# Proof to show function f satisfies Lipschitz condition when derivatives f' exist and are continuous

The question is as follows:

Given a function $f$, 2 known information:
(1) $f'(x)$ exist
(2) $f'(x)$ are continuous

Goal: 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 .

1) Continuous function $g$ in a compact interval $[a,b]$ implies $g$ bounded in $[a,b]$. (This part I make a bit more general than yours)

Consider the set $X=\{x\in[a,b]: g|[a,x] \text{ is bounded }\}$. Show that $a\in X$ and $\sup X\in X$ then $b=\sup X$ so $b\in X$.

2) Bounded derivative implies Lipschitz.

If $|f'(x)|\leq K$ for all $x\in[a,b]$, in particular for any $x,y\in[a,b]$ such that $x\neq y$ we have

$$\left|\dfrac{f(x)-f(y)}{x-y}\right|=|f'(c)|\leq K$$

for some $c\in[x,y]$ by MVT. Then $|f(x)-f(y)|\leq K|x-y|$.

• may be i am missing a point ,but wont g|[a,x] for a<x<b be bounded ? if g|[a,b] is bdd? – rohit May 6 '13 at 4:21
• @rohit yes, $g|[a,b]$ bounded iff for each $x\in[a,b]$ $g|[a,x]$ is bounded. – Gaston Burrull May 6 '13 at 4:24
• @Gastón Burrull What if $|f'(x)| \leqslant K$ for all $x \in (a, b)$ instead of $[a, b]$. Is the proof then still correct? – Peter May 28 '15 at 14:24
• yes, youre right – Gaston Burrull May 30 '15 at 20:38