Testing Contraction Conditions

What Must Be Tested?

A contraction is a map that brings every pair of points closer by at least one uniform factor strictly below $1$. The word “uniform” matters: it is not enough that each individual pair is reduced by some factor less than $1$ if those factors can approach $1$. Testing the condition therefore means finding a single bound that works for every pair in the domain.

There is a second check for a map to be iterated: it must take its domain into itself. We will keep that requirement separate from calculating distances. Neither completeness of the metric space nor any conclusion about fixed points is needed for the calculations here.

Definition: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $T:X\to Y$. The map $T$ is Lipschitz if there is a finite $L\ge 0$ such that $d_Y(Tx,Ty)\le Ld_X(x,y)$ for all $x,y\in X$. If $X$ has at least two points, its Lipschitz constant is $$ \operatorname{Lip}(T)=\sup_{\substack{x,y\in X\\x\ne y}}\frac{d_Y(Tx,Ty)}{d_X(x,y)}. $$ If $X$ has fewer than two points, set $\operatorname{Lip}(T)=0$. When the supremum is finite, it is the least possible Lipschitz bound. A contraction on a metric space $(X,d)$ is a self-map $T:X\to X$ for which there is one number $q$ with $0\le q<1$ such that $d(Tx,Ty)\le qd(x,y)$ for all $x,y\in X$.

Thus a map on a space with at least two points is a contraction exactly when its Lipschitz constant is strictly less than $1$. The strict inequality is important: a Lipschitz constant equal to $1$ does not meet the definition. For a space with zero or one point, every self-map satisfies the contraction inequality, for example with $q=0$, because there are no distinct pairs to test.

When $X$ is an interval with the usual metric, a useful way to calculate the constant is to simplify the quotient $\lvert T(x)-T(y)\rvert/\lvert x-y\rvert$ for $x\ne y$. The supremum of these quotients, not merely a bound checked at a few selected pairs, determines the best Lipschitz constant.

Affine Maps and Scalar Iteration

Worked Example: An Affine Self-Map of an Interval

Consider $T:[0,1]\to\mathbb{R}$ given by $T(x)=x/3+1/4$. First verify that this is a self-map of $[0,1]$. If $0\le x\le1$, then $1/4\le x/3+1/4\le1/3+1/4=7/12$, and $7/12\le1$. So $T(x)\in[0,1]$.

For distinct $x,y\in[0,1]$, calculate the distance ratio:

\frac{|T(x)-T(y)|}{|x-y|} =\frac{|(x/3+1/4)-(y/3+1/4)|}{|x-y|} =\frac{|x-y|/3}{|x-y|} =\frac13.

Every distinct pair gives the same ratio, so $\operatorname{Lip}(T)=1/3$, and $T$ is a contraction. As a check, the pair $x=0,y=1$ gives $T(0)=1/4$, $T(1)=7/12$, and $|T(1)-T(0)|=7/12-1/4=7/12-3/12=1/3$, while $|1-0|=1$. Its ratio is $1/3$, as calculated.

More generally, on $\mathbb{R}$ with its standard metric, an affine map $T(x)=ax+b$ has distance ratio $|a|$ whenever $x\ne y$, since $T(x)-T(y)=a(x-y)$. Its Lipschitz constant is $|a|$, and it is a contraction if and only if $|a|<1$. This affine formula defines a self-map of $\mathbb{R}$ for every real $a,b$. On a restricted domain, however, self-mapping must be checked separately, as in the interval example.

Proposition: If $T:X\to X$ is a contraction with bound $q$, where $0\le q<1$, then for every integer $n\ge1$, its $n$-fold iterate $T^n$ satisfies $d(T^n x,T^n y)\le q^n d(x,y)$ for all $x,y\in X$.

Proof. The claim for $n=1$ is the contraction inequality. Suppose it holds for some $n\ge1$. Apply the contraction inequality to the two points $T^n x,T^n y\in X$; this is permitted because $T$ maps $X$ into $X$. It gives $d(T^{n+1}x,T^{n+1}y)\le qd(T^n x,T^n y)$. The induction hypothesis then gives $d(T^{n+1}x,T^{n+1}y)\le q(q^n d(x,y))=q^{n+1}d(x,y)$. Induction proves the claim for every $n\ge1$. $\square$

Worked Example: Iterates of a Scalar Affine Map

On $\mathbb{R}$ with the standard metric, let $T(x)=3x/5+2$. This is a self-map of $\mathbb{R}$. For all $x,y$, $|T(x)-T(y)|=|3(x-y)/5|=(3/5)|x-y|$, so its contraction bound is $3/5$. Compute its second iterate directly:

T^2(x)=T(T(x))=\frac35\left(\frac35x+2\right)+2 =\frac9{25}x+\frac65+2 =\frac9{25}x+\frac{16}{5}.

