# Condition for Lipschitz Condition to be Satisfied

Jump to: navigation, search

## Theorem

Let $f$ be a real function.

Then $f$ satisfies the Lipschitz condition on a closed real interval $\closedint a b$ if:

$\forall y \in \closedint a b: \exists A \in \R: \size {\map {\phi'} y} \le A$

## Proof

Integrating both sides of $\size {\map {\phi'} y} \le A$ gives us:

 $\displaystyle \size {\map {\phi'} y}$ $\le$ $\displaystyle A$ $\displaystyle \leadsto \ \$ $\displaystyle -A$ $\le$ $\displaystyle \map {\phi'} y \le A$ $\displaystyle \leadsto \ \$ $\displaystyle \int {-A} \rd y$ $\le$ $\displaystyle \map \phi y \le \int A \rd y$ $\displaystyle \leadsto \ \$ $\displaystyle -A y$ $\le$ $\displaystyle \map \phi y \le y$ $\displaystyle \leadsto \ \$ $\displaystyle \size {\map \phi y}$ $\le$ $\displaystyle A y$

On the interval $\closedint a b$ it follows that $\size {\map \phi y}$ is bounded by the greater of $A a$ and $A b$.

Hence the result.

$\blacksquare$