Problem set 5, Real Analysis I, Spring, 2015.
(5) Consider the function on R defined by
f (x) =
1
|x|(log 1/|x|)2
0
if |x| ≤ 1/2,
otherwise.
(a) Verify that f is integrable.
Solution: Compute since f is even,
Z
|f | =
Z
1/2
−1/2
R
= 2
Z
= 2
Z
1
dx
|x|(log 1/|x|)2
1/2
1
dx
|x|(log 1/|x|)2
1/2
1
dx
x(log x)2
0
0
Z
− log 2
1
du
u2
−∞
− log 2
1 = 2 − u
= 2
(u = log x, du =
1
dx)
x
−∞
1
2
2
+ 2 lim
=
<∞
=
p→−∞ p
log 2
log 2
So f is integrable.
(b) Establish the inequality
f ∗ (x) ≥
c
|x| log 1/|x|
for some c > 0 and all |x| ≤ 1/2,
to conclude that the maximal function f ∗ is not locally
integrable.
1
2
Solution: Compute for x ∈ (0, 1/2]
∗
f (x) =
≥
=
=
=
Z
1
sup
|f (y)| dy
(I an interval)
I3x |I| I
Z
1 x
|f (y)| dy
(choose I = [0, x])
x 0
Z
1 log x 1
du
(u = log y, as above)
x −∞ u2
1
1
· (−
)
x
log x
1
x log 1/x
Since x is even, a similar computation holds for x ∈ [− 12 , 0).
It is also straightforward to check f ∗ (0) = ∞, as the computation above shows f ∗ (0) ≥ x log1 1/x for all x ∈ (0, 1/2].
This shows
f ∗ (x) ≥
1
,
|x| log 1/|x|
for all |x| ≤ 1/2.
To show f ∗ is not locally integrable, compute
Z
1/2
∗
f (x) dx ≥
0
Z
1
2
0
Z
1
2
1
dx
x log 1/x
1
dx
x log x
0
Z − log 2
1
=
− du
u
−∞
− log 2
= (− log |u|)−∞
=
−
(u = log x, du =
1
dx)
x
= − log log 2 + lim log p = ∞.
p→∞
Thus f ∗ is not locally integrable.
(12) Consider the function F (x) = x2 sin(1/x2 ), x 6= 0, with F (0) =
0. Show that F 0 (x) exists for every x, but F 0 is not integrable
on [−1, 1].
3
Solution: For x 6= 0, it is clear that F 0 (x) exists. For x = 0,
compute
F (h) − F (0)
h→0
h
2
h sin(1/h2 )
= lim
h→0
h
= lim h sin(1/h2 ) = 0.
F 0 (0) = lim
h→0
The last equality is by the squeeze theorem, as −|h| ≤ h sin(1/h2 ) ≤
|h|.
To show F 0 is not integrable, compute for x 6= 0
F 0 (x) = 2x sin(1/x2 ) − (2/x) cos(1/x2 ).
Now 2x sin(1/x2 ) is clearly integrable over [−1, 1]. Therefore,
F 0 is integrable if and only if the second term is. Compute
Z 1
Z 1
2
|(2/x) cos(1/x )| dx = 4
(1/x) | cos(1/x2 )| dx
−1
0
1
= 4
Z
= 2
Z∞∞
| cos u| [−1/(2u) du]
(1/u)| cos u| du
1
for the substitution u = 1/x2 . Now estimate | cos u| ≥ 1/2 if
u ∈ Ik = [kπ − π/3, kπ + π/3] for k a positive integer. For
u ∈ Ik , 1/u ≥ 1/(kπ + π/3). Therefore,
Z ∞
∞
X
1
1
(1/u)| cos u| du ≥
· .
kπ + π/3 2
1
k=1
This sum
the limit comparison test to the
P∞ is 1infinite by using
0
series k=1 k = ∞. So F is not integrable on [−1, 1].
(32) Let f : R → R. Prove that if f satisfies the Lipschitz condition
|f (x) − f (y)| ≤ M |x − y|
for some M and all x, y ∈ R, if and only if f satisfies the
following two properties
(i) f is absolutely continuous.
(ii) |f 0 (x)| ≤ M for a.e. x.
Hint to show that if f is Lipschitz with constant M , then
|f 0 (x)| ≤ M for almost every x: First prove that f is absolutely continuous, and conclude that f is differentiable almost
4
everywhere. Then apply the definition of the derivative and the
Lipschitz inequality to show that |f 0 (x)| ≤ M at each x where
f is differentiable.
