
a path joining (p, f p) and (q, f q) in the graph of a continuous function f : M → R, then
F1 : [a, b] → R is a path joining p and q in M. Therefore, M is a path-connected set if
the graph of f is path-connected. However, f doesn’t have to be continuous if both M
and Gf are path-connected. To show this, one may exploit the same counterexample
as in part (b).
5. Consider a function f : R → R. Suppose that
(i) f is continuous for x ≥ 0.
(ii) f 0 (x) exists for x > 0.
(iii) f (0) = 0.
5
(iv) f 0 is monotonically increasing.
Let
g(x) =
f (x)
,
x
x > 0.
Prove that g is monotonically increasing.
Solution: We have
g 0 (x) =
xf 0 (x) − f (x)
.
x2
Hence, it suffices to show that
xf 0 (x) − f (x) ≥ 0
for all x > 0.
(3)
To this end, observe that by the mean value, for any x ∈ R there exists yx ∈ (0, x) such that
f 0 (yx ) =
f (x)
f (x) − f (0)
=
x−0
x
Hence
xf 0 (x) − f (x) = x f 0 (x) − f 0 (yx ) ,
and (3) follows from the monotonicity of the derivative (condition (iv) in the statement of
the problem).
6. Let X be the set of all bounded real-valued functions on a non-empty set S.
(a) For x, y ∈ X set d(x, y) = supt∈S |x(t) − y(t)|. Show that d is a metric on X.
(b) For x ∈ X, let
f (x) = inf x(t)
t∈S
and
g(x) = sup x(t).
t∈S
Show that f and g are uniformly continuous functions from (X, d) to R.
Solution:
(a) Notice that, since functions in X are bounded, d(x, y) < +∞ for any x, y ∈ X.
Furthermore, d(x, y) ≥ 0 and d(x, y) = 0 if and only if x(t) = y(t) for all t ∈ S.
Clearly, d(x, y) = d(y, x). Finally, for any t ∈ S,
|x(t) − z(t)| ≤ |x(t) − y(t)| + |y(t) − z(t)| ≤ d(x, y) + d(y, z),
and hence
d(x, z) = sup |x(t) − z(t)| ≤ d(x, y) + d(y, z).
t∈S
6
(b) First, observe that for any x, y ∈ X we have
inf x(t) = inf x(t) − y(t) +y(t) ≥ inf [x(t) − y(t)] + inf y(t).
t∈S
t∈S
t∈S
t∈S
(4)
Indeed, for any s ∈ S,
inf [x(t) − y(t)] + inf y(t) ≤ [x(s) − y(s)] + y(s) = x(s),
t∈S
t∈S
showing that inf t∈S [x(t) − y(t)] + inf t∈S y(t) is a lower bound for {x(t) : t ∈ S}.
It follows from (4) that
f (x) − f (y) = inf x(t) − inf y(t) ≥ inf [x(t) − y(t)]
t∈S
t∈S
t∈S
= − sup[y(t) − x(t)] ≥ − sup |y(t) − x(t)|.
t∈S
t∈S
Similarly,
f (y) − f (x) ≥ − sup |y(t) − x(t)|.
t∈S
Combining these two inequalities, we obtain
|f (y) − f (x)| ≤ sup |y(t) − x(t)| = d(x, y).
t∈S
Thus, for any ε > 0, |f (y) − f (x)| ≤ ε whenever d(x, y) ≤ ε.
Since g(x) = −f (−x), the claim about g follows directly from its counterpart for f.
7