He defined a strict contraction f:X→X on a metric space (X,d) as requiring d(f(x),f(y))≤cd(x,y) for c∈(0,1) and for all x,y∈X. Your proposed solution doesn’t fix such a c; in fact, as x→0;x∈(0,1], c→1, which is why f′(0)=1.
Claim: You can’t solve the exercise
Proof (thanks to TheMajor). Let {xn}∞n=1 be a sequence in the domain converging to x∈[a,b] such that f′(x)=limn→∞f(xn)−f(x)xn−x. Since f is a strict contraction with contraction constant c, ∀n∈N+:|f(xn)−f(x)||xn−x|≤c<1. Since the absolute value is continuous, we conclude that |f′(x)|≤c<1. ◻️
I think we are working off different editions. According to the errata, the condition for strict contraction was changed to |f(x)−f(y)|<|x−y| for all distinct x,y∈[a,b].
Can you say more about why exercise 17.6.3 is wrong?
If we define f:[0,1]→R by f(x):=x/(1+x) then for distinct x,y∈[0,1] we have |f(x)−f(y)|=∣∣∣x1+x−y1+y∣∣∣=∣∣∣x−y(1+x)(1+y)∣∣∣<|x−y|
We also have f′(0)=1 since limx→0;x∈(0,1]f(x)−f(0)x−0=limx→0;x∈(0,1]x(x+1)x=1
In general, the derivative is f′(x)=1/(1+x)2, which is continuous on [0,1].
He defined a strict contraction f:X→X on a metric space (X,d) as requiring d(f(x),f(y))≤cd(x,y) for c∈(0,1) and for all x,y∈X. Your proposed solution doesn’t fix such a c; in fact, as x→0;x∈(0,1], c→1, which is why f′(0)=1.
Claim: You can’t solve the exercise
Proof (thanks to TheMajor). Let {xn}∞n=1 be a sequence in the domain converging to x∈[a,b] such that f′(x)=limn→∞f(xn)−f(x)xn−x. Since f is a strict contraction with contraction constant c, ∀n∈N+:|f(xn)−f(x)||xn−x|≤c<1. Since the absolute value is continuous, we conclude that |f′(x)|≤c<1. ◻️
I think we are working off different editions. According to the errata, the condition for strict contraction was changed to |f(x)−f(y)|<|x−y| for all distinct x,y∈[a,b].
Then you can solve it, yeah.