Functie lipschitziana cu integrala nula

[omc]

Fie \( f: [0,1] \to \mathbb{R} \) ai \( |f(x) - f(y)| \leq |x - y| \) si \( \int_0^1f(x)dx = 0 \). Sa se arate ca \( \int_0^xf(t)dt \geq \frac {1}{2}x(x - 1) \) pt orice x in [0,1].

[Ciprian Oprisa]

Luam \( g(x)=f(x)-x+\frac{1}{2} \).
Avem \( \int\limits_0^x{g(t)}dt=\int\limits_0^x{f(t)}dt-\int\limits_0^x{(t-\frac{1}{2})}dt=\int\limits_0^x{f(t)}dt-\frac{x(x-1)}{2} \)
Pentru \( x=1 \) avem \( \int\limits_0^1{g(t)}dt=0 \).
Inlocuind in prima conditie, avem
\( |g(x)-g(y)+x-y|\leq|x-y| \)
\( \Rightarrow -|x-y|\leq g(x)-g(y)+x-y\leq|x-y| \)
Alegem \( x<y \).
\( \Rightarrow x-y \leq g(x)-g(y)+x-y\leq y-x \)
\( \Rightarrow 0\leq g(x)-g(y) \)
\( \Rightarrow g(x)\geq g(y) \), deci \( g \) e descrescatoare.
Presupunem ca \( \int\limits_0^x{g(t)}dt<0 \).
Conform teoremei de medie, \( \exists c<x \) astfel incat \( g(c)<0 \), si cum \( g \) e descrescatoare obtinem ca pentru orice \( t>c \Rightarrow g(t)<0 \), deci \( \int\limits_0^1{g(t)}dt<0 \), absurd. Asadar
\( \int\limits_0^x{g(t)}dt\geq 0\Rightarrow\int\limits_0^x{f(t)}dt\geq\frac{x(x-1)}{2} \).