TargetIIT

f is a continuous function that maps the closed unit interval I = [0,1] into itself. Prove that if f(f(x)) = x for all x ε I, then f is monotonic

f must be bijective


and every bijective continuous  function from [0,1] to [0,1] is monotonic

ya. but we only need it to be injective and continuous, and for that note that if f(x[ss]1[/ss]) = f(x[ss]2[/ss]) then f(f(x[ss]1[/ss])) = f(f(x[ss]2[/ss])) which implies x[ss]1[/ss]=x[ss]2[/ss]

Which theorem allows us to relate this fact to f being monotonic?

we have f'(f(x))f'(x)=1

and f'(f(x))=[im]http://codecogs.izyba.com/gif.latex?%5Clim_%7Bh%5Crightarrow%200%7D%5Cfrac%7Bf%28f%28x+h%29%29-f%28f%28x%29%29%7D%7Bh%7D[/im]

and f(f(x))=x for all values x in the given interval so f'(f(x))=1

so from the first eqn we have f'(x)=1

careful, nothing has been said about differentiability.

The last result in #3 easily follows from intermediate value property.