[Apronus Home] [Mathematics] [Topological Methods In Real Analysis]
play piano online
Play Piano Online

Continuous Nowhere Differentiable Function

THEOREM

There exists a function f:[0,1]->[0,1]
that is continuous and nowhere differentiable.

Proof

C[0,1] denotes the set of all continuous functions f:[0,1]->R.
With the sup metric, it is a complete metric space.

Let X be a subset of C[0,1] such that it contains only those functions
for which f(0)=0 and f(1)=1 and f([0,1]) c [0,1].

X is a closed subset of C[0,1] hence it is complete.
(X,sup) is a complete metric space.

For every f:-X define f^ : [0,1] -> R by
f^(x) = 3/4 * f(3x) for 0 <= x <= 1/3,
f^(x) = 1/4 + 1/2 * f(2 - 3x) for 1/3 <= x <= 2/3,
f^(x) = 1/4 + 3/4 * f(3x - 2) for 2/3 <= x <= 1.

Verify that f^ belongs to X.

Verify that the mapping X-:f |-> f^:-X is a contraction
with Lipschitz constant 3/4.

By the Contraction Principle, there exists h:-X such that h^ = h.

This h is continuous.
Now we want to show that it is nowhere differentiable.

Verify the following for n:-N and k:-{1,2,3,...,3^n}.

1 <= k <= 3^n   ==>   0 <= (k-1) / 3^(n+1) < k / 3^(n+1) <= 1/3.

3^n < k <= 2 * 3^n   ==>   1/3 <= (k-1) / 3^(n+1) < k / 3^(n+1) <= 2/3.

2 * 3^n < k <= 3 * 3^n   ==>   2/3 <= (k-1) / 3^(n+1) <  k / 3^(n+1) <= 1.

Using the above, prove by induction that
/\n:-N /\k:-{1,2,3,4,...,3^n} |h( (k-1) / 3^n ) - h( k / 3^n )| >= 2^(-n).

Take any A, 0<=A<=1. We will show that h is not differentiable at A.

Let's construct a sequence t[n] approaching A.
Take any n:-N.
We can choose k:-{1,2,3,4,...,3^n} such that
(k-1) * 3^(-n) <= a <= k * 3^(-n).

By the triangle inequality, we have

|h( (k-1) / 3^n ) - h(A)| + |h(A) - h( k / 3^n )|
>= |h( (k-1) / 3^n ) - h( k / 3^n )|
>= 2^(-n)

Let t[n] be equal to (k-1)/3^n or to k/3^n
so that the following condition is fulfilled:

|h(t[n]) - h(A)| = max{ |h( (k-1) / 3^n ) - h(A)| , |h(A) - h( k / 3^n )| }.

Now we have t[n] != A and 2*|h(t[n]) - h(A)| >= 2^(-n).
Also |t[n]-A| <= 3^(-n).

We constructed {t[n]} contained in [0,1] converging to A, never equal to A.

Notice that for every n:-N we have

|h(t[n]) - h(A)|  /  |t[n] - A|  >=  1/2 * (3/2)^n. 

Hence lim |h(t[n]) - h(A)|  /  |t[n] - A| = oo as n approaches oo.

This means that h is not differentiable at A.

The proof is complete.

You can also see a proof of the existence of a nowhere monotonic continuous function.

Both proofs have been adopted from the book Topological Spaces: From Distance to Neighborhood.

[Apronus Home] [Contact Page]