An integral inequality [3]

Question

Let $f$ be a continuously differentiable function and decreasing on the interval $[a,b].$ Let $g$ be a continuous function on $[a,b].$ Define $G:[a,b]\to \mathbb{R}$ by the relation $$G(x)=\int_{a}^{x}g(t)dt.$$ Prove that $$m(f(a)-f(b))\leq -\int_{a}^{b}f'(t)G(t)dt\leq M(f(a)-f(b)).$$ Where $m=\min_{x\in [a,b]} G(x)$ and $M=\max_{x\in [a,b]}G(x).$

Answer 1

By the Mean Value Theorem there is $s \in [a,b]$ such that

$$\int_{a}^{b}f'(t)G(t)dt=G(s)\int_{a}^{b}f'(t)dt.$$

Can you proceed ?

Answer 2

By using $G(x)=\int_a^x g(t) dt\in [m, M]$ we get $\int_{a}^{b}f'(t)md t\leq \int_{a}^{b}f'(t)G(t)dt\leq \int_{a}^{b}f'(t)Mdt$. Since $\int_a^b f'(t)dt=f(b)-f(a)$, we get the desired inequality.