THEOREMThere exists a function f:[0,1]->[0,1]that is continuous and nowhere differentiable.ProofC[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 functionsfor 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 byf^(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 contractionwith 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^nso 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.