Real Analysis 509differ fromf(x)by at most 2 m^1 + 1. We conclude again that asn→∞,f(xn)→f(x).
This proves the continuity off.
Let us show next thatfdoes not have a finite derivative from the left at any point
x∈( 0 , 1 ]. For suchxconsider the ternary expansionx= 0 .a 1 a 2 a 3 ...that has infinitely
many nozero digits, and, applying the definition off forthisexpansion, letf(x)=
0 .b 1 b 2 b 3 ....Now consider an arbitrary positive numbern, and letkn≥nbe such that
akn =0. Construct a numberx′∈( 0 , 1 )whose firstkn−1 digits are the same as those
ofx, whoseknth digit is zero, and all of whose other digits are equal to 0 ifbkn+ 1 = 1
and to 1 ifbkn+ 1 =0. Then
0 <x−x′< 2 · 3 −kn+ 0. (^00) ︸ ︷︷... (^0) ︸
kn
22 ..., 0 = 3 −kn+^1 ,
while in the first case,
|f(x)−f(x′)|≥ 0. (^00) ︸ ︷︷... (^0) ︸
kn
bkn+ 1 = 0. (^00) ︸ ︷︷... (^0) ︸
kn
1 ,
and in the second case,
|f(x)−f(x′)|≥ 0. (^00) ︸ ︷︷... (^0) ︸
kn
11 ... 1 − 0. (^00) ︸ ︷︷... (^0) ︸
kn
0 bkn+ 2 ...,
and these are both greater than or equal to 2−kn−^1. Sincekn≥n, we have 0<x−x′<
3 −n+^1 and
∣
∣∣
∣
f(x)−f(x′)
x−x′
∣
∣∣
∣>
2 −kn−^1
3 −kn+^1=
1
6
(
3
2
)kn
≥1
6
(
3
2
)n
.Lettingn→∞, we obtain
x′→x, while∣∣
∣
∣
f(x)−f(x′)
x−x′∣∣
∣
∣→∞.
This proves thatfdoes not have a derivative on the left atx. The argument thatfdoes
not have a derivative on the right atxis similar and is left to the reader.
Remark.S. Banach has shown that in some sense, there are far more continuous functions
that are not differentiable at any point than continuous functions that are differentiable at
least at some point.
394.We apply the intermediate value property to the functiong :[a, b]→[a, b],
g(x)=f(x)−x. Becausef(a)≥aandf(b)≤b, it follows thatg(a)≤0 and
g(b)≥0. Hence there isc∈[a, b]such thatg(c)=0. Thiscis a fixed point off.