Lipschitz Continuity

analysis

Definition

Lipschitz Continuity

A function $f:D\to \mathbb{R}$ is Lipschitz continuous on $D$ if one constant bounds all output changes by the corresponding input changes. Formally, there exists $L\ge 0$ such that, for all $x_1,x_2\in D$,

$$|f(x_1)-f(x_2)|\le L|x_1-x_2|.$$

Any such $L$ is called a Lipschitz constant. The inequality says that no secant line of the graph can have absolute slope larger than $L$. The function may bend, but it may not become arbitrarily steep.

Uniform control

Lipschitz continuity is a global version of bounded steepness. The same constant $L$ must work for every pair of points in the domain. Rearranging the defining inequality gives

$$\frac{|f(x_1)-f(x_2)|}{|x_1-x_2|}\le L(x_1\ne x_2),$$

so every secant slope is controlled at once.

This immediately implies uniform continuity. Given $\epsilon >0$, choose $\delta =\epsilon /L$ when $L>0$. Then $|x_1-x_2|<\delta$ forces $|f(x_1)-f(x_2)|<\epsilon$. If $L=0$, the function is constant and the claim is trivial.

Relation to derivatives

On an interval, a differentiable function with bounded derivative is Lipschitz continuous. If

$$|f^{\prime }(x)|\le L\text{for all }x,$$

then the mean value theorem gives

$$|f(x_1)-f(x_2)|\le L|x_1-x_2|.$$

Conversely, if a Lipschitz function is differentiable at a point, then the derivative at that point has absolute value at most its Lipschitz constant. A Lipschitz function need not be differentiable everywhere, but Rademacher’s theorem says that Lipschitz functions on Euclidean space are differentiable almost everywhere.

Examples and non-examples

• Every constant function is Lipschitz continuous with $L=0$.
• The function $f(x)=ax+b$ is Lipschitz continuous on $\mathbb{R}$ with $L=|a|$.
• The function $f(x)=x^2$ is not Lipschitz continuous on all of $\mathbb{R}$, because its slope grows without bound. It is Lipschitz continuous on every bounded interval.
• The function $f(x)=\sqrt{x}$ is uniformly continuous on $[0,1]$, but it is not Lipschitz continuous there because its slope becomes unbounded near $0$.