# 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.