A Uniform Control on Change
Continuity at a point says that sufficiently small changes in the input produce small changes in the output. A Lipschitz estimate gives more: it controls the output change by a fixed multiple of the input change, using the same multiple throughout the domain. This makes continuity bounds easy to calculate and gives a useful way to compare maps between different metric spaces.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A map \(f:X\to Y\) is continuous at \(x\in X\) if for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(d_X(x,y)<\delta\) implies \(d_Y(f(x),f(y))<\varepsilon\) for every \(y\in X\). It is continuous if it is continuous at every point.
The inequality must hold for every pair of points in the domain. In particular, a bound that works only near one point is not, by itself, a global Lipschitz bound. When \(L=0\), the inequality requires \(f(x)=f(y)\) for all \(x,y\in X\), so \(f\) is constant (with the statement vacuous if \(X\) is empty).
Why a Lipschitz Map Is Continuous
Proof. Suppose \(f\) has Lipschitz bound \(L\ge0\), and fix \(x\in X\) and \(\varepsilon>0\). If \(L>0\), set \(\delta=\varepsilon/L\), which is positive. For any \(y\in X\) with \(d_X(x,y)<\delta\), the Lipschitz inequality gives $$ d_Y(f(x),f(y))\le Ld_X(x,y)<L\delta=\varepsilon. $$ This is continuity at \(x\). If \(L=0\), then \(d_Y(f(x),f(y))\le 0\) for every \(y\). Distances are nonnegative, so \(d_Y(f(x),f(y))=0\), and the metric axiom gives \(f(x)=f(y)\). Thus the output distance is \(0<\varepsilon\) for every \(y\), and any positive \(\delta\), for example \(\delta=1\), works. Since \(x\) was arbitrary, \(f\) is continuous on \(X\). \(\square\)
The calculation also shows the strength of the estimate: when \(L>0\), the same choice \(\delta=\varepsilon/L\) works at every point of \(X\). Continuity only asks for some suitable \(\delta\) at each point and for each \(\varepsilon\); Lipschitz continuity supplies a uniform formula.
Worked Example: An Affine Map of the Real Line
Give a Lipschitz bound for \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=3x-4\), where both copies of \(\mathbb{R}\) have the standard metric, and obtain an explicit continuity bound.
For any \(x,y\in\mathbb{R}\), subtract the outputs: $$ |f(x)-f(y)|=|(3x-4)-(3y-4)|=|3x-3y|=3|x-y|. $$ Thus \(L=3\) is a Lipschitz bound. For any \(\varepsilon>0\), choose \(\delta=\varepsilon/3\). If \(|x-y|<\delta\), then $$ |f(x)-f(y)|=3|x-y|<3\delta=3(\varepsilon/3)=\varepsilon. $$ For a direct check of the factor, take \(x=2\) and \(y=1\): \(f(2)=3(2)-4=2\), \(f(1)=3(1)-4=-1\), so the output distance is \(|2-(-1)|=3\), while the input distance is \(|2-1|=1\). The ratio is \(3/1=3\), consistent with the bound.
Contractions and Their Bounds
A contraction is a particular kind of Lipschitz map. The word is often used for a self-map, so the domain and codomain must be the same metric space in this definition.
A Lipschitz map from \(X\) to a different metric space \(Y\) can still have a bound less than \(1\); it is a map with a Lipschitz bound less than \(1\), but is not a contraction under the self-map definition above. The inequality concerns distances, not whether the map takes each point closer to a particular point. No fixed-point conclusion is needed to use the definition or to calculate continuity bounds.
Worked Example: A Contraction of the Real Line
Let \(\mathbb{R}\) have its standard metric and define the self-map \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x/4+3\). Find a contraction bound and the continuity parameter corresponding to \(\varepsilon>0\).
