1549901369-Elements_of_Real_Analysis__Denlinger_

558 Chapter 9 • Sequences and Series of Functions

(b) Let S = [O, l] and fn(x) = xn. As we saw in Example 9.1.7 (a), {fn}

converges pomtw1se.. on [O, l] to the function. f(x) = {^0 if. -^0 < x < 1 }. Here,

1, 1f x = 1

lim (lim fn(x)) = lim (lim xn) = 1, whereas lim ( lim fn(x)) =

n-+oo X--tO n-+oo x-+O x-+O n-+oo

lim f(x) = 0. Thus, the answer to the second part of Q#l is "no." 0

x--+0

Example 9.3.2

In Example 9.3.l (a) above, each function fn is continuous on [O, l] but the

limit function f is not. Hence, the answer to Q#2 is "no."

Examples 9.3.3

(a) In Example 9.1.7 (d), each function fn is integrable over [O, l], since it

differs from the constant function 0 at only finitely many points (See Theorem

7.4.9). But the limit function f is not integrable over [O, l], as shown in Example

7.2. 10. Hence, the answer to the first part of Q#3 is "no."

The graph of a typical f n is shown in Figure
9.10. It is clear that the pointwise limit of {fn}
on [O, l] is f(x) = 0. In this case, f 0
1
fn = 1, but
f 0
1
f = O; thus, f 0
1
f f n ft f 0
1
f. Therefore, the an-

swer to the second part of Q#3 is "no." 0

Examples 9.3.4

y

2n

I

ll

Figure 9.10

1 x

(a) In Example 9.1.7 (c), each function fn is differentiable at 0 but the limit
function is not. Hence, the answer to the first part of Q#4 is "no." But notice
that in this case the convergence is uniform, so that even uniform convergence
will not guarantee the differentiability of the limit function.

(b) Let S = [O, l] and define fn: [O, l]---+ JR

{

x + .!. if - 1 < x < _ l.. }
by fn(x) = 0 n if - ~-< x-< f;' ·

x - .!. n if .!. n < - x < - 1