Condition for Lipschitz Condition to be Satisfied
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Theorem
Let $\phi$ be a real function.
Then $\phi$ 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:
\(\ds \size {\map {\phi'} y}\) | \(\le\) | \(\ds A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -A\) | \(\le\) | \(\ds \map {\phi'} y \le A\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int {-A} \rd y\) | \(\le\) | \(\ds \map \phi y \le \int A \rd y\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -A y\) | \(\le\) | \(\ds \map \phi y \le y\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {\map \phi y}\) | \(\le\) | \(\ds A y\) |
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.
$\blacksquare$