📂Analysis

Local Lipschitz Condition

Definition

If $E$ is open in $\mathbb{R}^{n}$ and let’s denote it $\mathbf{f} : E \to \mathbb{R}^{n}$. If for all $\mathbf{x} _{0} \in E$, there exists a $\varepsilon > 0$ that satisfies $B \left( \mathbf{x} _{0} ; \varepsilon \right) \subset E$ and for all $\mathbf{x} , \mathbf{y} \in B \left( \mathbf{x} _{0} ; \varepsilon \right)$, there exists a $K >0$ that satisfies $| \mathbf{f} ( \mathbf{x} ) - \mathbf{f} ( \mathbf{y} ) | \le K | \mathbf{x} - \mathbf{y} |$, then $\mathbf{f}$ is said to be locally Lipschitz in $E$.

In this case, the following relationships hold.

Strong Lipschitz condition $\implies$ Lipschitz condition $\implies$ Local Lipschitz condition

Theorem

If $\mathbf{f} \in C^{1} (E)$, then $\mathbf{f}$ is locally Lipschitz in $E$.

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• Euclidean space $\mathbb{R}^{n}$’s ball is represented 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 a differential operator.

Proof

Since $E$ is open, there exists an open ball $B \left( \mathbf{x} _{0} ; \varepsilon \right) \subset E$ for the given $\mathbf{x}_{0} \in E$. Saying $\mathbf{f} \in C^{1} (E)$ means that there exists $D \mathbf{f}$, and we can set $\displaystyle K : = \max_{ \mathbf{x} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right] } \left\| D \mathbf{f} ( \mathbf{x} ) \right\|$.

If for each $\mathbf{x} , \mathbf{y} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right]$ we denote $\mathbf{u} := \mathbf{y} - \mathbf{x}$, as $B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right]$ is convex, for all $s \in [0,1]$

$$\mathbf{x} + s \mathbf{u} \in B \left[ \mathbf{x} _{0} ; {{\varepsilon} \over {2}} \right]$$

If we define the function $F : [0,1] \to \mathbb{R}^{n}$ as $F (s) := \mathbf{f} ( \mathbf{x} + s \mathbf{u} )$ then

$$F ' (s) = D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \mathbf{u}$$

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 obtained equation gives

$$\begin{align*} &\left| \mathbf{f} ( \mathbf{y} ) - \mathbf{f} ( \mathbf{x} ) \right| \\ \le & \int_{0}^{1} \left| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \mathbf{u} \right| ds \\ \le & \int_{0}^{1} \left\| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \right\| \left| \mathbf{u} \right| ds \end{align*}$$

Then, by the properties of operators,

$$\begin{align*} & \int_{0}^{1} \left\| D \mathbf{f} ( \mathbf{x} + s \mathbf{u} ) \right\| \left| \mathbf{u} \right| ds \\ \le & K | \mathbf{u} | \\ \le & K | \mathbf{y} - \mathbf{x} | \end{align*}$$

To summarize

$$\left| \mathbf{f} ( \mathbf{y} ) - \mathbf{f} ( \mathbf{x} ) \right| \le K | \mathbf{y} - \mathbf{x} |$$

Therefore, $\mathbf{f}$ is locally Lipschitz in $E$.

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