# Derivatives and continuity / Lipschitz equation

1. Dec 3, 2012

### Felafel

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

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

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
$$x_0.$$

3. 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 $$x_0$$

Last edited: Dec 3, 2012
2. Dec 3, 2012

### micromass

Staff Emeritus
How exactly did you obtain these inequalitities??

Also, you should type # instead of \$. Typing

Code (Text):

$$...$$

automatically places everything on a separate line. The command

Code (Text):

$...$

does not do this.

3. Dec 4, 2012

### Felafel

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

4. Dec 4, 2012

### micromass

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

5. Dec 4, 2012

### 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?

6. Dec 4, 2012

### micromass

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

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