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mahler1

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Homework Statement .

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)?

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)?

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