6
$\begingroup$

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 .

Thank you in advance.

$\endgroup$
4
$\begingroup$

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 $|f(x)-f(y)|\leq K|x-y|$.

$\endgroup$
  • 1
    $\begingroup$ may be i am missing a point ,but wont g|[a,x] for a<x<b be bounded ? if g|[a,b] is bdd? $\endgroup$ – rohit May 6 '13 at 4:21
  • 1
    $\begingroup$ @rohit yes, $g|[a,b]$ bounded iff for each $x\in[a,b]$ $g|[a,x]$ is bounded. $\endgroup$ – Gaston Burrull May 6 '13 at 4:24
  • $\begingroup$ @Gastón Burrull What if $|f'(x)| \leqslant K$ for all $x \in (a, b)$ instead of $[a, b]$. Is the proof then still correct? $\endgroup$ – Peter May 28 '15 at 14:24
  • $\begingroup$ yes, youre right $\endgroup$ – Gaston Burrull May 30 '15 at 20:38

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.