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The Question:

Let $f$ be a function defined on an interval $[a,b]$. What conditions could you place on $f$ to guarantee that

$$ \min f'\leq \frac{f(b)-f(a)}{b-a} \leq \max f' $$

My Answer

We must require a $\min f'$ and a $\max f'$ before we can proceed further. That means $f'$ must be defined in $[a,b]$, that is, $f$ should be differentiable in $[a,b]$. So that's one condition.

Now, if $f$ is continuous in $[a,b]$ and differentiable in $(a,b)$ then, per the mean value theorem $$ \exists \text{ } c\in(a,b) \text{ | } f'(c)=\frac{f(b)-f(a)}{b-a} $$

and obviously $$ \min f'\leq f'(c) \leq \max f' $$

So the only condition would be for $f$ to be differentiable in $[a,b]$.

Now, I'll tell what's bothering me. The textbook I'm reading currently has the following definition for absolute extrema.

Let $f$ be a function with domain $D$. Then $f$ has an absolute maximum value on $D$ at a point $c$ if $$ f(x) \leq f(c) \text{ } \forall \text{ } x \in D $$

and an absolute minimum value on $D$ at $c$ if $$ f(x) \geq f(c) \text{ } \forall \text{ } x \in D $$

The reason, I've considered $f$ to be differentiable in $[a,b]$ and not just $(a,b)$ is because of the definition of absolute extrema.

Did I miss anything?

Abhishek A Udupa
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    Looks good to me. – Zuy Sep 03 '21 at 06:50
  • A function can be differentiable on $[a,b}$, but its derivative needs not be bounded. – egreg Sep 03 '21 at 07:20
  • @egreg That is exactly the doubt I'm having.

    The textbook that I'm reading right now, where I came across this problem has the following definition for extrema: If $f$ is a function with domain $D$, then $f$ has an absolute extremum value on $D$ at a point $c$ if $f(x)\leq f(c)$ or $f(x)\geq f(c)$ (as applicable) for all $x$ in $D$.

    That is why I considered $f'$ to be bounded so that it can have extrema, though, I can make up functions that are unbounded and have extrema. I'm confused.

     – Abhishek A Udupa Sep 03 '21 at 07:30
  • @AbhishekAUdupa If a function is upper and lower unbounded, it cannot have extrema. – egreg Sep 03 '21 at 07:38
  • @egreg so then, if $f'$ is unbounded then $\max f'$ and $\min f'$ wouldn't exist right? – Abhishek A Udupa Sep 03 '21 at 07:43
  • @AbhishekAUdupa Exactly my point! – egreg Sep 03 '21 at 07:44
  • @egreg So that means $f$ should be differentiable in $[a,b]$ and my answer attempt is correct? – Abhishek A Udupa Sep 03 '21 at 07:55

1 Answers1

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You need to assume that the derivative is bounded if you want to actually bound the ratios $\frac{f(b)−f(a)}{b−a}$.

If you look at $f$ defined as:

  • $f(x) = \sin(1/x^2)\times x^2$ if $x\in(0,1]$
  • $f(0) = 0$

then $f$ is differentiable over all $[0,1]$, but its derivative is not bounded, as for $x>0$:

$$f'(x)=2x\sin\left(\frac1{x^2}\right) -\frac2x\times\cos\left(\frac1{x^2}\right)$$

and since the derivative values are limits of ratios $\frac{f(b)−f(a)}{b−a}$, those ratios are not bounded either.

Chris
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