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Unformatted text preview: a positive integer n such that
p log p n.
Since x log x n for x sufficiently large x, we can use the Bolzano intermediate value theorem to choose
a number a p such that a log a n. Now since
a log a a log a log a
a log a
n
2
2
we can use the Bolzano intermediate value theorem again to choose a number b
a, a such that
b log b n .
2
We now observe that
1
f a f b  sin n sin n
2
and so the proof is complete.
5. a. A function f is said to be Lipschitzian on a set S if there exists a number k such that the inequality
f t f x  kt x 
holds for all numbers t and x in S. Prove that every Lipschitzian function is uniformly continuous.
Suppose that f is a function defined on a set S, that k is a positive number and that the
inequality
f t f x  kt x 
holds for all numbers t and x in the set S. Suppose that 0. We define /k and observe
that, whenever t and x belong to S and t x  we have
f t
b. Given that f x x for all x fx  kt x k k . 0, 1 prove that f is uniformly continuous but not lipschitzian on 0, 1 . Solution: The fact that f is uniformly continuous on the closed bounded set 0, 1 follows at once
from the fact that f is continuous there. Now, to prove that f fails to be Lipschitzian, suppose that k is
any positive number. Given x
0, 1 we see that
f x f 0 
1
x 0 
x
and this exceeds k whenever x 1/k 2 . 6. a. Suppose that f is uniformly continuous on a set S, that x n is a sequence in the set S and that x n has a
partial limit x R. Prove that it is impossible to have f x n
Ý as n Ý.
Using the fact that f is uniformly continuous on S we choose a number 0 such that the
inequality
f t f s  1
holds whenever t and s belong to S and t s  . Since x is a partial limit of the sequence x n
we know that the condition
xn
x ,x
2
2
holds for infinitely many positive integers n. Choose a positive integer N such that
xN
x ,x .
2
2
For every one of the infinitely many positive integers n for which the condition
x ,x
xn
2
2
holds, since 202 x n xN  x  x x n x N  ,
2
2 we have
1 f xN .
f xn
Therefore the sequence of numbers f x n is frequently in the interval
cannot approach Ý.
Of course, this sequence cannot approach Ý either. Ý, 1 f x N and so it b. Did you assume that x S in Part a? If you did, go back and do the exercise again. You have no
information that x S. If you didn’t assume x S, you can sit this question out.
c. Suppose that f is uniformly continuous on a bounded set S and that x n is a sequence in S. Prove that it
is impossible to have f x n
Ý as n Ý.
The assertion follows at once from Part a and the fact that every bounded sequence of
numbers has a partial limit in R.
d. Prove that if f is uniformly continuous on a bounded set S then the function f is bounded. Solution: To obtain a contradiction, assume that f is unbounded above. Choose a sequence xn in S such that f x n n for each n. Now use part c...
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 Fall '08
 STAFF
 Math, Calculus

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