Lipschitz Implies Uniform Continuity

One Bound for Every Pair of Inputs

In the previous tutorial, Lipschitz Continuity, we measured the change in function values by a fixed multiple of the distance between inputs. Uniform continuity also requires one distance threshold to work throughout the domain, but its definition does not specify how to find that threshold. A Lipschitz bound supplies it directly: once the desired output tolerance is given, the Lipschitz constant tells us how close the inputs must be.

This implication does not depend on the domain being an interval, bounded, or closed. It works for every subset of \(\mathbb{R}\), including an unbounded set or one with gaps. The reason is that the Lipschitz inequality already controls every pair of points in the domain, no matter where those points are.

Theorem (Lipschitz Implies Uniform Continuity): Let \(E\subseteq\mathbb{R}\), and suppose \(f:E\to\mathbb{R}\) is Lipschitz with constant \(L\geq0\). Then \(f\) is uniformly continuous on \(E\). If \(L>0\), then for every \(\varepsilon>0\), the choice \(\delta=\varepsilon/L\) works. If \(L=0\), any \(\delta>0\) works.

Proof. First suppose \(L>0\). Fix \(\varepsilon>0\), and choose \(\delta=\varepsilon/L\), which is positive. Let \(x,y\in E\) satisfy \(|x-y|<\delta\). The Lipschitz inequality gives

|f(x)-f(y)|\leq L|x-y|<L\delta =L\frac{\varepsilon}{L} =\varepsilon.

The same \(\delta\) applies to every pair \(x,y\in E\), so \(f\) is uniformly continuous. Now suppose \(L=0\). For every \(x,y\in E\), the Lipschitz inequality gives

0\leq |f(x)-f(y)|\leq 0|x-y|=0.

Therefore \(f(x)=f(y)\) for every pair in \(E\). In particular, for any \(\varepsilon>0\) and any \(\delta>0\), the condition \(|x-y|<\delta\) implies \(|f(x)-f(y)|=0<\varepsilon\). Thus \(f\) is uniformly continuous in this case as well. \(\square\)

The theorem gives a quantitative choice, not merely an existence proof. A larger Lipschitz constant means that the function is allowed to change more rapidly, so the radius \(\varepsilon/L\) becomes smaller. Any positive \(\delta\leq\varepsilon/L\) also works when \(L>0\), because then \(L|x-y|<L\delta\leq\varepsilon\). The endpoint in that choice causes no difficulty: the hypothesis uses the strict inequality \(|x-y|<\delta\).

Using the Estimate on Different Domains

Worked Example: A Reciprocal Function on an Unbounded Domain

Let \(f(x)=1/(x+2)\) for \(x\in[0,\infty)\). For \(x,y\geq0\), both \(x+2\) and \(y+2\) are at least \(2\). Subtracting the function values gives

|f(x)-f(y)| =\left|\frac{1}{x+2}-\frac{1}{y+2}\right| =\frac{|y-x|}{(x+2)(y+2)} \leq \frac{|x-y|}{4}.

Thus \(f\) is Lipschitz with constant \(1/4\). Given \(\varepsilon>0\), choose \(\delta=4\varepsilon\). If \(x,y\in[0,\infty)\) and \(|x-y|<\delta\), then

|f(x)-f(y)| \leq\frac{|x-y|}{4} <\frac{\delta}{4} =\varepsilon.

This proves uniform continuity on the whole unbounded domain. No upper bound on \(x\) or \(y\) was needed; the denominators only become larger as the inputs increase.

Worked Example: A Cubic on a Bounded Interval

Consider \(g(x)=x^3\) on \([-1,1]\). For \(x,y\in[-1,1]\), factor the difference:

$$ |g(x)-g(y)| =|x^3-y^3| =|x-y||x^2+xy+y^2|. $$

The bounds \(|x|\leq1\) and \(|y|\leq1\) imply \(x^2\leq1\), \(y^2\leq1\), and \(|xy|\leq1\). Hence

|x^2+xy+y^2| \leq x^2+|xy|+y^2 \leq 3,

so \(|g(x)-g(y)|\leq3|x-y|\). Given \(\varepsilon>0\), take \(\delta=\varepsilon/3\). Whenever \(x,y\in[-1,1]\) and \(|x-y|<\delta\), it follows that

|g(x)-g(y)|\leq3|x-y|<3\delta=\varepsilon.

The factorization supplies a Lipschitz bound, and the theorem converts it into a uniform continuity radius valid across the interval.

Worked Example: A Lipschitz Function on a Disconnected Set

Let \(E=(-\infty,-2]\cup[2,\infty)\), and define \(h(x)=0\) for \(x\leq-2\) and \(h(x)=1\) for \(x\geq2\). If \(x\) and \(y\) lie in the same piece, then \(h(x)=h(y)\), so \(|h(x)-h(y)|=0\). If they lie in different pieces, we may label them so that \(x\leq-2\) and \(y\geq2\). Then \(y-x\geq4\), and

|h(x)-h(y)|=1 \leq\frac{1}{4}|x-y|.

Thus \(h\) is Lipschitz on \(E\) with constant \(1/4\). For any \(\varepsilon>0\), the choice \(\delta=4\varepsilon\) proves uniform continuity by the theorem. In fact, if \(\delta\leq4\), two points in different pieces cannot satisfy \(|x-y|<\delta\), so nearby pairs necessarily have equal function values. This illustrates how the shape of the domain can contribute to the behavior of a function.

A Shared Bound for a Family of Functions

The same reasoning can be applied to several functions at once. The key requirement is a common Lipschitz constant. If each function has its own constant but those constants have no finite common bound, the calculation above does not provide one radius that works for all of them.

