Derivatives and continuity / Lipschitz equation

  • Thread starter Felafel
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  • #1
Felafel
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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 $$x_0$$
 
Last edited:

Answers and Replies

  • #2
micromass
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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

Code:
$$ ... $$

automatically places everything on a separate line. The command

Code:
## ... ##

does not do this.
 
  • #3
Felafel
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How exactly did you obtain these inequalitities??

[itex] ##|f(x)−f(y)|≤L|(x−y)|##[/itex]
this is the definition of lipschitz equation,
while
[itex] ##|f(x)−f(x0)|≤f′+|(x−x0)|##[/itex]
is from lagrange's theorem
 
  • #4
micromass
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[itex] ##|f(x)−f(y)|≤L|(x−y)|##[/itex]
this is the definition of lipschitz equation,
while
[itex] ##|f(x)−f(x0)|≤f′+|(x−x0)|##[/itex]
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?
 
  • #5
Felafel
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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
micromass
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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?
 

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