Uniform Continuity of Polynomials on Bounded Sets

One Estimate for Every Pair of Inputs

The previous tutorial used unbounded images and rapid growth to detect failures of uniform continuity. Polynomials provide a useful contrast: although a polynomial of degree at least two need not be uniformly continuous on the whole real line, its changes can be controlled uniformly whenever its inputs stay in a bounded set. The key is to estimate the difference of the polynomial's values directly, using the distance between the inputs as a factor.

Recall that a function $f:E\to\mathbb{R}$ is Lipschitz on $E$ with constant $L\geq0$ if $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in E$. The theorem “Lipschitz Implies Uniform Continuity” from “Lipschitz Implies Uniform Continuity” then gives uniform continuity immediately. We will find such a constant for every polynomial on every bounded subset of $\mathbb{R}$, without requiring the set to be an interval or to be closed.

A Coefficient-Based Lipschitz Bound

Theorem (Polynomials Are Lipschitz on Bounded Sets): Let $E\subseteq\mathbb{R}$ be bounded, and let $$ p(x)=c_0+c_1x+\cdots+c_nx^n $$ be a polynomial. Then $p$ is Lipschitz on $E$. In particular, $p$ is uniformly continuous on $E$.

Proof. If $E$ is empty, the Lipschitz inequality holds vacuously. Otherwise, boundedness gives an $M\geq0$ such that $|x|\leq M$ for every $x\in E$. Set $R=\max\{1,M\}$. Thus $R\geq1$, and every $x\in E$ satisfies $|x|\leq R$.

For an integer $k\geq1$, the difference-of-powers identity gives

x^k-y^k=(x-y)\left(x^{k-1}+x^{k-2}y+\cdots+xy^{k-2}+y^{k-1}\right).

There are $k$ terms in parentheses. If $x,y\in E$, each term has absolute value at most $R^{k-1}$, because both $|x|$ and $|y|$ are at most $R$. The triangle inequality therefore yields

|x^k-y^k|\leq kR^{k-1}|x-y|.

For $x,y\in E$, the constant term cancels when we subtract $p(y)$ from $p(x)$. Applying the preceding estimate to each remaining term gives

\begin{aligned} |p(x)-p(y)| &\leq \sum_{k=1}^{n}|c_k|\,|x^k-y^k|\\ &\leq \left(\sum_{k=1}^{n}|c_k|\,kR^{k-1}\right)|x-y|. \end{aligned}

The finite sum $L=\sum_{k=1}^{n}|c_k|\,kR^{k-1}$ is nonnegative and finite, so it is a Lipschitz constant for $p$ on $E$. If $p$ is constant, this sum is zero and the function is Lipschitz with constant zero. In all cases, the theorem “Lipschitz Implies Uniform Continuity” gives uniform continuity on $E$. $\square$

The estimate is explicit: once a bound $R$ for the inputs is chosen, the coefficients of the polynomial give a usable Lipschitz constant. It need not be the smallest possible constant. Its purpose is to work for every pair of inputs in the set, rather than to provide an optimal value.

Worked Example: A Quadratic on a Bounded Interval

Let $p(x)=3x^2-2x+5$ on $E=[-2,3]$. Every $x\in E$ has $|x|\leq3$, so we may take $R=3$. The coefficient estimate gives

L=|3|\cdot 2\cdot 3^{1}+|-2|\cdot 1\cdot 3^{0}=18+2=20.

The resulting bound can also be checked directly. For $x,y\in[-2,3]$,

\begin{aligned} |p(x)-p(y)| &=|3x^2-2x+5-(3y^2-2y+5)|\\ &=|3(x-y)(x+y)-2(x-y)|\\ &=|x-y|\,|3(x+y)-2|\\ &\leq |x-y|\bigl(3|x|+3|y|+2\bigr)\\ &\leq20|x-y|. \end{aligned}

Given any $\varepsilon>0$, choose $\delta=\varepsilon/20$. If $x,y\in[-2,3]$ and $|x-y|<\delta$, then

|p(x)-p(y)|\leq20|x-y|<20\delta=\varepsilon.

