Lipschitz Maps

A Fixed Linear Bound on Changes

Uniform continuity asks for one input tolerance that controls output changes throughout the domain. Lipschitz continuity gives a more quantitative form of global control: it requires every output distance to be bounded by a fixed multiple of the corresponding input distance. The multiple does not change with the pair of points.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f:X\to Y$. A Lipschitz condition compares $d_Y(f(x),f(y))$ directly with $d_X(x,y)$. Unlike an epsilon-delta condition, it supplies an explicit rate of control. As established in “Epsilon-Delta Continuity in Metric Spaces,” every Lipschitz map is continuous; in particular, it is uniformly continuous by the same global estimate.

Definition (Lipschitz Map): The map $f:X\to Y$ is Lipschitz if there is a finite constant $L\geq0$ such that, for every $x,y\in X$, $$ d_Y(f(x),f(y))\leq Ld_X(x,y). $$ Any such $L$ is called a Lipschitz constant for $f$. If one can choose $L<1$, then $f$ is called a contraction.

The Lipschitz constant is an upper bound on how much distances can expand. A map can have many Lipschitz constants: if $L$ works, every $L'\geq L$ works as well. The smallest possible constant, when the domain has at least two points, can be found by taking the supremum of all output-to-input distance ratios. This observation gives a precise way to distinguish an available bound from the best one.

Theorem (Best Lipschitz Constant): Suppose $X$ has at least two points. Define $$ S_f=\left\{\frac{d_Y(f(x),f(y))}{d_X(x,y)}:x,y\in X,\ x\ne y\right\}. $$ Then $f$ is Lipschitz if and only if $S_f$ is bounded above. When $f$ is Lipschitz, its least Lipschitz constant is $\sup S_f$.

Proof. For distinct $x,y$, the metric property gives $d_X(x,y)>0$. Thus the Lipschitz inequality for this pair is equivalent to \[ \frac{d_Y(f(x),f(y))}{d_X(x,y)}\leq L. \] If $f$ is Lipschitz with constant $L$, every element of $S_f$ is at most $L$, so $S_f$ is bounded above. Conversely, if $S_f$ is bounded above, let $C=\sup S_f$. Then $C\geq0$, and every distinct pair satisfies $d_Y(f(x),f(y))\leq C d_X(x,y)$. For $x=y$, both distances are zero, so the same inequality holds. Hence $C$ is a Lipschitz constant. Finally, any Lipschitz constant is an upper bound for $S_f$, and therefore is at least $\sup S_f=C$. This proves that $C$ is the least one. $\square$

If $X$ has exactly one point, every map from $X$ is Lipschitz with least constant $0$, since there are no distinct pairs to compare. If $X$ is empty, the inequality is vacuously true for every $L\geq0$, again with least constant $0$. The theorem uses at least two points precisely so its set of ratios is nonempty.

Calculating Lipschitz Constants

Worked Example: A Linear Map on the Plane with the Maximum Metric

Give $\mathbb{R}^2$ the maximum metric $d_\infty((x_1,x_2),(y_1,y_2))=\max\{|x_1-y_1|,|x_2-y_2|\}$, and define $f:\mathbb{R}^2\to\mathbb{R}$ by $f(x_1,x_2)=3x_1-4x_2$. For $x=(x_1,x_2)$ and $y=(y_1,y_2)$, the triangle inequality gives

|f(x)-f(y)| =|3(x_1-y_1)-4(x_2-y_2)| \leq3|x_1-y_1|+4|x_2-y_2| \leq7d_\infty(x,y).

Thus $7$ is a Lipschitz constant. To see that no smaller constant works, take $x=(1,-1)$ and $y=(0,0)$. Then $d_\infty(x,y)=1$, while

$$ |f(x)-f(y)|=|3(1)-4(-1)|=7. $$

The ratio of output distance to input distance is exactly $7$, so every Lipschitz constant must be at least $7$. Therefore $7$ is the best Lipschitz constant.

