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 $\left[{a .. b}\right]$ if:

$\forall y \in \left[{a .. b}\right]: \exists A \in \R: \left|{\phi^{\prime} \left({y}\right)}\right| \le A$


Proof

Integrating both sides of $\left|{\phi^{\prime} \left({y}\right)}\right| \le A$ gives us:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\) \(\displaystyle \left\vert{\phi^{\prime} \left({y}\right)}\right\vert \le A\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\implies\) \(\displaystyle -A \le \phi^{\prime} \left({y}\right) \le A\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\implies\) \(\displaystyle \int {-A} \ \mathrm dy \le \phi \left({y}\right) \le \int A \ \mathrm dy\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\implies\) \(\displaystyle -A y \le \phi \left({y}\right) \le A y\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\implies\) \(\displaystyle \left\vert{\phi \left({y}\right)}\right\vert \le A y\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    


On the interval $\left[{a .. b}\right]$ it follows that $\left|{\phi \left({y}\right)}\right|$ is bounded by the greater of $A a$ and $A b$.

Hence the result.

$\blacksquare$

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