# Let $g$ be a differentiable, continuous function $[0,1]$ and $a≤g'(x)≤b$ for all $x\in [0,1]$

Let $$g$$ be a differentiable, continuous function $$[0,1]$$ and $$a≤g'(x)≤b$$ for all $$x\in [0,1]$$

Then prove that :

$$\frac{b^2}{12}≥\int_0^{1}g^{2}(x)dx-\left(\int_0^{1}g(x)dx\right)^{2}≥\frac{a^2}{12}$$

I'm trying using Hölder ? But I don't know how.

I don't have any ideas to prove this inequality ?

I think this is related with measure theory.

If any one have idea please tell me.

• You mean $f$ instead of $g$, right? Jul 20, 2019 at 8:16
• It seems like $a^{2}/10$ should be $a^{2}/12$, and maybe $a, b$ are both positive? Jul 20, 2019 at 8:30
• Do we have any condition on $a$? In this setting if we let $a\to -\infty$ we will obviously get a contradiction.
– Feng
Jul 20, 2019 at 8:31

I'll assume that $$a, b>0$$ and the RHS is $$a^{2}/12$$, not $$a^{2}/10$$. (Otherwise, there is a counterexample.) By the Mean Value Theorem, we have $$a\leq \frac{g(y) - g(x)}{y-x} = g'(c) \leq b\Rightarrow a(y-x) \leq g(y)-g(x) \leq b(y-x)$$ for any $$y>x$$. By the way, $$\int_{0}^{1}\int_{0}^{1} (g(y)- g(x))^{2} dydx = \int_{0}^{1} \int_{0}^{1} g(x)^{2} - 2g(x)g(y) + g(y)^{2} dydx = 2 Var(g)$$ where $$Var(g) = \int_{0}^{1} g(x)^{2} dx - \left( \int_0^1 g(x)dx\right)^{2}$$ which implies $$\frac{a^{2}}{12} = a^{2}\int_{0}^{1}\int_{0}^{1}(x-y)^{2}dydx \leq Var(g) \leq b^{2}\int_{0}^{1}\int_{0}^{1}(x-y)^{2}dydx = \frac{b^{2}}{12}$$