Graphics Reference
In-Depth Information
39 Can you besurethat a function which is uniformly continuous on
A
is necessarily continuous at each point of
A
?
In qn 37, wehaveproved that thefunction
x
x
is
uniformly
continuous on [
10, 10]. In qn 38, wehaveproved that thefunction
x
1/
x is not
uniformly continuous on (0, 1).
40 By considering
x
n
1/
n
and
y
n
, and taking
1, provethat
the function defined by
f
(
x
)
x
is not
uniformly continuous on R.
Notice the effect of changing the domain from question 37.
41 By considering
x
1/2
n
and
y
1/(2
n
,
prove that the function defined by
f
(
x
)
sin(1/
x
)
is not
uniformly
continuous on theintrval (0, 1). This shows that continuous
functions (see qn 6.48) may fail to be uniformly continuous even
when they are bounded.
)
, and taking
42 If a function
f
:
A
R satisfies a
Lipschitz condition
, namely that
there is a constant real number
L
such that
f
(
x
)
f
(
y
)
L
ยท
x
y
for any
x
,
y
A
, show that
f
is uniformly continuous on
A
,by
finding an appropriate
. Verify that the function of
qn 37, satisfies a Lipschitz condition with
L
20.
for a given
After studying the Mean Value Theorem in chapter 9 it will be
clear that any function with bounded derivatives necessarily satisfies a
Lipschitz condition, with an upper bound on the absolute value of the
derivatives as the Lipschitz constant.
43 We seek to prove that a continuous function
f
:[
a
,
b
]
is
necessarily uniformly continuous, and we do so by supposing that
it is not and establishing a contradiction.
When a function
is not
uniformly continuous, it means that, for
some
R
implies that
f
(
x
)
f
(
y
)
. That is, for each
there are points
x
and
y
in the
domain such that
x
y
but
f
(
x
)
f
(
y
)
.
Supposethat
f is not
uniformly continuous and construct two
sequences (
x
0, no
can befound for which
x
y
) and (
y
) for which the difference
x
y
1/
n
but
for which
f
(
x
)
f
(
y
)
.
(i) Is there any reason why the sequence (
x
) should be
convergent?
(ii) Is the sequence (
x
) bounded?
