698 Appendix C • Answers & Hints for Selected Exercises
( <=) Suppose the hypotheses. Let c: > 0. Then 3no E N 3 m, n ?: no =?
f fk is bounded and II f fkll < c:, so llSm -Snll < c:. By Thm 9.2.7,
k=m+l k=m+l
{Sn} converges uniformly on S.
co- Suppose L ak(x - c)k has radius of convergence p > 0. Let 0 < r < p.
k=O co co
By Thm 8.6.2, L ak(x - c)k converges absolutely in [c - r, c + r], so L akrk
k=O k=O
co
converges absolutely. On [c - r,c + r], llak(x - c)kil ~ laklrk, and I: laklrk
co k=O
converges. :. By Weierstrass M-test, L ak(x - c)k converges uniformly on
k=O
[c -r, c + r].
- (a) Let fk(x) = (-l)kxk and gk(x) = l/k. First we prove that \I 0 ~ x ~ 1,
n
0 < L fk < 1, using math induction. The case n = 1 is trivial. To prove the
k=l n n
general induction step, assume 0 < L fk < 1. Then 0 < 1 - L fk < 1,
k=l k=l
i.e., 0 < 1 - f (-l)kxk < 1, so 0 < x [1- f (-l)kxk] < 1, so 0 < x +
k=l k=l
n n+l n+l
L (-l)k+^2 xk+l < l; i.e., 0 < L (-l)k+lxk+l < l ; i.e., 0 < L fk(x) < 1.
k=l k=l k=l
co
By Dirichlet's test (9.2.16), L fk9k converges uniformly on [O, l]. But the
k=l
convergence is not absolute at x = 1 since L fc diverges.
- (a), (d), and (e) converge uniformly on the indicated sets.
EXERCISE SET 9.3
n co- Let 0 < a < 1. Let Sn = L xk. Since the radius of convergence of L xk
k=O k=O
is 1, Sn ---+ f(x) = l~x uniformly on [O, a], by Thm. 9.2.15. For contradiction,
n
suppose Sn---+ f uniformly on [O, 1). Now, \In EN, lim Sn(x) = lim L xk =
x--.1- x--.1-k=O
n
L lim xk = n + 1. Thus, each Sn has a finite limit as x ---+ 1-, but the limit
k=O x--.1-
function f does not. This would contradict Thm. 9.3.5. - Let fn(x) = l/n if x is rational, 0 if x is irrational. Then, just like the
Dirichlet function, each fn is discontinuous everywhere, but llfnll = l/n---+ 0,
so fn---+ 0 uniformly on (-00,00).