ASYMPTOTIC BEHAVIOR AND BOUNDED VARIATIONS
MINGFENG ZHAO
May 23, 2011
Theorem 1. If f ∈ C 1 (R), and
R
R
(|f 0 (x)| + |f (x)|)dx < ∞,then
lim f (x) = 0.
|x|→∞
Proof. Since f 0 ∈ L1 (R), then for every x ∈ R, we have
Z
f (x) − f (0) =
x
f 0 (t)dt.
0
Since f 0 ∈ L1 (R), so
R∞
0
f 0 (t)dt < ∞, and
R −∞
0
f 0 (t)dt < ∞, in particular (by dominated conver-
gence theorem), we have f (∞), f (−∞) exist.
Let f (∞) = A, f (−∞) = B. Since f ∈ L1 (R), then there exists some R > 0 such that
Z
|f (x)|dx < 1.
|x|≥R
W.L.O.G, we assume A 6= 0, we can find some R0 > R > 0 such that |f (x)| ≥
A
2,
when x ≥ R0 , so
we have
Z
1≥
Z
|f (x)|dx
∞
≥
|f (x)|dx
R0
|x|≥R
Z
∞
≥
R0
A
dx = ∞.
2
Contradiction.
1
2
MINGFENG ZHAO
Example 1. Some one says that if f (x) → A for some A ∈ R, as x → ∞, then
lim f 0 (x) = 0.
x→∞
But this statement is wrong. Here I give a counterexample: Let
sin(x4 )
,
x
f (x) =
for all x ≥ 1.
Then we have
lim f (x)
x→∞
=
=
lim
x→∞
sin(x4 )
x
0.
But for the derivative, we have
f 0 (x)
=
cos(x4 ) · (4x3 ) · x − sin(x4 ) · 1
x2
=
4x4 cos(x4 ) − sin(x4 )
x2
=
4x2 cos(x4 ) −
sin(x4 )
.
x2
We know
sin(x4 )
= 0.
x→∞
x2
lim
But for 4x2 cos(x4 ),
lim 4x2 cos(x4 ) does not exists.
x→∞
1
In fact, take xk = (2kπ) 4 , we have
1
1
4x2k cos(x4k ) = 4(2kπ) 2 cos(2kπ) = 4(2kπ) 2 → ∞, as k → ∞.
And we can find that f (x) =
sin(x4 )
x
vibrates very rapidly at infinity.
ASYMPTOTIC BEHAVIOR AND BOUNDED VARIATIONS
Figure 1. f (x) =
3
sin(x4 )
x
Maybe you say that if f is an increasing function, then the statement is true. But also it is wrong.
Here is another example: Let f (x) be a function defined as below, for any k ≥ 1, for any k ≤ x ≤ k +1,
we define
f (x) =




 −k 2 2k+1 x − k −



 0,
1 k2k+1
+ k, if k ≤ x ≤ k +
if k +
1
k2k
Figure 2. f (x) on [k, k + 1]
1
k2k
< x ≤ k + 1.
4
MINGFENG ZHAO
For x ≤ 1, we let f (x) ≡ 0. If we let fk (x) = f (x) · 1[k,k+1] (x), then we have
f (x) =
∞
X
for all x ∈ R.
fk (x),
k=1
And also we know fk (x) ≥ 0 for all x ∈ R, and
Z
Z
fk (x)dx
k+1
=
fk (x)dx
R
k
=
1
1
·k· k
2
k2
=
1
2k+1
Hence, we have
Z
f (x)dx
=
R
Z X
∞
fk (x)dx
R k=1
=
∞ Z
X
k=1
=
∞
X
k=1
=
fk (x)dx
By the Monotone Convergence theorem
R
1
2k+1
1
.
2
That means that f ∈ L1 (R). Now we can define
Z
x
g(x) =
f (t)dt,
for all x ∈ R.
−∞
As x → ∞, we know
Z
∞
lim g(x) =
x→∞
f (t)dt =
−∞
1
.
2
But for the derivative of g, since f is continuous, by the Fundamental Theorem of Calculus, we
have
g 0 (x) = f (x) ≥ 0.
ASYMPTOTIC BEHAVIOR AND BOUNDED VARIATIONS
When take xk = k +
1
,
k2k+1
we know
g 0 (xk ) = f (xk ) = k → ∞,
as k → ∞.
Figure 3. g(x)
Theorem 2. Let f ∈ C 1 (R), and for some A ∈ R, we have
lim f (x) = A.
x→∞
Then there exists some sequence {xk }∞
k=1 such that xk → ∞, as k → ∞, and
lim f 0 (xk ) = 0.
k→∞
This implies that, if limx→∞ f 0 (x) exists, then
lim f 0 (x) = 0.
x→∞
In particular, if f 00 (x) ≤ 0 for all x ∈ R, or f 00 (x) ≥ 0, and kf 0 kL∞ (R) < ∞, then we can get
lim f 0 (x) = 0.
x→∞
5
6
MINGFENG ZHAO
Proof. For any k ≥ 1, since f ∈ C 1 (R), by the Lagrange’s Intermediate Value Theorem, we know
there exists some xk ∈ (k, k + 1) such that
f (k + 1) − f (k) = f 0 (xk ).
Since limx→∞ f (x) = A, then we have
lim f 0 (xk )
k→∞
=
lim [f (k + 1) − f (k)]
k→∞
=
A−A
=
0.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu