Uniform Continuity of the Square Root

A Uniform Estimate on an Unbounded Domain

The previous tutorial showed that polynomials are Lipschitz on bounded sets by controlling the input sizes in a difference estimate. The square-root function presents a different situation. Its domain \([0,\infty)\) is unbounded, so a bound based on keeping inputs in a fixed bounded set would not establish uniform continuity on the whole domain. Instead, a direct comparison of two square roots gives an estimate that works for every pair of nonnegative inputs.

The function \(s(x)=\sqrt{x}\) is continuous at every point of its domain, as established in “Continuity of the Square-Root Function.” Uniform continuity asks for more: for each output tolerance, one input tolerance must work regardless of where the two inputs lie. The estimate below provides exactly that common control.

The Square-Root Difference Estimate

For nonnegative \(x\) and \(y\), the square roots are nonnegative as well. If \(x\geq y\), then \(\sqrt{x}\geq\sqrt{y}\), and

The last inequality follows because \(\sqrt{x}+\sqrt{y}\geq\sqrt{x}-\sqrt{y}\geq0\). Thus the square of the change in the square root is no larger than the change in the input. Reversing the roles of \(x\) and \(y\) when necessary gives the estimate for every pair.

Proof. If \(x\geq y\), both sides are nonnegative and the calculation above gives \(|\sqrt{x}-\sqrt{y}|^2\leq x-y=|x-y|\). Taking nonnegative square roots yields the claimed inequality. If \(y\geq x\), the same argument with \(x\) and \(y\) interchanged gives \(|\sqrt{x}-\sqrt{y}|^2\leq y-x=|x-y|\), and taking nonnegative square roots again gives the result. These two cases include equality of the inputs, so the estimate holds for every \(x,y\in[0,\infty)\). \(\square\)

The estimate has the form of a positive-power bound: the output difference is at most the square root of the input difference. In particular, small input distances force small output distances uniformly, even when the inputs are very large or very close to zero.

Uniform Continuity on the Whole Domain

Proof. Fix \(\varepsilon>0\) and choose \(\delta=\varepsilon^2\), which is positive. Let \(x,y\in[0,\infty)\) satisfy \(|x-y|<\delta\). By the Square-Root Difference Estimate,

This \(\delta\) depends only on \(\varepsilon\), not on \(x\) or \(y\). Therefore the defining condition for uniform continuity holds on the whole domain. \(\square\)

The choice \(\delta=\varepsilon^2\) may be much smaller than necessary for a particular pair of inputs, but it is a single valid choice for all pairs. The domain's unboundedness causes no problem because the square-root estimate contains no factor that grows with the size of the inputs.

The Estimate Is Useful Near Zero

A familiar rationalization identity for \(x\ne y\) is

This identity is correct for nonnegative \(x,y\) with \(x\ne y\), but the denominator can be arbitrarily small when both inputs are close to zero. Consequently, it does not by itself provide a single Lipschitz constant on \([0,\infty)\). The square-root difference estimate avoids dividing by a quantity that may approach zero.

Uniform Continuity Does Not Require a Lipschitz Bound

A function is Lipschitz if there is a finite constant \(L\geq0\) such that \(|f(x)-f(y)|\leq L|x-y|\) for every pair in its domain. A Lipschitz function is uniformly continuous, but the square-root function on \([0,\infty)\) illustrates that uniform continuity can hold without a global Lipschitz bound.

Proof. Suppose such an \(L\) existed. For each positive integer \(n\), choose \(x=1/n^2\) and \(y=0\). The proposed Lipschitz inequality would imply

Multiplying by \(n^2>0\) gives \(n\leq L\) for every positive integer \(n\). No finite real number \(L\) bounds all positive integers, a contradiction. Thus no global Lipschitz constant exists. \(\square\)

There is no conflict between this result and uniform continuity. The square-root estimate gives a bound involving \(\sqrt{|x-y|}\), rather than a constant times \(|x-y|\). Since \(\sqrt{h}\) tends to zero as \(h\) tends to zero through nonnegative values, this weaker rate of control is still enough for uniform continuity.

Reading the Two Estimates Correctly

The two estimates answer different questions. The global bound \(|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}\) applies to every pair of nonnegative inputs and proves uniform continuity on the entire domain. The rationalized bound can give a stronger, linear estimate when the inputs are known to stay away from zero. Near zero, its denominator can become small, while the square-root estimate remains valid without an additional condition.

A common pitfall is to see that the square-root function is continuous at every point and conclude, without further argument, that it must be uniformly continuous. Pointwise continuity alone does not give a common radius across an unbounded domain. Here the uniform conclusion follows from a specific estimate whose constants do not depend on the location of the inputs. Another pitfall is to treat failure of a global Lipschitz inequality as failure of uniform continuity. The square root shows why those properties must be distinguished: it is uniformly continuous but not globally Lipschitz.