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Uniform Continuity · Tutorial 390 of 1000

Continuity Mastery Examination

Work through representative uniform continuity problems and learn a compactness argument that turns continuity of a periodic function into uniform continuity on the real line.

Intermediate 11 min read

What You'll Learn

  • Choose a common input scale when proving uniform continuity directly
  • Use sequences of nearby inputs to disprove uniform continuity
  • Apply compact-domain results without overlooking their hypotheses
  • Prove that every continuous periodic function on the real line is uniformly continuous
  • Show that every continuous periodic function on the real line is bounded
  • Distinguish pointwise continuity from a uniform estimate across the domain

A Cumulative Examination in Uniform Continuity

Uniform continuity problems often turn on the same few decisions: whether to construct one input scale that works everywhere, whether a sequence can expose failure, and whether compactness is actually available. This examination brings those decisions together. The aim is not to repeat the definitions in isolation, but to select an appropriate tool, check its hypotheses, and make each estimate explicit.

Recall that \(f:E\to\mathbb{R}\) is uniformly continuous if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x,y\in E\) and \(|x-y|<\delta\) imply \(|f(x)-f(y)|<\varepsilon\). The same \(\delta\) must work for every pair in the domain. The Sequential Criterion for Uniform Continuity gives a complementary test: failure can be demonstrated by pairs of domain points whose input distances tend to zero while their output distances do not tend to zero.

The problems below use estimates, sequences, compactness, and periodicity. In each proof, the choice of method follows from the shape of the domain and the function. An estimate that works on a bounded interval need not work on the entire real line, and continuity alone does not supply one common radius on an unbounded domain.

Worked Problems: Building Uniform Estimates

Worked Example: A Polynomial on a Bounded Interval

Show directly that \(f(x)=x^2\) is uniformly continuous on \([-2,3]\). For \(x,y\) in this interval, factor the difference:

$$ |x^2-y^2|=|x-y||x+y|. $$

Since \(-2\leq x,y\leq3\), we have \(|x|\leq3\) and \(|y|\leq3\). Therefore \(|x+y|\leq|x|+|y|\leq6\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/6\). Whenever \(x,y\in[-2,3]\) and \(|x-y|<\delta\),

$$ |f(x)-f(y)|=|x-y||x+y|<6\delta=\varepsilon. $$

This proves uniform continuity on the interval. The bound comes from the restricted domain: it controls the factor \(|x+y|\) uniformly. It is not a proof that \(x^2\) is uniformly continuous on all of \(\mathbb{R}\), where no such bound on \(|x+y|\) is available.

Worked Example: A Reciprocal on an Unbounded Domain

Show that \(f(x)=1/(1+x)\) is uniformly continuous on \([0,\infty)\). For \(x,y\geq0\), both denominators are at least \(1\), so

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

Given \(\varepsilon>0\), take \(\delta=\varepsilon\). If \(x,y\in[0,\infty)\) and \(|x-y|<\delta\), the displayed estimate gives \(|f(x)-f(y)|\leq|x-y|<\varepsilon\). Thus \(f\) is uniformly continuous. The domain is unbounded, but that is not an obstacle: the denominator provides a global estimate independent of the locations of \(x\) and \(y\).

Worked Example: A Sequential Disproof on an Unbounded Domain

Show that \(f(x)=x^2\) is not uniformly continuous on \([0,\infty)\). Define \(x_n=n\) and \(y_n=n+1/n\) for positive integers \(n\). Both sequences lie in the domain, and

$$ |y_n-x_n|=\frac{1}{n}\longrightarrow0. $$

However, direct calculation gives

$$ |f(y_n)-f(x_n)| =\left(n+\frac{1}{n}\right)^2-n^2 =2+\frac{1}{n^2} \geq2. $$

The output differences therefore do not tend to zero. By the Sequential Criterion for Uniform Continuity, \(f\) is not uniformly continuous on \([0,\infty)\). The choice of inputs is important: their separation shrinks, while the factorization \(|y_n^2-x_n^2|=|y_n-x_n||y_n+x_n|\) shows how the increasing size of the inputs offsets that shrinkage.

The first two examples prove uniform continuity by finding a global estimate. The third disproves it by finding pairs that violate the sequential test. These are complementary strategies. When a difference factors into an input-distance term and another term, ask whether the second term can be bounded on the domain. If it cannot, sequences with increasingly large inputs may reveal the failure.

Continuous Periodic Functions on the Real Line

Compactness supplies another route to uniform continuity. A continuous function on a compact interval is uniformly continuous by the Heine-Cantor Theorem for Intervals. A periodic function repeats the behavior on a bounded interval throughout the real line. The following theorem makes that observation precise; it is a useful way to establish uniform continuity when a global algebraic estimate is inconvenient.

Theorem (Continuous Periodic Functions Are Uniformly Continuous): Suppose \(f:\mathbb{R}\to\mathbb{R}\) is continuous and has period \(T>0\), meaning \(f(x+T)=f(x)\) for every \(x\in\mathbb{R}\). Then \(f\) is uniformly continuous on \(\mathbb{R}\).

Proof. The function is continuous on the compact interval \([-1,T+1]\), so the Heine-Cantor Theorem for Intervals implies that its restriction to this interval is uniformly continuous. Given \(\varepsilon>0\), choose \(\delta_0>0\) such that \(u,v\in[-1,T+1]\) and \(|u-v|<\delta_0\) imply \(|f(u)-f(v)|<\varepsilon\). Set \(\delta=\min(\delta_0,1)\).

Take any \(x,y\in\mathbb{R}\) with \(|x-y|<\delta\). There is an integer \(k\) such that \(u=x-kT\in[0,T)\). Set \(v=y-kT\). Since the same quantity is subtracted from both inputs, \(|u-v|=|x-y|<\delta\leq1\). Thus \(v>u-1\geq-1\) and \(v<u+1<T+1\). Also \(u\in[0,T)\subseteq[-1,T+1]\), so both \(u\) and \(v\) lie in the compact interval. The choice of \(\delta_0\) gives \(|f(u)-f(v)|<\varepsilon\).

Periodicity implies \(f(x-kT)=f(x)\) and \(f(y-kT)=f(y)\). For positive integers \(k\), the first identity follows by applying \(f(z-T)=f(z)\) repeatedly; for negative integers it follows by applying \(f(z+T)=f(z)\) repeatedly. Hence \(f(u)=f(x)\) and \(f(v)=f(y)\), so \(|f(x)-f(y)|<\varepsilon\). The same \(\delta\) works for every pair \(x,y\in\mathbb{R}\), which proves uniform continuity. \(\square\)

The compact interval in the proof is slightly wider than one period. That margin is essential to the argument: after shifting \(x\) into \([0,T)\), the nearby point \(y\) may lie just below \(0\) or just above \(T\). The condition \(\delta\leq1\) ensures that both shifted points remain in \([-1,T+1]\), where the compact-interval estimate applies.

Worked Example: The Distance to the Integers

Define \(d(x)=\inf_{m\in\mathbb{Z}}|x-m|\), the distance from \(x\) to the set of integers. First, this function is continuous. For any \(x,y\in\mathbb{R}\) and any integer \(m\), the triangle inequality gives \(|x-m|\leq|x-y|+|y-m|\). Taking the infimum over integers \(m\) yields \(d(x)\leq|x-y|+d(y)\). Interchanging \(x\) and \(y\) gives

$$ |d(x)-d(y)|\leq|x-y|. $$

This estimate directly proves continuity. Also, translating by an integer does not change the distance to the integers: for \(m\in\mathbb{Z}\),

$$ d(x+1)=\inf_{n\in\mathbb{Z}}|x+1-n| =\inf_{m\in\mathbb{Z}}|x-m| =d(x), $$

where \(m=n-1\) ranges over all integers as \(n\) does. Thus \(d\) has period \(1\). The Continuous Periodic Functions Theorem now proves that \(d\) is uniformly continuous on \(\mathbb{R}\). In this example the distance estimate is an even more direct proof, but checking continuity and periodicity illustrates how the theorem can be applied.

A Second Consequence of Periodicity

The compact-interval reduction also gives a global size bound. This is not true of every uniformly continuous function on the real line; for example, affine functions can be unbounded. Periodicity is the extra structure that restricts all values to those attained during one period.

Theorem (Continuous Periodic Functions Are Bounded): If \(f:\mathbb{R}\to\mathbb{R}\) is continuous and has period \(T>0\), then \(f\) is bounded on \(\mathbb{R}\).

Proof. By continuity and the Extreme Value Theorem, \(f\) is bounded on the compact interval \([0,T]\). Thus there is an \(M\geq0\) such that \(|f(r)|\leq M\) for every \(r\in[0,T]\). For any \(x\in\mathbb{R}\), choose an integer \(k\) so that \(r=x-kT\in[0,T)\). Periodicity gives \(f(x)=f(r)\), and therefore \(|f(x)|\leq M\). This holds for every real \(x\), proving that \(f\) is bounded on \(\mathbb{R}\). \(\square\)

The theorem is a global conclusion obtained from one compact interval. In the same way, the uniform continuity theorem used a compact interval to control pairs of nearby inputs after a common translation. The two arguments share the reduction to one period, but their conclusions differ: boundedness controls individual output values, while uniform continuity controls differences between outputs.

Choosing the Right Argument

A common mistake is to infer uniform continuity simply because a function is continuous at every point. The Heine-Cantor Theorem applies on compact domains, not arbitrary domains. Another mistake is to try to find a single \(\delta\) by repeating a pointwise continuity argument without checking whether the resulting radii can be chosen uniformly. On an unbounded domain, estimates that depend on \(|x|\) or \(|y|\) may not give one common radius.

For a proof, begin by looking for a direct bound on \(|f(x)-f(y)|\) in terms of \(|x-y|\). If the domain is compact and the function is continuous, Heine-Cantor may provide a shorter route. If the claim is that uniform continuity fails, use the sequential criterion and calculate both the input and output differences. For a continuous periodic function on the real line, reduce nearby inputs by the same period shift and then use compactness on a slightly enlarged period interval.

The examination’s central discipline is to verify the scope of each tool. A bound proved only on a bounded interval cannot be used on the whole real line. A sequential disproof needs output differences that genuinely fail to approach zero. A compactness argument needs a compact interval containing the shifted points. Once these checks are explicit, the proof strategy becomes easier to select and harder to misuse.

Check Your Understanding

For each question, identify the hypothesis or estimate that makes the argument valid.

  1. In the proof for \(x^2\) on \([-2,3]\), which factor is bounded uniformly, and why does that estimate not apply on all of \(\mathbb{R}\)?
  2. For \(f(x)=1/(1+x)\) on \([0,\infty)\), verify the difference estimate and state a valid choice of \(\delta\) for a given \(\varepsilon>0\).
  3. For the sequences \(x_n=n\) and \(y_n=n+1/n\), what happens to the input distances and the output distances under \(f(x)=x^2\)?
  4. Why does the proof for a continuous periodic function shift both inputs by the same integer multiple of the period?
  5. Where is compactness used in each of the two periodic-function theorems, and what conclusion does it provide?

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