Worked Example: Squaring on a Bounded Interval

Define $f:[-2,2]\to\mathbb{R}$ by $f(x)=x^2$, using the usual metrics. For any $x,y\in[-2,2]$,

|f(x)-f(y)|=|x^2-y^2|=|x-y||x+y|\leq4|x-y|,

because $|x+y|\leq|x|+|y|\leq4$. Hence $4$ is a Lipschitz constant. To verify that it is the least one, choose $x=2$ and $y=2-1/n$ for an integer $n\geq1$. Both points lie in the interval, and $x\ne y$. Their ratio is

\frac{|f(x)-f(y)|}{|x-y|} =\frac{|4-(2-1/n)^2|}{1/n} =\frac{4/n-1/n^2}{1/n} =4-\frac1n.

These ratios approach $4$ as $n$ increases. Any Lipschitz constant must be at least every ratio $4-1/n$, and therefore must be at least $4$. Combined with the upper bound, this shows that the best constant is $4$. Notice that no distinct pair in this example needs to give ratio exactly $4$: the supremum of the ratios can be the best constant even when it is not attained.

Worked Example: A Uniformly Continuous Function That Is Not Lipschitz

Consider $g:[0,1]\to\mathbb{R}$, defined by $g(x)=\sqrt{x}$. The estimate proved for the square-root function in “Uniform Continuity Between Metric Spaces” gives

|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|} \qquad (x,y\geq0).

It follows that $g$ is uniformly continuous on $[0,1]$: given $\varepsilon>0$, the choice $\delta=\varepsilon^2$ ensures that $|x-y|<\delta$ implies $|g(x)-g(y)|<\varepsilon$. However, $g$ is not Lipschitz there. For $x=1/n^2$ and $y=0$, the output-to-input ratio is

\frac{|g(1/n^2)-g(0)|}{|1/n^2-0|} =\frac{1/n}{1/n^2} =n.

These ratios are unbounded as $n$ increases. By the Best Lipschitz Constant Theorem, no finite Lipschitz constant exists. This example shows that uniform continuity does not, by itself, give a linear bound on output changes.

How Lipschitz Bounds Combine

Lipschitz estimates are useful partly because they can be passed through constructions. For a composition, the first map controls the distance at its output, and the second map controls the resulting distance in the next space. The constants therefore multiply.

Theorem (Lipschitz Constants Under Composition): Let $f:X\to Y$ be Lipschitz with constant $L$, and let $g:Y\to Z$ be Lipschitz with constant $M$, where $X,Y,Z$ are metric spaces. Then $g\circ f:X\to Z$ is Lipschitz with constant $ML$.

Proof. For any $x_1,x_2\in X$, apply the Lipschitz inequality for $g$ to $f(x_1),f(x_2)$, and then apply the inequality for $f$:

d_Z(g(f(x_1)),g(f(x_2))) \leq M d_Y(f(x_1),f(x_2)) \leq MLd_X(x_1,x_2).

Since $M,L\geq0$, the product $ML$ is a finite nonnegative constant. The displayed inequality is exactly the Lipschitz condition for $g\circ f$. $\square$

The product $ML$ is a guaranteed constant, but it need not be the best one. For instance, one of the maps may compress distances much more than its stated bound suggests. The theorem asserts a reliable bound from the chosen constants, not necessarily an equality between best constants.

Worked Example: A Bound for a Two-Stage Map

Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=2x+1$, and let $g:\mathbb{R}\to\mathbb{R}$ be $g(u)=u^2$, with usual metrics. For all $x,y\in\mathbb{R}$,

$$ |f(x)-f(y)|=2|x-y|, $$

so $f$ is Lipschitz with constant $2$. The map $g$ is not Lipschitz on all of $\mathbb{R}$, since its distance ratios are unbounded. The composition theorem cannot be applied to these two maps on their full stated domains. But if we restrict $f$ to $[0,1]$, then $f([0,1])=[1,3]$, and the square map on $[1,3]$ satisfies

|u^2-v^2|=|u-v||u+v|\leq6|u-v| \qquad (u,v\in[1,3]).

