Condition for Lipschitz Condition to be Satisfied

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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.

$\blacksquare$