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

Then prove that :


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.

  • 1
    $\begingroup$ You mean $f$ instead of $g$, right? $\endgroup$
    – InsideOut
    Jul 20, 2019 at 8:16
  • $\begingroup$ It seems like $a^{2}/10$ should be $a^{2}/12$, and maybe $a, b$ are both positive? $\endgroup$
    – Seewoo Lee
    Jul 20, 2019 at 8:30
  • 1
    $\begingroup$ Do we have any condition on $a$? In this setting if we let $a\to -\infty$ we will obviously get a contradiction. $\endgroup$
    – Feng
    Jul 20, 2019 at 8:31

1 Answer 1


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 $$ \frac{a^{2}}{12} = a^{2}\int_{0}^{1}\int_{0}^{1}(x-y)^{2}dydx \leq Var(g) \leq b^{2}\int_{0}^{1}\int_{0}^{1}(x-y)^{2}dydx = \frac{b^{2}}{12} $$


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