Derivatives and continuity / Lipschitz equation

[Felafel]

Hi! I think I've managed to solve this problem, but I'd like it to be checked

Homework Statement

show that if $$f : A\subset \mathbb{R}\to \mathbb{R}$$ and has both right derivative:
$$f_{+}'(x_0),$$

and left derivative
$$f_{-}'(x_0)$$
in $$x_0\in A$$, then $$f$$
is continuos in
$$x_0.$$

The Attempt at a Solution

Let's assume $$f_{+}' > f_{-}'$$, as the derivative exists, it means it is $$< \infty$$.
Therefore, $$|f(x)-f(y)|≤f_{+}'|(x-y)|$$ is a Lipschitz equation, and for
$$|f(x)-f(x_0)|≤f_{+}'|(x-x_0)|$$
it is a lipschitz equation in x_0.
Thus, for the lipschitz equation properties, the function is continuos in $$x_0$$

 

[micromass]

Therefore, $$|f(x)-f(y)|≤f_{+}'|(x-y)|$$ is a Lipschitz equation, and for
$$|f(x)-f(x_0)|≤f_{+}'|(x-x_0)|$$

How exactly did you obtain these inequalitities??

Also, you should type # instead of $. Typing

$$ ... $$

automatically places everything on a separate line. The command

## ... ##

does not do this.

 

[Felafel]

How exactly did you obtain these inequalitities??

$|f(x)−f(y)|≤L|(x−y)|$
this is the definition of lipschitz equation,
while
$|f(x)−f(x0)|≤f′+|(x−x0)|$
is from lagrange's theorem

 

[micromass]

$|f(x)−f(y)|≤L|(x−y)|$
this is the definition of lipschitz equation,
while
$|f(x)−f(x0)|≤f′+|(x−x0)|$
is from lagrange's theorem

What does Lagrange's theorem say? Are you really allowed to apply it in this case? Are all the conditions satisfied?

 

[Felafel]

uhm.. ok, i guess i don't have all the conditions to apply lagrange actually.
any hint about how else i can solve it?

 

[micromass]

uhm.. ok, i guess i don't have all the conditions to apply lagrange actually.
any hint about how else i can solve it?

Did you prove already that every differentiable functions is continuous? Can you try to adapt that proof?