Local Lipschitz Condition
Definition
Let $E$ be open in $\mathbb{R}^{n}$ and $\mathbf{f} : E \to \mathbb{R}^{n}$. If for every $\mathbf{x} _{0} \in E$ there exist an $\varepsilon > 0$ satisfying $B \left( \mathbf{x} _{0} ; \varepsilon \right) \subset E$ and a $K >0$ satisfying $| \mathbf{f} ( \mathbf{x} ) - \mathbf{f} ( \mathbf{y} ) | \le K | \mathbf{x} - \mathbf{y} |$ for all $\mathbf{x} , \mathbf{y} \in B \left( \mathbf{x} _{0} ; \varepsilon \right)$, then $\mathbf{f}$ is said to be locally Lipschitz on $E$.
In this case, the following relationship holds.
Strong Lipschitz condition $\implies$ Lipschitz condition $\implies$ Local Lipschitz condition
Theorem
If $\mathbf{f} \in C^{1} (E)$, then $\mathbf{f}$ is locally Lipschitz on $E$.
- The balls of the Euclidean space $\mathbb{R}^{n}$ are denoted as $$ B \left( \mathbf{x}_{0} ; d \right) := \left\{ \mathbf{x} \in \mathbb{R}^{n} \mid | \mathbf{x}_{0} - \mathbf{x} | < d \right\} \\ B \left[ \mathbf{x}_{0} ; d \right] := \left\{ \mathbf{x} \in \mathbb{R}^{n} \mid | \mathbf{x}_{0} - \mathbf{x} | \le d \right\}$$ and $D$ is the differential operator.
Proof
Since $E$ is open, for a given $\mathbf{x}_{0} \in E$ there exists an open ball $B \left( \mathbf{x} _{0} ; \varepsilon \right) \subset E$. Saying $\mathbf{f} \in C^{1} (E)$ means that $D \mathbf{f}$ exists, so we may set $\displaystyle K : = \max_{ \mathbf{x} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right] } \left\| D \mathbf{f} ( \mathbf{x} ) \right\|$.
For $\mathbf{x} , \mathbf{y} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right]$, let $\mathbf{u} := \mathbf{y} - \mathbf{x}$. Since $B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right]$ is convex, for all $s \in [0,1]$
$$ \mathbf{x} + s \mathbf{u} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right] $$
Defining the function $F : [0,1] \to \mathbb{R}^{n}$ by $F (s) := \mathbf{f} ( \mathbf{x} + s \mathbf{u} )$ gives
$$ F ' (s) = D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \mathbf{u} $$
Therefore
$$ \mathbf{f} ( \mathbf{y} ) - \mathbf{f} ( \mathbf{x} ) = F (1) - F(0) = \int_{0}^{1} F ' (s) ds = \int_{0}^{1} D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \mathbf{u} ds $$
Taking the absolute value of both sides of the equation obtained above yields
$$ \begin{align*} &\left| \mathbf{f} ( \mathbf{y} ) - \mathbf{f} ( \mathbf{x} ) \right| \\ \le & \int_{0}^{1} \left| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \mathbf{u} \right| ds \\ \le & \int_{0}^{1} \left\| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \right\| \left| \mathbf{u} \right| ds \end{align*} $$
Then, by the properties of operators,
$$ \begin{align*} & \int_{0}^{1} \left\| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \right\| \left| \mathbf{u} \right| ds \\ \le & K | \mathbf{u} | \\ \le & K | \mathbf{y} - \mathbf{x} | \end{align*} $$
In summary,
$$ \left| \mathbf{f} ( \mathbf{y} ) - \mathbf{f} ( \mathbf{x} ) \right| \le K | \mathbf{y} - \mathbf{x} | $$
and therefore $\mathbf{f}$ is locally Lipschitz on $E$.
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