analysis – $ f $ increases monotonically, $ 0 the f le 1 $ and $ int_0 ^ 1 f (x) – x dx = 0 $ so $ int_0 ^ 1 | f (x) -x | dx le frac {1} {2} $.

$ f (x) $ monotonically increases $[0,1]$, $ 0 the $ 1 and $ int_0 ^ 1 f (x) x mathrm {d} x = 0 $. Prove it $ int_0 ^ 1 | f (x) -x | mathrm {d} x the frac {1} {2} $.

It's easy if $ f (x) ge x $ in $[0,1]$. And even in $[a,b]$ we have $ int_a ^ b | f (x) -x | mathrm {d} x the frac {(b-a) ^ 2} {2} $. But the zero points of $ f (x) – x $ can be infinitely many. It is there that the difficulty exists.