2.5. COMPACTNESS AND CONTINUOUS FUNCTIONS 59

The latter part of this theorem is called the Weierstrass M test. As a very interestingapplication, consider the question of nowhere differentiable functions. This considers thesimple case of functions having values in the Banach space R.

Consider the following description of a function. The following is the graph of thefunction on [0,1] .

1The 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 ofperiod 1. This means f (x+n) = f (x) for all x ∈ R and n ∈ Z, the integers. In other wordsto find the graph of f on [1,2] you simply slide the graph of f on [0,1] a distance of 1 to getthe same tent shaped thing on [1,2] . Continue this way. The following picture illustrateswhat a piece of the graph of this function looks like. Some might call it an infinite sawtooth.

Now define g(x) ≡ ∑∞k=0( 3

4

)kf(4kx). Letting Mk = (3/4)k , an application of the

Weierstrass M test, Theorem 2.5.42 shows g is everywhere continuous. This is becauseeach function in the sum is continuous and the series converges uniformly on R. However,this function is nowhere differentiable. This is shown next.

Let δ m =± 14 (4

−m) where we assume m > 2. That of interest will be m→ ∞.

g(x+δ m)−g(x)δ m

=∑

∞k=0( 3

4

)k (f(4k (x+δ m)

)− f

(4kx))

δ m

If you take k > m,

f(

4k (x+δ m))− f

(4kx)

= f(

4k(

x± 14(4−m)))− f

(4kx)

= f

4kx±

integer︷ ︸︸ ︷14

4k−m

− f(

4kx)= 0

Therefore,

g(x+δ m)−g(x)δ m

=1

δ m

m

∑k=0

(34

)k(f(

4k (x+δ m))− f

(4kx))

The absolute value of the last term in the sum is∣∣∣( 3

4

)m( f (4m (x+δ m))− f (4mx))

∣∣∣ and wechoose the sign of δ m such that both 4m (x+δ m) and 4mx are in some interval which is ofthe form [k/2,(k+1)/2) which is certainly possible because the distance between thesetwo points is 1/4 and such half open intervals include all of R. Thus, since f has slope ±1on the interval just mentioned,∣∣∣∣(3

4

)m

( f (4m (x+δ m))− f (4mx))∣∣∣∣= (3

4

)m

4m |δ m|= 3m |δ m|

2.5. COMPACTNESS AND CONTINUOUS FUNCTIONS 59The latter part of this theorem is called the Weierstrass M test. As a very interestingapplication, consider the question of nowhere differentiable functions. This considers thesimple case of functions having values in the Banach space R.Consider the following description of a function. The following is the graph of thefunction on [0,1].The height of the function is 1/2 and the slope of the rising line is 1 while the slope ofthe falling line is —1. Now extend this function to the whole real line to make it periodic ofperiod 1. This means f (x+-n) = f (x) for all x € R and n € Z, the integers. In other wordsto find the graph of f on [1,2] you simply slide the graph of f on [0, 1] a distance of 1 to getthe same tent shaped thing on [1,2]. Continue this way. The following picture illustrateswhat a piece of the graph of this function looks like. Some might call it an infinite sawtooth.Now define g(x) = YZ (3)* f (44x). Letting My = (3/4)*, an application of theWeierstrass M test, Theorem 2.5.42 shows g is everywhere continuous. This is becauseeach function in the sum is continuous and the series converges uniformly on R. However,this function is nowhere differentiable. This is shown next.Let 5,, = +4 (4-™) where we assume m > 2. That of interest will be m — c.g(xt+ 5m) g(x) _ Lido ($)! Uf (4+ Sm) —F (442))Om OmIf you take k > m,ile) -r(05) = r(0 (eaten) 9integer——a— f|4&et wae _f (4'x) ~0Therefore,m k— ee a sw iy @ (F (4 e+ 8m) —F (4%) )(3)"" (6 (4" (x4 5m) — f(4"x))] and wechoose the sign of 6,, such that both 4” (x+ 6,,) and 4’"x are in some interval which is ofthe form [k/2,(k+1)/2) which is certainly possible because the distance between thesetwo points is 1/4 and such half open intervals include all of R. Thus, since f has slope +1on the interval just mentioned,(3) a+ uy) —s04"%)| =F) 4"1dnl = 3" 5™ k=0The absolute value of the last term in the sum is