# Showing Bounded Derivative $\implies$ Lipschitz Function (Uniformly Continuous)

Let $$I$$ be an interval in $$\mathbb{R}$$ (bounded or unbounded). Let $$f: I \rightarrow \mathbb{R}$$ be differentiable, with the property that $$|f'(x)| \le M, \forall x \in I$$.

Show that $$|f(x) - f(y)| \le M |x - y|$$, for all $$x \in I$$.

Essentially, I need to show that a function whose derivative is bounded is uniformly continuous.

I've seen the answer posted at Prove that a function whose derivative is bounded is uniformly continuous., but it seems that answer assumes the conclusion I need to arrive at.

I'm trying to understand the solution posted above, but I don't quite see how they assume the conclusion I need to arrive at. Is the case similar with my question, where we can simply use the Mean Value Theorem? If not, how should I approach this problem? A bounded derivative means that the rate of change for the function will decrease and decrease, which means the function itself must also be decreasing (monotonically?). Is my line of thinking correct, or on the right track? I can't seem to put these thoughts into a formal proof. Any help on this problem would be very helpful. Thank you!

Mean Value Theorem: $|f(x)-f(y)|=|f'(\xi_{x,y})||x-y|\leq M|x-y|$.
• I don't see how we jump from $|f'(x)| \le M, \forall x \in I$ to that, though. Could you explain how? Apr 26, 2018 at 3:27
• Realizing $x$ to that specific $\xi_{x,y}$. Apr 26, 2018 at 3:28