Suppose $\sup_{x \in \mathbb{R}} f'(x) \le M$.

I am trying to show that this is true if and only if $$\frac{f(x) - f(y)}{x - y} \le M$$

for all $x, y \in \mathbb{R}$.


$\text{sup}_{x \in \mathbb{R}} f'(x) \le M$

$f'(x) \le M$ for all $x \in \mathbb{R}$

$\lim_{y \to x} \frac{f(y) - f(x)}{y - x} \le M$

$\lim_{y \to x} \frac{f(x) - f(y)}{x - y} \le M$

I can see geometrically why this property holds, but how do I get rid of the limit here? Or am I approaching it wrong in general?

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Mean_value_theorem $\endgroup$ – Martin R Dec 4 '15 at 8:50
  • $\begingroup$ If something is true of $f(x)$ for all x, then it's also true for $\lim_{x \to k} f(x)$, right? $\endgroup$ – Justin L. Dec 4 '15 at 8:50
  • 1
    $\begingroup$ @JustinL. But if the inequality holds for the limit it might not hold for the function. $\endgroup$ – Scientifica Dec 4 '15 at 8:52
  • $\begingroup$ @Riggs As Martin R said use the Mean Value Theorem. $\endgroup$ – Scientifica Dec 4 '15 at 8:52
  • 1
    $\begingroup$ Try to show both directions individually, do not only write down a list of statements, but put words between them to indicate their logical relationship (what implies what). $\endgroup$ – Carsten S Dec 4 '15 at 8:56

well, first of all, we have to presume f is continuous and differentiable on R. This statement isn't true otherwise.

1) Suppose $\frac{f(x) - f(y)}{x - y} > M$ for some $x, y \in \mathbb R$.

By the mean value theorem, there exist a $c; x <c < y$ where $f'(c) = \frac{f(x) - f(y)}{x - y}$.

So $f'(c) > M$.

So $\sup f'(x) \le M \implies f'(c) \le M$ for all $c \in \mathbb R \implies$ $\frac{f(x) - f(y)}{x - y} \le M$ for all $x, y \in \mathbb R$.

2) Suppose $\frac{f(x) - f(y)}{x - y} \le M$ for some $x, y \in \mathbb R$.

Then $\lim_{y \rightarrow x}\frac{f(x) - f(y)}{x - y} = f'(x) \le M$ for all $x \in \mathbb R$. So {$f'(x)|x \in \mathbb R$} is bounded above by M so $\sup_{x \in \mathbb{R}} f'(x) \le M$.


The 'if' part follows from the definition of the derivative as a limit. If some expression is always less than or equal to $M,$ and the limit exists, then the limit also satisfies that inequality. That, in its turn, follows from the epsilon-delta definition of a limit.

The 'only if' part is the really interesting part. As commenters have pointed out it is (a direct consequence of) the Mean Value Theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.