1549901369-Elements_of_Real_Analysis__Denlinger_

9.4 *Two Results of Weierstrass 571

. ~ cos(l0k7rx)
Proof. Define the function f(x) = L.., fk(x), where fk(x) =
2

k.

k=l

Claim #1: f is continuous everywhere.

Proof: Use the Weierstrass M-test and Corollary 9.3.7.

Claim #2: f is differentiable nowhere.

Proof: Let x E R We shall prove that f is not differentiable at x by using

the sequential criterion. We shall produce a sequence { Xn} such that Xn --+ x

b u t { f(xn)-f(x)} d" iverges.
Xn -X
First, represent x using a "decimal":

(^00) dk
x = N + O.d1 d2d3 ... dk ... = N + L -k '
k=l 10
where N is an integer and \:/k E N, dk E {O, 1, 2, 3, · · · , 9}.
Part 1: Let n be a fixed positive integer, and let
Then an ::; x ::; bn.
1
an= N + O.d1d2d3 · · · dn and bn =an+ -
0
1 n.
From lf(an) - f(bn)I = lfn(an) - fn(bn) + L [fk(an) - fk(bn)]I show
kf=.n
that
Show that fork;::::: n , lfk(an) - fk(bn)I = 0, and hence,
n-1
IJ(an) - f(bn)I;::::: lfn(an) - fn(bn)I - L lfk(an) - fk(bn)I. (16)
k=l
Also show that lfn(an) - fn(bn)I =
2
n
1
_ 1. Apply this to Inequality (16) to get
(17)
Show that, for 1 ::; k ::; n, the mean value theorem guarantees that
10k7r 1 5k-n7r
lfk(an) - fk(bn)I::; 2k · 10n = ~-
Apply this to Inequality (17) to get
1 n-1 5k-n7r 1 7r 1 1
IJ(an) - J(bn)I;::::: 2 n-l - k~l ~ > 2n-l - 2n · 4 > 2n ·