Solution: First assume f satisfies the Lipschitz condition. To
prove f is absolutely continuous, let > 0 and δ = /M . Then
if {In = (an , bn )}N
n=1 are a disjoint set of intervals in R with
N
X
(bn − an ) < δ,
n=1
then compute using the Lipschitz condition
N
X
n=1
|f (bn ) − f (an )| ≤
N
X
M (bn − an ) = M
n=1
N
X
(bn − an ) < M δ = .
n=1
Thus f is absolutely continuous.
To prove (ii) assuming f is Lipschitz, note since f is absolutely continuous, it is differentiable almost everywhere. If x is
a point where f 0 (x) exists, compute
f (x) − f (y)
.
y→x
x−y
The Lipschitz condition then implies for y 6= x, the difference
quotient
f (x) − f (y)
∈ [−M, M ].
x−y
Therefore, the limit f 0 (x) ∈ [−M, M ] as well. This shows
|f 0 (x)| ≤ M for a.e. x.
Now assume (i) and (ii). Since f is absolutely continuous, f 0
exists almost everywhere and for a < b
Z b
f (b) − f (a) =
f 0 (x) dx
f 0 (x) = lim
a
Condition (ii) then implies
Z b
Z b
|f (b) − f (a)| = f 0 (x) dx ≤
|f 0 (x)| dx ≤ M (b − a).
a
a
This is the Lipschitz condition for a < b. The remaining cases
a = b and a > b follow easily.
(10) For f : R → R which is continuous with continuous derivative,
define kf kC 1 = kf kC 0 + kf 0 kC 0 . Let C 1 (R) be the set of all such
functions f so that kf kC 1 < ∞. Show that C 1 (R) is a Banach
space:
5
(a) Show that k · kC 1 is a norm.
Solution: Let α ∈ R and f ∈ C 1 (R). Then
kαf kC 1 =
=
=
=
kαf kC 0 + k(αf )0 kC 0
|α| · kf kC 0 + kαf 0 kC 0
|α| · kf kC 0 + |α| · kf 0 kC 0
|α| · kf kC 1 .
Now let f, g ∈ C 1 (R). Then compute by the Triangle Inequality for C 0
kf + gkC 1 = kf + gkC 0 + k(f + g)0 kC 0
≤ kf kC 0 + kgkC 0 + kf 0 kC 0 + kg 0 kC 0
= kf kC 1 + kgkC 1
Finally, assume kf kC 1 = 0. Then kf kC 0 + kf 0 kC 0 = 0, and
so f and f 0 are both identically 0. Clearly f = 0 in C 1 (R).
Thus k · kC 1 is a norm.
(b) Let {fn } be a Cauchy sequence in C 1 (R). Show that both
{fn } and {fn0 } are Cauchy sequences in C 0 (R).
Solution: For all , there is an N so that if n, m ≥ N ,
then
kfn − fm kC 0 + k(fn − fm )0 kC 0 = kfn − fm kC 1 < .
0
This shows kfn − fm kC 0 and kfn0 − fm
kC 0 are both < , and
0
so {fn } and {fn } are Cauchy sequences in C 0 (R).
(c) Conclude that there are uniform limits f, g ∈ C 0 (R) of fn
and fn0 respectively.
Solution: Since C 0 (R) is complete, then by part (b), fn
and fn0 are convergent to limits f, g respectively in C 0 (R).
This is the topology of uniform
Rx
R x 0 convergence.
(d) For all x ∈ R, show that 0 fn (y) dy → 0 g(y) dy as n →
∞.
Solution: Since g ∈ C 0 (R), |g(y)| ≤ M = kgkC 0 for all
y ∈ R. We know fn0 → g uniformly, and so there is an N so
that if n ≥ N , then kfn0 − gkC 0 = supy∈R |fn0 (y) − g(y)| < 1.
For these n ≥ N , then
sup |fn0 (y)| ≤ sup |fn0 (y) − g(y)| + sup |g(y)| ≤ 1 + M.
y∈R
y∈R
y∈R
So on each interval [0, x] (or [x, 0] for x < 0), and n ≥ N ,
the functions |fn0 | are uniformly bounded by M + 1, which
has integral (M + 1)|x| < ∞. Also fn0 → g everywhere,
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and the Dominated Convergence Theorem
R x (or the Bounded
Rx
Convergence Theorem) applies to show 0 fn0 (y) dy → 0 g(y) dy.
(e) Show that f 0 = g everywhere and that fn → f in C 1 (R).
