There is continuity of f' (x) = 2π cos (2π x), and f(0) = f( 1).

Therefore | f (x) f' (x) | =| sin (2π x) 2π cos (2π x) | = π| sin (4π x) | The period is 1/4k.

∫{0， 1} |f(x)f'(x)|dx = 4k ∫{0， 1/4} π sin(4πx)dx = ∫{0，π} sin(t)dt = 2。

On the other hand, ∫{0, 1} f(x)? dx = ∫{0， 1} sin？ (2πx)dx = ∫{0， 1 }( 1-cos(4πx))/2 dx = 1/2。

(1) and (2) are invalid.

As far as (1) is concerned, f(x) = x is a counterexample.