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Theorem. There exists an everywhere-continuous, nowhere-differentiable function.

n

Let h(x) = |x − bx + 1/2c|, and fn (x) = h(44n x) . Then fn (x) is continuous for every integer n ≥ 0.

We have

1

1

< n

0 ≤ fn (x) ≤

n

2·4

4

.

Setting Fm (x) =

Pm

n=0

fn (x) for integral m ≥ 0 we have

0 ≤ Fm (x) ≤ Fm+1 (x)

Since

m

1

X

1 − 4m+1

1

4

Fm (x) <

=

<

n

4

1

−

1/4

3

n=0

, for any fixed x there exists

lim Fm (x) = f (x)

m→∞

We will prove first that f is continuous for every x.

For integral m, k with m > k ≥ 0 and every x, we have

0 ≤ Fm (x) − Fn (x) =

m

X

m

X

fn (x) <

n=k+1

n=k+1

1

1

< k

n

4

4

Let > 0 be given. We choose an integer k ≥ 0 such that

1

<

4k

3

Then for every x and for integral m > k,

0 ≤ Fm (x) − Fk (x) <

3

so that

0 ≤ f (x) − Fk (x) ≤

3

Hence for every ξ and every h, we have

|f (ξ + h) − Fk (ξ + h)| ≤

3

and

3

Now let ξ be fixed. Fk (x) is continuous at ξ; thus, for suitable δ > 0, for |h| < δ we have

|f (ξ) − Fk (ξ)| ≤

|Fk (ξ + h) − Fk (ξ)| <

3

so that

|f (ξ + h) − f (ξ) < |f (xi + h) − Fk (ξ + h)| + |f (ξ) − Fk (ξ)| + |Fk (ξ + h) − Fk (ξ)| <

.

Finally we shew that f (x) is not differentiable for any x.

(ξ)

If we had f 0 (ξ) = t for some value ξ, then for suitable δ > 0 we would have f (x)−f

−

t

<

x−ξ

0 < |x − ξ| < δ.

1

1

2

for

For every sequence ξk , for which ξk 6= ξ, ξk → ξ, there would exist a k0 such that for k ≥ k0 we would

have

f (ξk ) − f (ξ)

1

− t <

ξk − xi

2

and therefore also

f (ξk+1 ) − f (ξ)

1

− t <

ξk+1 − xi

2

so that

f (ξk ) − f (ξ) f (ξk+1 ) − f (ξ)

<1

−

ξk − xi

ξk+1 − ξ

Thus to obtain a contradiction, it suffices to produce a sequence ξk , k ≥ 1 and integral with ξk 6= ξ,

ξk → ξ such that

f (ξk ) − f (ξ)

ξk − ξ

is an integer for all k and in fact is even if k is even and odd if k is odd.

And this can and will be done..

For integral k ≥ 1 we set

ξk =

Evidently, we have ξk 6= ξ, |ξk − ξ| = 4−k → 0, ξk → ξ.

If n is an integer ≥ k then

4n ξk = 4n ξ ± 4n−k

so that fn (ξk ) = fn (ξ). Thus, for integers m, k with m ≥ k ≥ 1,

Fm (ξk ) − Fm (ξ) =

m

X

(fn (ξk ) − fn (ξ)) =

n=0

k−1

X

(fn (ξk ) − fn (ξ))

n=0

so that for integers k ≥ 1

f (ξk ) − f (ξ) =

k−1

X

(fn (ξ) − fn (ξ))

n=0

If n is an integer such that 0 ≤ n ≤ k − 1 then

2 · 2−k = 2−2n−1

Thus

4−k ≤ 2−2n−1 − 2−k

Setting

a = b22n+1 ξc

we have

a ≤ 22n+1 ξ < a + 1

2−2n−1 a ≤ ξ < 2−2n−1 (a + 1)

2

I assert that we have

2−2n−1 a ≤ ξk < 2−2n−1 (a + 1)

If a is even, then it follows that fn (ξk ) − fn (ξ) = ξk − ξ, and if a is odd, it follows that fn (ξk ) − fn (ξ) =

−(ξk − ξ).

n (ξ)

= ±1.

In either case fn (ξξkk)−f

−ξ

Tus if k ≥ 1 we have

k−1

fn (ξk ) − fn (ξ) X

=

±1

ξk − ξ

n=0

and so is even for even k, and odd for odd k.

3

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