The question goes as follows

Suppose that $f'(x)\geq M>0$ $\forall x\in [0,1]$. Show that there is an interval of length $\frac{1}{4}$ on which $|f|\geq M/4$.

and the answer book states

Note that $f$ is increasing. If $f(1/2)\geq 0$, then $f(3/4)\geq M/4$, so certainly $f\geq M/4$ on the interval $[3/4,1]$. On the other hand, if $f(1/2)\leq 0$, then $f(1/4)\leq -M/4$, so $f\leq -M/4$ on the interval $[0,1/4]$.

How does $f(1/2)\geq 0\Rightarrow f(3/4)\geq M/4$ given that $f$ is increasing on $[0,1]$? (the same question goes for $f(1/2)\leq 0$).

  • 3
    $\begingroup$ Do you have access to mean value theorem at this point of the book? In that case, the inequality should be a straightforward application of mvt. $\endgroup$
    – minimalrho
    Commented Mar 14, 2020 at 4:05
  • 2
    $\begingroup$ Hint: As $f$ is differentiable and continuous on $[0,1]$, apply the mean value theorem. First, observe that $f(\frac{3}{4})-f(\frac{1}{4})=f'(c)(\frac{3}{4}-\frac{1}{4})\geq \frac{M}{2}$ and that $f$ is increasing on $[0,1]$. Consider two cases for and $f(\frac{1}{4})\leq-\frac{M}{4}$ and $f(\frac{1}{4})>-\frac{M}{4}$. $\endgroup$
    – Axion004
    Commented Mar 14, 2020 at 4:15
  • 1
    $\begingroup$ Thank you, this answers my question. $\endgroup$ Commented Mar 14, 2020 at 4:26

1 Answer 1


Essentially, you can "integrate" the expression $f'(x)\geq M$ to deduce for $x>a$ $$f(x) \geq M \cdot (x-a) + f(a) \space \space [*].$$ To prove this, use the mean value theorem: assuming the usual conditions are met if $x>a$ then $\exists c $ with $a \leq c \leq x$ such that $$\frac {f(x)-f(a)}{x-a} = f'(c)$$ which rearranges to $$f(x)=f'(c) \cdot (x-a) + f(a).$$ Given $f'(c) \geq M$ and $x-a > 0$, the result $[*]$ follows.

Applying $[*]$ to the question, we have:

using $a = 1/2$, for $x \geq 3/4$, $x-a > 1/4$ so $$f(x) \geq M \cdot \frac 14 + f(1/2)$$ or using $x=1/2$, for $a \leq 1/4$, $x-a>1/4$ so $$f(1/2) \geq M \cdot \frac 14 + f(a)$$ From these, we can deduce respectively that

if $f(1/2) \geq 0$ then $f(x) \geq \frac M4$ for $x \geq 3/4$

if $f(1/2) < 0$ then $f(a) < - \frac M4$ for $a \leq 1/4$.


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