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

Proof

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

$f'(x) \le M$ 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?

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

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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.

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