Continuity Proof Workshop

Proofs Need One Radius for the Whole Domain

Uniform continuity proofs often turn on a single question: does the estimate control every relevant pair of inputs, or only pairs near one fixed point? Ordinary continuity at a point permits the radius to depend on that point. Uniform continuity requires a radius that works for all pairs in the domain. Keeping this distinction visible is especially important when working with sequences of functions: a uniform approximation can transfer a global estimate, but pointwise continuity estimates alone cannot.

Earlier in this course, we established that a uniform limit of uniformly continuous functions is uniformly continuous. Here we focus on a useful quantitative refinement and on the proof habits that make such arguments reliable. The main tools will be the triangle inequality, an error bound independent of the input, and a global estimate for increments.

A Uniform Perturbation Estimate

Suppose two functions are uniformly close, and one already has a useful estimate for how much its values can change between nearby inputs. The triangle inequality compares the change in the other function with this known change plus the approximation errors at the two inputs. The approximation error appears twice, which is why the bound must account for both endpoints.

Theorem: Let \(E\subseteq\mathbb{R}\), and let \(f,g:E\to\mathbb{R}\). Suppose \(|f(x)-g(x)|\leq\eta\) for every \(x\in E\), where \(\eta\geq0\). Then, for every \(x,y\in E\), \[ |f(x)-f(y)|\leq 2\eta+|g(x)-g(y)|. \]

Proof. Fix \(x,y\in E\). Insert \(g(x)\) and \(g(y)\) between the two values of \(f\), and apply the triangle inequality:

|f(x)-f(y)| \leq |f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|.

The first and third terms are each at most \(\eta\), by the assumed uniform error bound. Therefore

|f(x)-f(y)|\leq \eta+|g(x)-g(y)|+\eta =2\eta+|g(x)-g(y)|.

Since \(x\) and \(y\) were arbitrary, the estimate holds for every pair in \(E\). \(\square\)

This estimate gives a practical proof method. To prove that \(f\) is uniformly continuous, one may approximate it uniformly by a function \(g\) whose changes are easy to control. Given a target error \(\varepsilon>0\), first make the two approximation errors together less than, for example, \(2\varepsilon/3\). Then use uniform continuity of \(g\) to make \(|g(x)-g(y)|<\varepsilon/3\) for every pair with \(|x-y|<\delta\). The resulting bound is less than \(\varepsilon\), with the same \(\delta\) working throughout the domain.

Worked Examples: Building Global Estimates

Worked Example: A Bounded Rational Function

Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=\frac{|x|}{1+|x|}\). We prove uniform continuity by finding a global Lipschitz estimate. Set \(u=|x|\) and \(v=|y|\). Since \(u,v\geq0\),

\left|\frac{u}{1+u}-\frac{v}{1+v}\right| =\frac{|u-v|}{(1+u)(1+v)} \leq |u-v|.

The denominator is at least \(1\), which justifies the inequality. Also, the reverse triangle inequality gives \(\big||x|-|y|\big|\leq|x-y|\). Combining these two facts,

|f(x)-f(y)|\leq\big||x|-|y|\big|\leq|x-y|.

Thus \(f\) is Lipschitz with constant \(1\), and hence uniformly continuous. Explicitly, given \(\varepsilon>0\), choosing \(\delta=\varepsilon\) ensures that \(|x-y|<\delta\) implies \(|f(x)-f(y)|<\varepsilon\), for all real \(x,y\).

Worked Example: A Polynomial on a Bounded Interval

Let \(p(x)=x^2\) on \(E=[-3,3]\). For \(x,y\in E\), factor the difference and bound both factors involving the inputs:

|p(x)-p(y)| =|x^2-y^2| =|x-y||x+y| \leq 6|x-y|,

because \(|x+y|\leq|x|+|y|\leq3+3=6\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/6\). Then \(|x-y|<\delta\) gives

|p(x)-p(y)|\leq6|x-y|<6\delta=\varepsilon.

This proves uniform continuity on the interval. The boundedness of the inputs is doing real work: the factor \(|x+y|\) has a single bound on this domain.

A Common Lipschitz Bound Passes to the Limit

Uniform convergence provides a bound on approximation error that holds at every input. If, in addition, the approximating functions all obey the same Lipschitz estimate, that estimate survives in the limit. This is more quantitative than knowing only that the limit is uniformly continuous: the limit inherits the very same Lipschitz constant.

Theorem: Let \(E\subseteq\mathbb{R}\) be nonempty, and suppose \(f_n:E\to\mathbb{R}\) converges uniformly to \(f:E\to\mathbb{R}\). If there is a constant \(L\geq0\) such that \(|f_n(x)-f_n(y)|\leq L|x-y|\) for every \(n\) and every \(x,y\in E\), then \(f\) is Lipschitz on \(E\) with constant \(L\).

Proof. By the uniform convergence error characterization from “Uniform Convergence Preview,” the numbers

e_n=\sup_{z\in E}|f_n(z)-f(z)|

tend to \(0\). Fix any \(x,y\in E\). For every positive integer \(n\), the triangle inequality and the common Lipschitz bound give

\begin{aligned} |f(x)-f(y)| &\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|\\ &\leq e_n+L|x-y|+e_n\\ &=2e_n+L|x-y|. \end{aligned}

The left-hand side and \(L|x-y|\) do not depend on \(n\), while \(e_n\to0\). Taking the limit as \(n\to\infty\) yields

|f(x)-f(y)|\leq L|x-y|.

The inputs \(x,y\) were arbitrary, so this inequality holds throughout \(E\). It is exactly the Lipschitz condition with constant \(L\). \(\square\)

The uniformity in the hypotheses matters twice: the error bound \(e_n\) controls both endpoints regardless of their location, and the Lipschitz constant \(L\) does not change with \(n\). If the Lipschitz constants grow without bound, this argument provides no fixed global estimate for the limit.

Worked Example: A Uniform Limit with a Shared Lipschitz Bound

For each positive integer \(n\), define \(f_n:\mathbb{R}\to\mathbb{R}\) by

f_n(x)=\frac{x}{1+x^2}+\frac{\sin(nx)}{n}.

We identify the uniform limit and verify a common Lipschitz bound. First, for all \(x\in\mathbb{R}\),

\left|f_n(x)-\frac{x}{1+x^2}\right| =\frac{|\sin(nx)|}{n} \leq\frac1n.

The bound is independent of \(x\) and tends to zero, so \(f_n\) converges uniformly to \(f(x)=\frac{x}{1+x^2}\).

To find a common Lipschitz constant, write \(q(x)=\frac{x}{1+x^2}\). Direct subtraction gives

|q(x)-q(y)| =\frac{|x-y||1-xy|}{(1+x^2)(1+y^2)}.

The denominator is at least \(1\), and \(|1-xy|\leq 1+|xy|\). Moreover, \(|xy|\leq(1+x^2)(1+y^2)\), while \(1\leq(1+x^2)(1+y^2)\). Hence \(|1-xy|\leq2(1+x^2)(1+y^2)\), and therefore \(|q(x)-q(y)|\leq2|x-y|\). The sine estimate \(|\sin u-\sin v|\leq|u-v|\) gives

\left|\frac{\sin(nx)}n-\frac{\sin(ny)}n\right| \leq\frac{|nx-ny|}{n}=|x-y|.

Consequently, for every \(n\) and every \(x,y\),

|f_n(x)-f_n(y)|\leq2|x-y|+|x-y|=3|x-y|.

The theorem shows that the limit \(f\) is Lipschitz with constant \(3\). In fact, the calculation already gives a constant \(2\) for \(q\); the theorem’s conclusion with \(3\) is valid, even though it is not the sharpest bound. The key point is that a single constant controls every \(f_n\), not merely that each separate \(f_n\) is continuous.

The Radius Must Come from the Right Hypothesis

A frequent error in global arguments is to fix an index \(n\), invoke continuity of \(f_n\), and then claim that its continuity estimate works for every pair in the domain. Ordinary continuity does not supply such a radius. At a point \(a\), continuity gives a radius that controls pairs involving \(a\) (or, in the usual definition, inputs near \(a\)); the radius may depend on \(a\). Uniform continuity is the hypothesis that supplies one radius for all pairs \(x,y\) in the domain.

For a concrete check, \(x^2\) is continuous on \(\mathbb{R}\), but is not uniformly continuous there. Given any \(\delta>0\), set \(h=\min\{\delta/2,1\}\), so \(0<h<\delta\). Choose \(x>1/h\) and let \(y=x+h\). Then \(|x-y|=h<\delta\), while

$$ |x^2-y^2|=|(x-y)(x+y)|=h(2x+h)>2. $$

Thus no positive \(\delta\) can force the change in \(x^2\) to be less than \(1\) for every pair of inputs at distance less than \(\delta\). In a proof involving a fixed \(f_n\), the global radius must come from uniform continuity of that \(f_n\), or from a separate estimate that is genuinely uniform in the inputs. A pointwise continuity radius cannot be substituted.

A reliable proof check is to identify what each choice may depend on. In a uniform continuity proof, \(\delta\) may depend on \(\varepsilon\) and on the function or domain, but not on the pair \(x,y\). In a uniform convergence estimate, the index may depend on \(\varepsilon\), but not on the input. Keeping these dependencies explicit helps catch arguments that establish only pointwise control.

Check Your Understanding

Use the estimates and proof principles in this workshop to answer the following questions.

Why does the uniform perturbation estimate contain two copies of the error bound \(\eta\)?
In the common-Lipschitz-limit theorem, where is uniform convergence used in the proof?
What role does the shared constant \(L\) play, and what goes wrong if the constants depend on \(n\) and grow without bound?
Why is continuity of a fixed function \(f_n\) not enough to produce one radius for every pair in its domain?
For \(p(x)=x^2\) on \([-3,3]\), which factorization gives the global estimate needed to choose \(\delta\)?

What You'll Learn