This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Lipschitz Functions

Definition: If $f : A \to \mathbb{R}$ is a function, then $f$ is said to be a Lipschitz function if $\exists L > 0$ such that $\forall x, y \in A$, we have that $\mid f(x) - f(y) \mid ≤ L \mid x - y \mid$.

We will now look at some important theorems regarding Lipschitz functions.

Proposition 1: The function $f : A \to \mathbb{R}$ is a Lipschitz function if the set of slopes between all points $x, y \in A$ is bounded.

• Proof: Let $f : A \to \mathbb{R}$ be a Lipschitz function. Then $\exists L > 0$ such that $\forall x, y \in A$ we have that $\mid f(x) - f(y) \mid ≤ L \mid x - y \mid$. Rearranging this inequality we obtain that:

(1)

\begin{align} \biggr \rvert \frac{f(x) - f(y)}{x - y} \biggr \rvert ≤ L \end{align}

• Let $M := \left \{ \frac{f(x) - f(y)}{x - y} : x, y \in A, x \neq y \right \}$ be the set of slopes between all of the points $x$ and $y$ from $A$. From the inequality above, we get that $M$ is bounded by $L$. $\blacksquare$

From proposition 1, we can construct a geometric representation of what it is meant for a function to be Lipschitz. We get that a function is Lipschitz if the set of slopes between any two points $x$ and $y$ from $A$ is bounded. Thus, the supremum of this set will be the largest slope between any two points and the infimum of this set will be the smallest slope of this set.

Theorem 1: If $f : A \to \mathbb{R}$ is a Lipschitz function on $A$, then $f$ is uniformly continuous on $A$.

• Proof: Let $f : A \to \mathbb{R}$ be a Lipschitz function on $A$. Then $\exists L > 0$ such that for all $x, y \in A$ we have that $\mid f(x) - f(y) \mid ≤ L \mid x - y \mid$. We want to show that $f$ is uniformly continuous, that is we want to show that $\forall \epsilon > 0$ then $\exists \delta > 0$ such that if $x, y \in A$ is such that $\mid x - y \mid < \delta$ then $\mid f(x) - f(y) \mid < \epsilon$.

• Let $\epsilon > 0$ be given and let $\delta_{\epsilon} = \frac{\epsilon}{L}$. Then for all $x, y \in A$ such that $\mid x - y \mid < \delta_{\epsilon} = \frac{\epsilon}{L}$ then $\mid f(x) - f(y) \mid ≤ L \mid x - y \mid < L \delta_{\epsilon} = L \cdot \frac{\epsilon}{L} = \epsilon$, and so $f$ is uniformly continuous on $A$. $\blacksquare$