Unformatted text preview: on f n is continuous on S then the
family f n n 1, 2, 3, is equicontinuous on S.
Suppose that x S and that 0. Choose an integer N such that the inequality
sup|f n f |
holds whenever n N. Using the fact that the function f N is continuous at the number x, choose a
number 1 0 such that the inequality 368 fN x | |f N t 5
holds whenever t S and |t x | 1 . Now we use the fact that there are only finitely many positive
integers n N to choose a number 2 0 such that the inequality
|f n t f n x |
holds whenever t S and |t x | 2 and 1 n N. Finally, we define to be the smaller of the two
numbers 1 and 2 . Now given any member t of S satisfying the condition |t x | , the condition
|f n t f n x |
is already known for 1 n N and if n N then we have
|f n t f n x | |f n t f t f t f N t f N t f N x f N x f x f x f n x |
|f n t f t | |f t f N t | |f N t f N x | |f N x f x | |f x fn x | . 4. Invent a meaning for equi-uniform continuity of a family on a set S and decide whether or not your
definition provides an analogue of the preceding exercise.
A family of functions on a set S is said to be equi-uniformly continuous on S if for every 0
there exists a number 0 such that whenever t and x belong to S and |t x | and f
|f t f x | .
The point of this exercise is to invite the student to show that if f n is a uniformly convergent
sequence of uniformly continuous functions on a set S then the family f n n 1, 2, 3, is
equi-uniformly continuous on S. The proof is similar to the one used in Exercise 3. Some Exercises on Bounded Convergence
1. Prove that the sequence f n in the bounded convergence theorem for derivatives actually converges
boundedly to the function f.
Using the fact that the sequence f n converges boundedly, choose a number p such that the
|f n x | p
holds for every positive integer n and every x
a, b . Using the fact that the sequence f n c ,
being convergent, is bounded, choose a number q such that
|f n c | q
for every positive integer n.
Now given any positive integer n and any member x of a, b we have
|f n x | f n c Þ f n t dt
x c |f n c | Þ c f n t dt
x qp b a. 2. Suppose that a n is a strictly increasing sequence of positive integers and that
f n x sin a n x sin a n1 x 2
for every positive integer n and every number x.
a. Work out the integral
2 Þ0 f n x dx and deduce that there must be at least one number x
converge to zero.
For each n and x we have 369 0, 2 such that the sequence f n x does not |f n x | sin a n x sin a n1 x 2 4. If the sequence f n converged pointwise (and therefore boundedly) on the interval 0, 2 to
the constant function 0 then the bounded convergence theorem would give us
lim nÝ 2 Þ0 f n x dx 2 Þ0 0 0. However,
2 Þ0 2 Þ0 f n x dx sin a n1 x 2 dx 2 sin a n x for each n and we conclude that the sequence f n cannot converge pointwise to 0.
b. Prove that there must be at least one number x...
View Full Document