Thus $g$ restricted to $[1,3]$ has Lipschitz constant $6$, and the composite $x\mapsto(2x+1)^2$ on $[0,1]$ has Lipschitz constant $12$. Directly, for $x,y\in[0,1]$, both $2x+1$ and $2y+1$ lie in $[1,3]$, so

|(2x+1)^2-(2y+1)^2| \leq6|(2x+1)-(2y+1)| =12|x-y|.

The domain of each stage matters: a map may be Lipschitz on the portion of its domain reached by the preceding map even if it is not Lipschitz on a larger set.

There is a parallel result for maps whose values have two coordinates. Equip $Y_1\times Y_2$ with the maximum metric, as in the product-metric results from “Examples of Metrics.” A coordinatewise estimate then gives a bound for the whole map.

Theorem (Coordinatewise Lipschitz Bound): Let $f_i:X\to Y_i$ be Lipschitz with constant $L_i$, for $i=1,2$. Give $Y_1\times Y_2$ the maximum metric, and define $F:X\to Y_1\times Y_2$ by $F(x)=(f_1(x),f_2(x))$. Then $F$ is Lipschitz with constant $\max\{L_1,L_2\}$.

Proof. For any $x,y\in X$, the maximum metric gives

d_{\max}(F(x),F(y)) =\max\{d_{Y_1}(f_1(x),f_1(y)),d_{Y_2}(f_2(x),f_2(y))\}.

The two coordinate estimates bound the entries in this maximum by $L_1d_X(x,y)$ and $L_2d_X(x,y)$, respectively. Consequently,

d_{\max}(F(x),F(y)) \leq\max\{L_1,L_2\}d_X(x,y).

This proves the stated Lipschitz bound. $\square$

What a Lipschitz Bound Does—and Does Not—Say

A Lipschitz constant controls every pair of points, not just pairs that happen to be close. In particular, if $X$ is bounded and nonempty, then the image of a Lipschitz map is bounded. To see this, choose $x_0\in X$. If $d_X(x,y)\leq R$ for all $x,y\in X$, and $L$ is a Lipschitz constant, then for every $x\in X$,

d_Y(f(x),f(x_0))\leq Ld_X(x,x_0)\leq LR.

Thus all points of $f(X)$ lie within distance $LR$ of $f(x_0)$. The nonempty-domain condition matters for this argument because it supplies a point $x_0$ around which to bound the image. Conversely, a Lipschitz map on an unbounded domain need not have bounded image: the identity map on $\mathbb{R}$ is Lipschitz and has unbounded image.

The main pitfall is to treat “uniformly continuous” and “Lipschitz” as interchangeable. Every Lipschitz map is uniformly continuous, as recalled from “Epsilon-Delta Continuity in Metric Spaces,” but the square-root example shows the converse fails. When proving a Lipschitz claim, it is not enough to find a tolerance $\delta$ for each $\varepsilon$; one must establish one finite constant that bounds every distance ratio.

Write the distance ratio.
For distinct inputs, compare the output distance with the input distance. A uniform upper bound for these ratios gives a Lipschitz constant.

Prove an upper bound.
Use algebraic identities, triangle inequalities, or bounds on the domain to control the output difference by a fixed multiple of the input difference.

Test whether the bound is best.
Find a pair attaining the proposed ratio, or a sequence of pairs whose ratios approach it. To disprove Lipschitz continuity, show that the ratios are unbounded.

Lipschitz maps provide quantitative control that is stable under composition and coordinatewise assembly. Their constants make it possible to track how much distance can expand at each step, while the best-constant criterion identifies the sharpest global bound available.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What does a Lipschitz constant bound, and how must it behave as the input pair varies?
Why can the best Lipschitz constant be a supremum that no individual pair of distinct points attains?
What Lipschitz constant is guaranteed for a composition when the two maps have constants $L$ and $M$?
Why does the square-root function on $[0,1]$ fail to be Lipschitz even though it is uniformly continuous?
For a map into a product with the maximum metric, what bound follows from coordinate constants $L_1$ and $L_2$?

What You'll Learn