$ f,g$ are monotonically increasing in $ [0,1]$ and $ 0\le f , g \le 1$ . $ \int_0^1 f – g \mathrm{d}x = 0$ . Prove that

$ $ \int_0^1 |f – g|\mathrm{d}x \le \frac{1}{2}$ $

In my previous question, $ g(x) = x$ . And my teacher said $ x$ can be replaced by $ g(x)$ . In fact, in previous question, we don’t need to use the condition $ \int_0^1 f – g \mathrm{d}x = 0 $ . But if we replace $ x$ with $ g$ , this condition becomes necessary.

Also, if $ g = x$ , we can replace the $ \frac{1}{2}$ with $ \frac{1}{4}$ ,that is

$ $ \int_0^1|f-g| \mathrm{d}x \le \frac{1}{4}$ $ I am wondering how to prove that.