The displayed formula gives $|T^2(x)-T^2(y)|=(9/25)|x-y|$, so $\operatorname{Lip}(T^2)=9/25$. This agrees with the bound $(3/5)^2=9/25$. To check the constant using a specific pair, take $x=1,y=0$. Then $T^2(1)=9/25+16/5=9/25+80/25=89/25$ and $T^2(0)=16/5=80/25$, so their distance is $9/25$; the original distance is $1$. The ratio is $9/25$.

The proposition gives a bound for iterates even when it does not identify their exact Lipschitz constants. In the scalar affine example, the exact constants follow from the explicit formulas. No fixed-point theorem is being used: this calculation concerns distances between images of pairs of points.

Nonlinear Maps: Use the Difference Quotient

Worked Example: Squaring on a Short Interval

Let $X=[0,1/4]$ with the usual metric, and define $T:X\to\mathbb{R}$ by $T(x)=x^2$. If $0\le x\le1/4$, then $0\le x^2\le1/16\le1/4$, so the image lies in $X$: this is a self-map.

For distinct $x,y\in X$, factor the difference of squares and divide by the nonzero distance:

\frac{|T(x)-T(y)|}{|x-y|} =\frac{|x^2-y^2|}{|x-y|} =\frac{|x-y||x+y|}{|x-y|} =x+y \le\frac14+\frac14=\frac12.

This proves $\operatorname{Lip}(T)\le1/2$. To see that no smaller bound works, take $x=1/4$ and $y=1/4-1/(4n)$ for integers $n\ge1$. These are distinct points of $X$, and their ratio is $x+y=1/2-1/(4n)$, which approaches $1/2$ as $n$ increases. Therefore the supremum is $1/2$, so $\operatorname{Lip}(T)=1/2$. The map is a contraction. For a numerical check, $n=2$ gives $x=1/4$, $y=1/8$, and ratio $x+y=1/4+1/8=3/8$, which is at most $1/2$.

The upper bound and the exact constant require different arguments: an inequality for every pair establishes an upper bound, while pairs whose ratios approach that bound establish that the supremum is no smaller. The supremum need not itself be achieved by a distinct pair.

Worked Example: A Nonlinear Map with Constant One

On $X=[0,\infty)$ with the usual metric, define $S(x)=x/(1+x)$. The denominator is positive for every $x\in X$. Also $0\le x/(1+x)\le1$, so $S(x)\in[0,1]\subseteq X$; thus $S$ is a self-map.

For distinct $x,y\ge0$, compute:

\frac{|S(x)-S(y)|}{|x-y|} =\frac{\left|\frac{x}{1+x}-\frac{y}{1+y}\right|}{|x-y|} =\frac{|x-y|}{(1+x)(1+y)|x-y|} =\frac{1}{(1+x)(1+y)}.

Since $x,y\ge0$, this ratio is at most $1$, so $S$ is Lipschitz with bound $1$. For $x=0$ and $y=1/n$, where $n\in\mathbb{N}$, the ratio is $1/(1+1/n)=n/(n+1)$. These values approach $1$, so no bound strictly below $1$ works for all pairs. Hence $\operatorname{Lip}(S)=1$, and $S$ is not a contraction. For example, $n=3$ gives ratio $3/(3+1)=3/4$; this is below $1$, but it does not provide a uniform bound below $1$ for every pair.

A Common Mistake: Pointwise Is Not Uniform

For the map $S$, the quotient $1/((1+x)(1+y))$ is strictly less than $1$ for every distinct pair $x,y\in[0,\infty)$: at least one of $x,y$ is positive. A tempting but incorrect argument is: “Every pair has ratio less than $1$, therefore $S$ is a contraction.” This does not follow. The contraction definition requires one fixed $q<1$ that bounds all the ratios, and the ratios $n/(n+1)$ approach $1$. For any proposed $q<1$, sufficiently large $n$ gives $n/(n+1)>q$, so that proposed uniform bound fails.

When testing a contraction, use this sequence of checks:

Confirm that the formula gives a self-map of the stated space.
For distinct points, calculate or bound the distance ratio.
Find one uniform upper bound $q<1$, or show that the supremum of the ratios is at least $1$.
If claiming the exact Lipschitz constant, justify both the upper bound and the matching supremum.

These calculations matter because the contraction condition is a precise hypothesis in fixed-point results. A map may be continuous and may decrease the distance for each individual pair without satisfying that hypothesis. Conversely, the definition itself requires no completeness assumption: completeness becomes relevant when drawing further conclusions about iterating a contraction, not when deciding whether its distance inequality holds.

Check Your Understanding

For each question, distinguish the self-map check from the Lipschitz-constant calculation.

On $\mathbb{R}$, find the Lipschitz constant of $T(x)=-2x+3$. Is it a contraction?
On $[0,1]$, does $T(x)=x/2+1/4$ map the interval into itself? If so, calculate its Lipschitz constant.
For $T(x)=x^2$ on $[0,1]$, determine its exact Lipschitz constant and decide whether it is a contraction.
If $T$ has contraction bound $q=2/3$, what bound does the proposition give for $T^3$?
Explain why knowing that every distinct pair has distance ratio strictly below $1$ does not, by itself, prove that a map is a contraction.

What You'll Learn