How bad can it get in terms of a continuous function not having a derivative at some points?
It turns out it can be the case the function is nowhere differentiable but everywhere
continuous. An example of such a pathological function different than the one I am about to
present was discovered by Weierstrass in 1872. Before showing this, here is a simple
observation.
Lemma 7.4.1Suppose f^{′}
(x)
exists and let c be a number. Then letting g
(x)
≡ f
(cx)
,
g′(x) = cf′(cx).
Here the derivative refers to either the derivative, the left derivative, or the right derivative.Also, if f
(x)
= a + bx, then
′
f (x) = b
where again, f^{′}refers to either the left derivative, right derivative or derivative. Furthermore,in the case where f
(x )
= a + bx,
f (x+ h)− f (x) = bh.
Proof: It is known from the definition that
′
f (x+ h)− f (x )− f (x) h = o(h)
Therefore,
g(x+ h)− g(x) = f (c(x+ h))− f (cx) = f ′(cx)ch+ o (ch)
and so
g (x + h)− g(x)− cf′(cx )h = o(ch) = o (h )
and so this proves the first part of the lemma. Now consider the last claim.
f (x+ h)− f (x) = a + b(x+ h)− (a+ bx) = bh
= bh + 0 = bh +o (h ).
Thus f^{′}
(x)
= b. ■
Now consider the following description of a function. The following is the graph of the
function on
[0,1]
.
PICT
The height of the function is 1/2 and the slope of the rising line is 1 while the slope of the
falling line is −1. Now extend this function to the whole real line to make it periodic of period
1. This means f
(x+ n)
= f
(x)
for all x ∈ ℝ and n ∈ ℤ, the integers. In other words to find
the graph of f on
[1,2]
you simply slide the graph of f on
[0,1]
a distance of 1 to get the
same tent shaped thing on
[1,2]
. Continue this way. The following picture illustrates
what a piece of the graph of this function looks like. Some might call it an infinite
sawtooth.
PICT
Now define
∑∞ ( 3)k ( k )
g(x) ≡ 4 f 4 x .
k=0
Letting M_{k} =
(3∕4)
^{−k}, an application of the Weierstrass M test shows g is everywhere
continuous. This is because each function in the sum is continuous and the series converges
uniformly on ℝ.
Let δ_{m} = ±
14
(4−m )
where we assume m > 2. That of interest will be m →∞.
( ( ))
f (4k(x+ δm))− f (4kx) = f 4k x ± 1(4−m ) − f (4kx)
4
( ◜int◞eg◟er◝)
| k 1 k− m| (k )
= f|(4 x± 4 4 |) − f 4 x = 0
Therefore,
m ( )
g(x+-δm-)−-g(x) 1-∑ 3 k( ( k ) ( k ))
δm = δm 4 f 4 (x + δm ) − f 4 x
k=0
The absolute value of the last term in the sum is
|( )m |
|| 3 (f (4m (x+ δ ))− f (4mx ))||
| 4 m |
and we choose the sign of δ_{m} such that both 4^{m}
(x+ δm)
and 4^{m}x are in some interval
[k∕2,
(k+ 1)
∕2) which is certainly possible because the distance between these two points is
1/4 and such half open intervals include all of ℝ. Thus, since f has slope ±1 on the interval
just mentioned,
|( )m | ( )m
|| 3 (f (4m (x+ δ ))− f (4mx ))||= 3 4m |δ | = 3m |δ |
| 4 m | 4 m m
As to the other terms, 0 ≤ f
(x)
≤ 1∕2 and so
| ( ) | ( ) ( )
||m∑−1 3 k( ( k ) ( k ))|| m∑−1 3 k 1-−-(3∕4)m- 3 m
|| 4 f 4 (x+ δm ) − f 4 x || ≤ 4 = 1∕4 = 4 − 4 4
k=0 k=0