# 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.