Epsilon Proof Workshop

From the Definition to a Proof Plan

The previous tutorial used sequences to prove and disprove uniform continuity. This tutorial returns to the epsilon–delta definition and treats it as a practical proof-writing tool. The central task is to begin with an arbitrary output tolerance \(\varepsilon>0\) and find one positive input radius \(\delta\) that works for every pair of points in the domain. The choice of \(\delta\) may depend on \(\varepsilon\) and on the function, but it must not depend on the particular points being compared.

A useful way to plan a direct proof is to work backward from the desired conclusion. First write the output difference and seek an estimate in terms of \(|x-y|\). Then decide how small \(|x-y|\) must be to make the estimate less than \(\varepsilon\). In a proof, these steps are presented in the logical order: choose \(\delta\), assume \(|x-y|<\delta\), and verify that the output difference is less than \(\varepsilon\).

Proof-planning principle: For each \(\varepsilon>0\), find a positive \(\delta\), independent of \(x\) and \(y\), such that \(x,y\in E\) and \(|x-y|<\delta\) imply \(|f(x)-f(y)|<\varepsilon\). If an estimate gives \(|f(x)-f(y)|\leq C|x-y|\) for a fixed \(C>0\), the choice \(\delta=\varepsilon/C\) works. If \(C=0\), the output difference is always zero, so any positive \(\delta\) works.

The estimate is often the main mathematical work. Once it has been found, the radius is usually determined by solving a simple inequality. For example, if \(|f(x)-f(y)|\leq 4|x-y|\), it is enough to require \(|x-y|<\varepsilon/4\). A strict input inequality then gives \(4|x-y|<\varepsilon\), as required.

Three Direct Epsilon–Delta Examples

Worked Example: An Affine Function on the Real Line

Let \(f:\mathbb{R}\to\mathbb{R}\) be defined by \(f(x)=7x-4\). We prove directly that \(f\) is uniformly continuous. Let \(\varepsilon>0\) be given and choose \(\delta=\varepsilon/7\), which is positive. If \(x,y\in\mathbb{R}\) and \(|x-y|<\delta\), then

|f(x)-f(y)| =|(7x-4)-(7y-4)| =7|x-y| <7\delta =\varepsilon.

The same radius works for every \(x,y\in\mathbb{R}\). In particular, the cancellation of the constant term matters: the estimate depends only on the distance between the inputs, not on their location.

Worked Example: A Rational Function on a Half-Line

Define \(f:[0,\infty)\to\mathbb{R}\) by \(f(x)=x/(1+x)\). For \(x,y\geq0\), direct subtraction gives

\begin{aligned} |f(x)-f(y)| &=\left|\frac{x}{1+x}-\frac{y}{1+y}\right|\\ &=\frac{|x(1+y)-y(1+x)|}{(1+x)(1+y)}\\ &=\frac{|x-y|}{(1+x)(1+y)} \leq |x-y|. \end{aligned}

The last inequality holds because both \(1+x\) and \(1+y\) are at least \(1\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon\). Whenever \(x,y\in[0,\infty)\) and \(|x-y|<\delta\), the estimate gives

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

Thus \(f\) is uniformly continuous on its unbounded domain. The proof does not need a bound on the inputs; the denominator in the difference estimate supplies the needed control.

Worked Example: A Rational Function on a Bounded Interval

Let \(h:[0,1]\to\mathbb{R}\) be defined by \(h(x)=x/(1+x^2)\). For \(x,y\in[0,1]\), we have

\begin{aligned} |h(x)-h(y)| &=\left|\frac{x}{1+x^2}-\frac{y}{1+y^2}\right|\\ &=\frac{|x(1+y^2)-y(1+x^2)|}{(1+x^2)(1+y^2)}\\ &=\frac{|x-y|\,|1-xy|}{(1+x^2)(1+y^2)}. \end{aligned}

Since \(0\leq xy\leq1\), we have \(|1-xy|\leq1\). Also, each factor in the denominator is at least \(1\). Therefore \(|h(x)-h(y)|\leq|x-y|\). Given any \(\varepsilon>0\), take \(\delta=\varepsilon\). If \(x,y\in[0,1]\) and \(|x-y|<\delta\), then

|h(x)-h(y)|\leq|x-y|<\varepsilon.

This verifies uniform continuity with an explicit radius. Notice that the domain restriction is used in the estimate \(|1-xy|\leq1\); without checking that restriction, the displayed bound would not have been justified.

Managing Epsilon for a Sum

When the output difference is a sum of two changes, a common technique is to split the allowed error between them. If the total must be less than \(\varepsilon\), it is enough to make each contribution less than \(\varepsilon/2\). The two functions may require different radii, so the radius for the sum is chosen to satisfy both requirements.

Theorem (Sums of Uniformly Continuous Functions): Let \(E\subseteq\mathbb{R}\), and let \(f,g:E\to\mathbb{R}\) be uniformly continuous on \(E\). Then \(f+g\) is uniformly continuous on \(E\).

