The question is as follows:
Given a function $f$, 2 known information:
(1) $f'(x)$ exist
(2) $f'(x)$ are continuousGoal: function $f$ satisfies Lipschitz condition on any bounded interval $[a,b]$
Here is my attempt:
1/ Recall Lipschitz condition: a function $f$ satisfies Lipschitz if there is a real number $N$ such that $|f(x) - f(y)| \leq N |x - y|$
2/ First, I plan to show this :
(*) Knowing $f'(x)$ is continuous at any point $x \implies f'(x)$ is bounded at some neighborhood about $x$
Using definition of continuity on $f'(x)$, I say that for any $\epsilon > 0$, there is $\delta > 0$ such that for any $y$, we have $|x - y| < \delta => |f'(x) - f'(y)| < \epsilon$.
After some works, I realize that $f'(y)$ is in the neighborhood of $(f'(x) - \epsilon, f'(x) + \epsilon)$.
So if I let my $N = max${$f'(x) + \epsilon, -f'(x) + \epsilon$}, I reach the conclusion that $f'(x)$ is bounded (at least below)
3/ Then, I plan to use (*) to say this :
If function $f$ has a derivative $f'$ such that $f'$ is bounded by some number K $ \implies f$ satisfies Lipschitz condition on any interval [a,b]
I plan to use the Mean Value Theorem, provided that by (*), there is a derivative $f'(x) < K$ where $x$ is in between some $x_1$ and $x_2$ in $[a,b]$
Would someone please check if my ideas are correct?
I feel very shaky about part 2 of my work. If the derivative is bounded, then I think the proof will be way easier. But to conclude that continuous $\implies $ bounded, I'm not sure if I can claim such thing .
Thank you in advance.