

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 <= (k1) / 3^(n+1) < k / 3^(n+1) <= 1/3. 3^n < k <= 2 * 3^n ==> 1/3 <= (k1) / 3^(n+1) < k / 3^(n+1) <= 2/3. 2 * 3^n < k <= 3 * 3^n ==> 2/3 <= (k1) / 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( (k1) / 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 (k1) * 3^(n) <= a <= k * 3^(n). By the triangle inequality, we have h( (k1) / 3^n )  h(A) + h(A)  h( k / 3^n ) >= h( (k1) / 3^n )  h( k / 3^n ) >= 2^(n) Let t[n] be equal to (k1)/3^n or to k/3^n so that the following condition is fulfilled: h(t[n])  h(A) = max{ h( (k1) / 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
Both proofs have been adopted from the book