For \(x,y\in\mathbb{R}\), $$ |f(x)-f(y)|=\left|\left(\frac{x}{4}+3\right)-\left(\frac{y}{4}+3\right)\right| =\left|\frac{x-y}{4}\right|=\frac14|x-y|. $$ Therefore \(L=1/4\) is a Lipschitz bound, and \(0\le1/4<1\), so this self-map is a contraction. The continuity theorem gives \(\delta=\varepsilon/(1/4)=4\varepsilon\). Indeed, if \(|x-y|<4\varepsilon\), then \(|f(x)-f(y)|=|x-y|/4<4\varepsilon/4=\varepsilon\).
Check: at \(x=4\) and \(y=0\), the inputs have distance \(|4-0|=4\), and the outputs are \(f(4)=4/4+3=4\) and \(f(0)=0/4+3=3\). Their distance is \(|4-3|=1\), which equals \((1/4)\cdot4=1\). This verifies the bound at this pair.
Distance to a Nonempty Set
Let \((X,d)\) be a metric space and let \(A\subseteq X\) be nonempty. The distance from \(x\in X\) to \(A\) is the real number $$ d_A(x)=\inf\{d(x,a):a\in A\}. $$ The set of distances is nonempty and bounded below by \(0\), so its infimum exists; it is finite because choosing any \(a_0\in A\) gives \(d_A(x)\le d(x,a_0)<\infty\). The infimum need not be attained by a point of \(A\).
Proof. Fix \(x,y\in X\). For every \(a\in A\), the triangle inequality gives \(d(x,a)\le d(x,y)+d(y,a)\). Taking the infimum over \(a\in A\) on both sides gives \(d_A(x)\le d(x,y)+d_A(y)\), since \(d(x,y)\) is constant as \(a\) varies. Interchanging \(x\) and \(y\) gives \(d_A(y)\le d(x,y)+d_A(x)\). The two inequalities imply $$ -d(x,y)\le d_A(x)-d_A(y)\le d(x,y), $$ and therefore \(|d_A(x)-d_A(y)|\le d(x,y)\). This is the Lipschitz inequality with \(L=1\). \(\square\)
Worked Example: Distance to Two Points
In \(\mathbb{R}\) with the standard metric, let \(A=\{-1,2\}\). Compute the distance function at \(0\) and \(1/2\), then check its Lipschitz inequality for that pair.
By the definition of infimum for this finite set, \(d_A(x)=\min\{|x-(-1)|,|x-2|\}\). At \(x=0\), the two distances are \(|0+1|=1\) and \(|0-2|=2\), so \(d_A(0)=\min\{1,2\}=1\). At \(x=1/2\), they are \(|1/2+1|=3/2\) and \(|1/2-2|=3/2\), so \(d_A(1/2)=\min\{3/2,3/2\}=3/2\).
The input distance is \(|0-1/2|=1/2\), and the output distance is \(|d_A(0)-d_A(1/2)|=|1-3/2|=1/2\). Hence the inequality reads \(1/2\le1\cdot(1/2)\), with equality. The general theorem guarantees the same inequality for every pair of real inputs, not just these two.
Bounds for Linear Maps
For vector spaces equipped with norms, the norm supplies a metric: \(d(u,v)=\|u-v\|\). A linear map can therefore be studied using the same Lipschitz inequality. Suppose \(V,W\) are normed vector spaces, \(T:V\to W\) is linear, and there is a finite \(C\ge0\) such that \(\|Tv\|_W\le C\|v\|_V\) for every \(v\in V\). Linearity gives \(T(u)-T(v)=T(u-v)\), so $$ \|T(u)-T(v)\|_W=\|T(u-v)\|_W\le C\|u-v\|_V. $$ Thus \(C\) is a Lipschitz bound for \(T\), and the continuity theorem applies. The needed hypothesis is the stated norm bound; linearity alone does not supply a bound that is uniform over the whole domain.
Worked Example: A Diagonal Linear Map on the Plane
Let \(\mathbb{R}^2\) have the Euclidean metric from the norm \(\|(x,y)\|=\sqrt{x^2+y^2}\). For \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(2x,y/2)\), find a Lipschitz bound and check whether it can be reduced.