Proof. Let \(\varepsilon>0\). Uniform continuity of \(f\) gives a number \(\delta_f>0\) such that, for \(x,y\in E\), \(|x-y|<\delta_f\) implies \(|f(x)-f(y)|<\varepsilon/2\). Uniform continuity of \(g\) gives \(\delta_g>0\) such that \(|x-y|<\delta_g\) implies \(|g(x)-g(y)|<\varepsilon/2\). Set \(\delta=\min\{\delta_f,\delta_g\}\), which is positive. If \(x,y\in E\) and \(|x-y|<\delta\), then both estimates apply. The triangle inequality yields

\begin{aligned} |(f+g)(x)-(f+g)(y)| &\leq |f(x)-f(y)|+|g(x)-g(y)|\\ &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. \end{aligned}

Thus \(f+g\) is uniformly continuous on \(E\). The choice of the minimum is essential: it ensures that the single radius \(\delta\) meets both conditions. \(\square\)

This argument is a model for proofs in which several estimates must hold at once. Each condition produces a positive radius; taking the minimum of finitely many such radii preserves positivity and ensures all the conditions hold together. The error split can be adjusted when there are more terms: for a finite sum of \(m\) functions, one can assign each term an error tolerance of \(\varepsilon/m\).

Working from the Outside In for a Composition

For a composition, the desired output is controlled first by the outer function. The inner function must then make its own outputs close enough to meet the outer function’s requirement. This gives a reliable order of choices: start with \(\varepsilon\), obtain an intermediate tolerance from the outer function, and then obtain an input radius from the inner function.

Theorem (Composition of Uniformly Continuous Functions): Let \(E,F\subseteq\mathbb{R}\), let \(f:E\to F\), and let \(g:F\to\mathbb{R}\). If \(f\) is uniformly continuous on \(E\) and \(g\) is uniformly continuous on \(F\), then \(g\circ f\) is uniformly continuous on \(E\).

Proof. Let \(\varepsilon>0\). By uniform continuity of \(g\) on \(F\), there exists \(\eta>0\) such that for all \(u,v\in F\),

|u-v|<\eta \quad\Longrightarrow\quad |g(u)-g(v)|<\varepsilon.

By uniform continuity of \(f\) on \(E\), there exists \(\delta>0\) such that for all \(x,y\in E\),

|x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\eta.

Now take any \(x,y\in E\) with \(|x-y|<\delta\). Since \(f(x),f(y)\in F\), the choice of \(\eta\) applies to these two points and gives

|(g\circ f)(x)-(g\circ f)(y)| =|g(f(x))-g(f(y))| <\varepsilon.

The same \(\delta\) works for every such pair \(x,y\), so \(g\circ f\) is uniformly continuous on \(E\). \(\square\)

Worked Example: A Composition on the Real Line

Consider \(H:\mathbb{R}\to\mathbb{R}\) defined by \(H(x)=\sin(x/(1+x^2))\). We verify uniform continuity by applying the composition theorem. Define \(q(x)=x/(1+x^2)\). For arbitrary real \(x,y\), the denominator is positive, and

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

To obtain a global estimate, use \(|1-xy|\leq1+|xy|\) and \((1+x^2)(1+y^2)\geq1+x^2y^2\). If \(t=|xy|\geq0\), then \(1+t\leq2(1+t^2)\), since \(2(1+t^2)-(1+t)=1-t+2t^2>0\); the final quadratic is positive because its discriminant is \(1-8<0\) and its leading coefficient is positive. Thus \(|1-xy|/((1+x^2)(1+y^2))\leq2\), and consequently \(|q(x)-q(y)|\leq2|x-y|\). Given \(\eta>0\), choosing \(\delta=\eta/2\) proves that \(q\) is uniformly continuous.

Sine is uniformly continuous on \(\mathbb{R}\), as established earlier in this course. Since \(q\) maps \(\mathbb{R}\) into \(\mathbb{R}\), the composition theorem applies and shows that \(H=\sin\circ q\) is uniformly continuous on \(\mathbb{R}\).

Common Errors in Epsilon Proofs

A direct proof can look convincing while failing one of the quantifiers. Check these points before considering an argument complete:

The radius must be uniform. A choice such as \(\delta=\varepsilon/(1+|x|)\) depends on an input point, so it does not by itself prove uniform continuity on the domain.
The radius must be positive. When combining several requirements, take the minimum of finitely many positive radii, not an unverified expression that might be zero.
Domain assumptions must support each estimate. For example, a denominator bound or a restriction such as \(0\leq xy\leq1\) must be established before it is used.
Keep strict inequalities clear. If the hypothesis is \(|x-y|<\delta\), an estimate \(|f(x)-f(y)|\leq C|x-y|\) gives a strict final bound when \(C>0\) and \(\delta=\varepsilon/C\).

A practical proof checklist is to state the arbitrary \(\varepsilon>0\), give the chosen \(\delta>0\), take arbitrary \(x,y\) in the domain satisfying \(|x-y|<\delta\), and show the required output inequality. For sums, allocate the error before choosing the common radius. For compositions, choose the outer tolerance before choosing the inner radius. These steps make the dependence of each choice visible and prevent a pointwise continuity argument from being mistaken for a uniform one.

Write the target inequality.
Identify the output difference that must be less than the given \(\varepsilon\).

Find an estimate.
Use algebra and the domain assumptions to bound the output difference using \(|x-y|\).

Choose a radius independent of the points.
For multiple conditions, take the minimum of the positive radii that meet them.

Verify the implication for arbitrary inputs.
Assume \(|x-y|<\delta\) and display the inequalities leading to an output difference less than \(\varepsilon\).

Check Your Understanding

Use the definition and the proof techniques in this tutorial to answer the following questions.

In the affine example, why must the chosen radius be independent of \(x\) and \(y\)?
For the sum theorem, why is \(\min\{\delta_f,\delta_g\}\) a suitable common radius?
In the composition theorem, why is the outer tolerance chosen before the input radius?
Which domain assumption justifies the estimate \(|1-xy|\leq1\) in the example on \([0,1]\)?
If a proof produces a radius that depends on the input point, what part of the uniform continuity requirement has not yet been met?

What You'll Learn