Theorem (Common Radius for a Lipschitz Family): Let \(E\subseteq\mathbb{R}\), let \(I\) be an index set, and suppose \(f_i:E\to\mathbb{R}\) is Lipschitz with the same constant \(L\geq0\) for every \(i\in I\). For every \(\varepsilon>0\), there is a \(\delta>0\) such that for every \(i\in I\) and every \(x,y\in E\), \(|x-y|<\delta\) implies \(|f_i(x)-f_i(y)|<\varepsilon\). If \(L>0\), one may take \(\delta=\varepsilon/L\); if \(L=0\), any positive \(\delta\) works.

Proof. If \(I\) is empty, the assertion has no functions to check, and any positive \(\delta\) works. Otherwise, if \(L>0\), fix \(\varepsilon>0\) and set \(\delta=\varepsilon/L\). Take any index \(i\in I\) and any \(x,y\in E\) with \(|x-y|<\delta\). The assumed common Lipschitz bound gives

|f_i(x)-f_i(y)| \leq L|x-y| <L\delta =\varepsilon.

The choice of \(\delta\) depends on \(\varepsilon\) and the shared constant \(L\), but not on \(i\), \(x\), or \(y\). If \(L=0\), each function has equal values at every pair of points in \(E\), by the zero-constant argument in the previous proof. Any positive \(\delta\) then works for every index. This proves the result. \(\square\)

Worked Example: Linear Functions with Bounded Slopes

For each parameter \(a\in[0,3]\), define \(f_a(x)=ax\) on \(\mathbb{R}\). For every \(x,y\in\mathbb{R}\),

|f_a(x)-f_a(y)| =|ax-ay| =|a||x-y| \leq3|x-y|.

All functions in this family share the Lipschitz constant \(3\). Consequently, for any \(\varepsilon>0\), \(\delta=\varepsilon/3\) works for every \(a\in[0,3]\). By contrast, the larger family \(q_n(x)=nx\), where \(n\) ranges over the positive integers, has no finite common Lipschitz constant. For \(\varepsilon=1\), consider the pair \(x=0\), \(y=1/n\). Its input distance tends to zero as \(n\) increases, but

$$ |q_n(1/n)-q_n(0)| =|1-0| =1. $$

Thus no single positive \(\delta\) works for every \(q_n\): given any \(\delta>0\), choose a positive integer \(n>1/\delta\). Then \(|1/n-0|<\delta\), while the output distance is \(1\), not less than \(\varepsilon=1\).

Why the Implication Does Not Reverse

Lipschitz continuity is a sufficient condition for uniform continuity, but it is stronger. A uniformly continuous function need not have its output changes bounded by a constant times the input changes. The square-root function on \([0,1]\) provides a direct example.

Worked Example: Uniform Continuity Without a Lipschitz Bound

Let \(s(x)=\sqrt{x}\) on \([0,1]\). For \(x,y\geq0\), assume first that \(x\geq y\). Then \(\sqrt{x}\geq\sqrt{y}\), and

|\sqrt{x}-\sqrt{y}|^2 =x+y-2\sqrt{xy} \leq x-y.

Indeed, the last inequality is equivalent to \(2y\leq2\sqrt{xy}\), which holds because \(y\leq x\) implies \(y^2\leq xy\), and hence \(y\leq\sqrt{xy}\). If \(y\geq x\), the same argument with the variables interchanged gives \(|\sqrt{x}-\sqrt{y}|^2\leq y-x\). Therefore, for all \(x,y\in[0,1]\),

|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}.

Given \(\varepsilon>0\), choose \(\delta=\varepsilon^2\). If \(|x-y|<\delta\), then

|\sqrt{x}-\sqrt{y}| \leq\sqrt{|x-y|} <\sqrt{\delta} =\varepsilon.

So \(s\) is uniformly continuous. But it is not Lipschitz on \([0,1]\). For each \(t\in(0,1]\), the pair \(x=t^2\), \(y=0\) gives the ratio

\frac{|s(t^2)-s(0)|}{|t^2-0|} =\frac{t}{t^2} =\frac{1}{t}.

These ratios are unbounded as \(t\) approaches zero. Therefore no finite Lipschitz constant can work, even though the explicit uniform continuity estimate does work.

A common mistake is to regard the choice \(\delta=\varepsilon/L\) as a formula that applies unchanged when \(L=0\). Division by zero is not defined; the zero case must be handled separately, as in the theorem. Another is to assume that a uniform continuity radius depends on the location of the inputs. Here it depends only on the output tolerance and the Lipschitz constant. This is precisely why the estimate works on an entire domain, including domains with no endpoints or finite bound.

When proving the implication, identify a valid Lipschitz constant first; it need not be the least one. Then use the same constant to choose a radius for each \(\varepsilon\), and verify the resulting chain of inequalities for arbitrary points in the domain. The estimate is stronger than pointwise continuity because it controls all pairs at once. The square-root example also shows the proper scope of the result: a linear bound on output change is sufficient for uniform continuity, but uniform continuity can hold with a weaker, nonlinear bound.

Check Your Understanding

Use the Lipschitz inequality and the examples in this tutorial to answer the following questions.

Why does \(\delta=\varepsilon/L\) work when \(L>0\), and where is the strict inequality used?
What does a Lipschitz bound with \(L=0\) imply about the values of the function on its domain?
In the reciprocal-function example, which feature of the domain ensures that the denominator product has a fixed positive lower bound?
Why can the piecewise constant function on the disconnected domain be Lipschitz even though it takes two different values?
What assumption lets one radius work for every function in a family?
How does the square-root example show that uniform continuity does not imply Lipschitz continuity?