# Locally Lipschitz function implications

1. Dec 2, 2013

### mahler1

The problem statement, all variables and given/known data.

Let $f:\mathbb R \to \mathbb R$, $x_0, α \in \mathbb R$. $f$ is locally Lipschitzof of order $α$ at the point $x_0$ if there are $ε, M>0$ such that

$|f(x)-f(x_0)|<M|x-x_0|^α$ for every $x :0< |x-x_0|<ε$

Prove that:
1)If $f$ is locally Lipschitz of order $α>0$ at $x_0 \implies f$ is continuous at $x_0$
2)If $f$ is locally Lipschitz of order $α>1$ at $x_0 \implies f$ is derivable at $x_0$

The attempt at a solution.

Point 1) I think I did it ok: Suppose $f$ is locally Lipschtiz with $α>0$ at $x_0$ and let $ε>0$. We know there is $δ_0$: 0< |x-x_0|<δ_0 \implies $|f(x)-f(x_0)|<M|x-x_0|^α$. Now let $δ_1=(\dfrac{ε}{M})^{\frac{1}{α}}$, if we consider $δ=min\{δ_0,δ_1\}$, then if $x : |x-x_0|<δ \implies |f(x)-f(x_0)|<ε$

I am stuck at point 2), I've tried to prove it using the definition of differentiable function:

I want to show that $lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists. Now, I might be asking something obvious here but is it true that if $lim_{x \to x_0} |\dfrac{f(x)-f(x_0)}{x-x_0} |$ exists $\implies lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists?. If this is true, then I could prove $lim_{x \to x_0} |\dfrac{f(x)-f(x_0)}{x-x_0} |$ exists:

$0\leq lim_{x \to x_0} |\dfrac{f(x)-f(x_0)}{x-x_0} | \leq \dfrac{M|x-x_0|^α}{|x-x_0|}=M|x-x_0|^{α-1} \to 0$ when $x \to x_0$. This means $lim_{x \to x_0} |\dfrac{f(x)-f(x_0)}{x-x_0} |=0 \implies lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0} =0$.

There is something strange here that leads me to think there is something wrong with my proof: as $x_0$ is arbitrary, one could deduce the function $f$ is constant since $f'=0 \forall x$, so the only differentiable functions that are locally Lipschitz with $α>1$ are constant functions... I don't think this is correct, but where is the mistake in my proof? How could I correctly prove point 2)?

Last edited: Dec 2, 2013
2. Dec 2, 2013

### Dick

I think that's all pretty much ok. Though you might want to take a second look at how you are defining $\delta_1$ in the first part.

3. Dec 2, 2013

### mahler1

Oh, sure, I've meant $δ_1=(\dfrac{\epsilon}{M})^{\frac{1}{α}}$. Thanks for checking. If you say so, then it's probably correct. But it just seemed weird to me that the only functions that satisfy the conditions of 2) are the constant functions.

4. Dec 2, 2013

### Dick

No, not weird. Your proof is pretty clear. It just says that it's not very interesting to discuss functions that are locally Lipschitz with $\alpha>1$ at ALL points on an interval.

5. Dec 2, 2013

### brmath

For $\alpha$ = 1 we have this result:

"A Lipschitz function g : R → R is absolutely continuous and therefore is differentiable almost everywhere, that is, differentiable at every point outside a set of Lebesgue measure zero. Its derivative is essentially bounded in magnitude by the Lipschitz constant, and for a < b, the difference g(b) − g(a) is equal to the integral of the derivative g′ on the interval [a, b]." I'm quoting Wikipedia, but this result is a standard one.

The locally Lipschitz hypothesis allows us to expect this result throughout R. However, this is not quite as strong as differentiability. Can you think of an f which fails to be differentiable at one point, although satisfying this Lipschitz condition?

For $\alpha$> 1, the function is called Holder continuous. You were absolutely on the right track: "Lipschitz functions are differentiable almost everywhere and functions that are H ̈older continuous of exponent α > 1 have zero as a derivative everywhere. " see this paper: http://educ.jmu.edu/~querteks/seniorthesis.pdf (page 20).

For $\alpha$ less than 1 the bound is meaningless since it is blowing up. So I dont think you can conclude anything about the differentiability of f. The paper I reference above does seem to imply that f would not ordinarily be differentiable.

Finally your question about absolute values: |a| < |b| means -b < a < b . So if you are not sure whether your reasoning applies, look again at what the inequality means. You can if necessary break your proof up into two cases.