Solution: The fundamental theorem of calculus applies to
the result of part (d) to show that
fn (x) − fn (0) → G(x) − G(0),
where G is an antiderivative of g. But since fn → f in
C 0 , we also know fn (x) − fn (0) → f (x) − f (0). So f (x) =
G(x) + C, where C = f (0) − G(0) is a constant. Since
G is an antiderivative of g, we see f 0 (x) = G0 (x) = g(x).
Therefore
kfn − f kC 1 = kfn − f kC 0 + kfn0 − f 0 kC 0 → 0
as n → ∞, and C 1 (R) is a Banach space.
(11) For continuous f : R → R, define
kf kC 0,1 = kf kC 0 + sup
x6=y
|f (x) − f (y)|
.
|x − y|
Define C 0,1 (R) to be the set of continuous f so that kf kC 0,1 <
∞.
(a) For f ∈ C 0,1 (R), show that kf kC 0,1 = kf kC 0 + kf 0 kL∞ .
Hint: use 32(ii).
Solution: Problem 32 says that
(1)
|f (x) − f (y)| ≤ M |x − y|
if and only if f is absolutely continuous and |f 0 (x)| ≤ M
for almost all x. So the quantity
|f (x) − f (y)|
sup
|x − y|
x6=y
is bounded by M , and in fact is the infimum of all M satisfying (1). Similarly, the L∞ norm kf 0 kL∞ is the infimum
of upper bounds for |f 0 (x)|. Problem 32 then says we are
taking the infima over the same set, and so
|f (x) − f (y)|
sup
= kf 0 kL∞
|x − y|
x6=y
for f ∈ C 0,1 (R). This shows kf kC 0,1 = kf kC 0 + kf 0 kL∞ .
(b) Repeat and modify the steps in problem (10) to show that
C 0,1 (R) is a Banach space.
Solution: Let {fn } be a Cauchy sequence in C 0,1 (R).
Then part (a) and a simple computation as in 10(b) shows
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that {fn } is a Cauchy sequence in C 0 (R), and {fn0 } is a
Cauchy sequence in L∞ (R). Since C 0 (R) and L∞ (R) are
both Banach spaces, there are limits fn → f in C 0 (R) and
fn0 → g in L∞ (R).
We would like to show fn → f in C 0,1 (R), which involves
two parts. First, we need to verify that f 0 = g almost everywhere, as by part (a), this will verify that kfn −f kC 0,1 →
0. Second, we need to verify that f ∈ C 0,1 (R).
To show f 0 = g almost everywhere, we note that by Problem 32, each fn is absolutely continuous, and so
Z x
fn (x) − fn (0) =
fn0 (y) dy.
0
Now let n → ∞. Since fn → f in C 0 , fn (x) → f (x) and
fn (0) → f (0). Similarly, since fn0 → g in L∞ , we may use
the Bounded Convergence Theorem as in 10(d) above to
show that
Z x
Z x
0
lim
fn (y) dy =
g(y) dy.
n→∞
0
0
This shows
f (x) − f (0) =
Z
x
g(y) dy.
0
Since g ∈ L∞ (R), it is locally integrable, and thus f is
absolutely continuous and we may apply the Fundamental
Theorem of Calculus to conclude that f 0 (x) = g(x) for
almost every x.
Finally, to verify that f ∈ C 0,1 (R), we know since f ∈
C 0 (R) that kf kC 0 < ∞. The remaining term is taken care
of by Problem 32: We know that f is both absolutely continuous and that its derivative f 0 = g is uniformly bounded
almost everywhere (since g ∈ L∞ ). So Problem 32 shows
that f is Lipschitz, and so f ∈ C 0,1 (R).
(c) Show that the identity map is an isometry from C 1 (R) to
C 0,1 (R). Hint: Use the corresponding fact about L∞ and
C 0.
Solution: For f ∈ C 1 (R), by part (a) and the fact that
the L∞ norm is the same as the C 0 norm for continuous
functions,
kf kC 0,1 = kf kC 0 + kf 0 kL∞ = kf kC 0 + kf 0 kC 0 = kf kC 1 .
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(d) Find, with proof, a function in C 0,1 (R) − C 1 (R).
Solution: Consider for example

 0 for x < 0,
x for 0 ≤ x ≤ 1,
f (x) =
 1 for x > 1.
Then kf kC 0 = 1, and it is easy to check that for x 6= y,
|f (x) − f (y)|
|x − y|
is < 1 if at least one of x, y ∈
/ [0, 1]. If x, y ∈ [0, 1], then
this quantity is 1. This shows kf kC 0,1 = 1 + 1 = 2, and so
f ∈ C 0,1 (R). On the other hand, f is not differentiable at
x = 0, 1, and so f ∈
/ C 1 (R).