For inputs \(u=(x_1,y_1)\) and \(v=(x_2,y_2)\), write \(a=x_1-x_2\) and \(b=y_1-y_2\). Then $$ \|T(u)-T(v)\|^2=\|(2a,b/2)\|^2=4a^2+\frac{b^2}{4}. $$ Since \(b^2/4\le4b^2\) and \(4a^2=4a^2\), this is at most \(4a^2+4b^2=4\|u-v\|^2\). Both sides are nonnegative, so taking square roots gives \(\|T(u)-T(v)\|\le2\|u-v\|\). Thus \(L=2\) is a valid bound.
To check that no smaller bound works, take \(u=(1,0)\) and \(v=(0,0)\). Then \(\|u-v\|=\sqrt{1^2+0^2}=1\), while \(T(u)=(2,0)\), \(T(v)=(0,0)\), and \(\|T(u)-T(v)\|=\sqrt{2^2+0^2}=2\). Any Lipschitz bound for this pair must be at least \(2/1=2\). Therefore \(2\) is the smallest possible bound.
Common Pitfall: A Local Estimate Is Not Automatically Global
A frequent error is to find a useful estimate near a point and then claim it is a global Lipschitz bound. For example, factoring a difference of squares gives \(|x^2-y^2|=|x-y||x+y|\). The factor \(|x+y|\) is not bounded as \(x,y\) range over all of \(\mathbb{R}\).
Wrong calculation: claiming that \(f(x)=x^2\) has global Lipschitz bound \(L=2\). At \(x=3\) and \(y=0\), the input distance is \(|3-0|=3\), but the output distance is \(|9-0|=9\); the proposed bound would require \(9\le2\cdot3=6\), which is false. In fact, no finite global Lipschitz bound \(L\) exists: using \(x=n\) and \(y=0\) for each \(n\in\mathbb{N}\), such a bound would require \(n^2\le Ln\). Since \(n>0\), this implies \(n\le L\) for every positive integer \(n\), which no finite \(L\) can satisfy.
Correction on a restricted domain: if \(x,y\in[-2,2]\), then \(|x+y|\le|x|+|y|\le2+2=4\). Consequently, $$ |x^2-y^2|=|x-y||x+y|\le4|x-y|. $$ So the restriction of \(x\mapsto x^2\) to \([-2,2]\), with the metric inherited from \(\mathbb{R}\), is Lipschitz with bound \(4\). The domain restriction matters: it provides the bound on the extra factor that the global calculation lacks.
Lipschitz estimates are useful whenever one needs quantitative control rather than only a continuity conclusion. For a particular map, first identify the two metrics, then seek a single finite multiplier that works for every pair of inputs. For a linear map, this amounts to bounding its effect on vector lengths; for distance-to-set functions, the triangle inequality provides the estimate directly. In each case, checking the domain and the quantifiers prevents a local or pair-specific calculation from being mistaken for a global bound.
Check Your Understanding
For each question, specify the metric spaces involved and verify the required inequality for all inputs, not only a selected pair.
- For \(f:\mathbb{R}\to\mathbb{R}\) given by \(f(x)=-5x+1\), find a Lipschitz bound and the corresponding choice of \(\delta\) for a given \(\varepsilon>0\).
- Is the map \(f(x)=x/2\) a contraction from \(\mathbb{R}\) to \(\mathbb{R}\)? What feature of the definition is important to your answer?
- Let \((X,d)\) be a metric space and \(a\in X\). Prove that \(x\mapsto d(x,a)\) is 1-Lipschitz by applying the distance-to-set result to \(A=\{a\}\).
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(x,3y)\) with Euclidean metrics, find a Lipschitz bound and test whether it is sharp.
- Explain why the factor \(|x+y|\) in \(|x^2-y^2|=|x-y||x+y|\) gives a finite bound on \([-2,2]\) but not on all of \(\mathbb{R}\).