A Proof of Corollary 5.1.7 (a) Starting from equation (7), use the fact that | f ( t ) | is bounded on [ 0 , M ] [ since f ( t ) is piecewise continuous] to show that if s > max ( a , 0 ) , then | F ( s ) | ≤ max 0 ≤ t ≤ M | f ( t ) | ∫ 0 M e − s t d t + K ∫ M ∞ e − ( s − a ) t d t = max 0 ≤ t ≤ M | f ( t ) | 1 − e − s M s + K s − a e − ( s − a ) M . (b) Argue that there is a constant K 1 such that K e − ( s − a ) M s − a < K 1 s for s sufficiently large. (c) Conclude that K e − ( s − a ) M s − a < K 1 s for s sufficiently large, | F ( s ) | ≤ L / s , where L = max 0 ≤ t ≤ M | f ( t ) | + K 1 .
A Proof of Corollary 5.1.7 (a) Starting from equation (7), use the fact that | f ( t ) | is bounded on [ 0 , M ] [ since f ( t ) is piecewise continuous] to show that if s > max ( a , 0 ) , then | F ( s ) | ≤ max 0 ≤ t ≤ M | f ( t ) | ∫ 0 M e − s t d t + K ∫ M ∞ e − ( s − a ) t d t = max 0 ≤ t ≤ M | f ( t ) | 1 − e − s M s + K s − a e − ( s − a ) M . (b) Argue that there is a constant K 1 such that K e − ( s − a ) M s − a < K 1 s for s sufficiently large. (c) Conclude that K e − ( s − a ) M s − a < K 1 s for s sufficiently large, | F ( s ) | ≤ L / s , where L = max 0 ≤ t ≤ M | f ( t ) | + K 1 .
(a) Starting from equation (7), use the fact that
|
f
(
t
)
|
is bounded on
[
0
,
M
]
[ since
f
(
t
)
is piecewise continuous] to show that if
s
>
max
(
a
,
0
)
, then
|
F
(
s
)
|
≤
max
0
≤
t
≤
M
|
f
(
t
)
|
∫
0
M
e
−
s
t
d
t
+
K
∫
M
∞
e
−
(
s
−
a
)
t
d
t
=
max
0
≤
t
≤
M
|
f
(
t
)
|
1
−
e
−
s
M
s
+
K
s
−
a
e
−
(
s
−
a
)
M
.
(b) Argue that there is a constant
K
1
such that
K
e
−
(
s
−
a
)
M
s
−
a
<
K
1
s
for
s
sufficiently large.
(c) Conclude that
K
e
−
(
s
−
a
)
M
s
−
a
<
K
1
s
for
s
sufficiently large,
|
F
(
s
)
|
≤
L
/
s
,
Q1
Theorem 18.1 states that "Let f be a continuous real-valued function on a closed interval [a, b]. Then f is a
bounded function.". Read the proof of Theorem 18.1 with [a, b] replaced by (a, b). Where does it break down?
Explain.
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