The radius depends on $\varepsilon$, but not on the location of either input in the interval. That common choice is exactly what uniform continuity requires.

The Set Need Not Be an Interval

The proof used only a bound on the absolute values of points in $E$. It did not use the presence of points between two elements of $E$, nor did it require that $E$ contain its limit points. The same coefficient estimate therefore applies to bounded sets with gaps or missing limit points.

Worked Example: A Polynomial on a Disconnected Set

Consider $q(x)=2x^4-x^2+7$ on $E=\{0\}\cup\{1/n:n\text{ is a positive integer}\}$. This set is bounded and every $x\in E$ satisfies $|x|\leq1$, so $R=1$. The coefficient estimate gives

L=|2|\cdot4\cdot1^3+|-1|\cdot2\cdot1^1=8+2=10.

Thus, for any $x,y\in E$, the theorem proves $|q(x)-q(y)|\leq10|x-y|$, and hence $q$ is uniformly continuous on $E$. For example, this bound also works when one point is $0$: if $y=1/n$, then

|q(1/n)-q(0)| =\left|\frac{2}{n^4}-\frac{1}{n^2}\right| \leq\frac{2}{n^4}+\frac{1}{n^2} \leq\frac{10}{n} =10|1/n-0|.

The general estimate handles every pair in the set, not just pairs involving zero. No interval argument or compactness assumption is needed.

Using a Larger Bound When It Simplifies the Work

A bounded set may have a convenient bound that is not its smallest possible one. Any valid $R\geq1$ can be used in the theorem. A larger $R$ may produce a less sharp Lipschitz constant, but the resulting estimate remains valid and is often easier to calculate.

Worked Example: Choosing a Simple Bound on a Bounded Set

Let $r(x)=-x^5+4x^2-3x+1$ on $E=[-1,2]$. Since $|x|\leq2$ on this interval, take $R=2$. The coefficient estimate gives

L=|-1|\cdot5\cdot2^4+|4|\cdot2\cdot2^1+|-3|\cdot1\cdot2^0 =80+16+3=99.

For all $x,y\in[-1,2]$, it follows that $|r(x)-r(y)|\leq99|x-y|$. For a requested output tolerance $\varepsilon>0$, choosing $\delta=\varepsilon/99$ gives

|x-y|<\delta \quad\Longrightarrow\quad |r(x)-r(y)|\leq99|x-y|<99\delta=\varepsilon.

This choice works throughout the interval, including near both endpoints. We did not need to find the best Lipschitz constant; the verified bound of $99$ is sufficient to give a uniform radius.

Why Boundedness Matters

In the proof, the factor controlling the difference of the $k$-th powers contains $R^{k-1}$. On a bounded set, one fixed $R$ controls that factor for every input. If inputs are allowed to grow without bound, this estimate may no longer give a finite Lipschitz constant for terms of degree at least two. This is a limitation of the estimate on the whole line, not a claim that every polynomial fails to be uniformly continuous there.

Theorem (Behavior of Polynomials on the Real Line): Every constant or affine polynomial is Lipschitz, and hence uniformly continuous, on $\mathbb{R}$. Every polynomial of degree $d\geq2$ is not uniformly continuous on $\mathbb{R}$.

Proof. A constant polynomial has zero output difference. If $p(x)=ax+b$, then for all real $x,y$,

$$ |p(x)-p(y)|=|a||x-y|, $$

so $p$ is globally Lipschitz with constant $|a|$.

Now let $p$ have degree $d\geq2$, with leading coefficient $a_d\ne0$, and write $p(x)=\sum_{k=0}^{d}a_kx^k$. For positive integers $n$, set $x_n=n$, $y_n=n+n^{-(d-1)}$, and $t_n=n^{-(d-1)}$. Then $|y_n-x_n|=t_n\to0$. By the binomial theorem,

p(n+t_n)-p(n) =\sum_{k=1}^{d}a_k\sum_{j=1}^{k}\binom{k}{j}n^{k-j}t_n^j.

For the leading term $k=d$, the $j=1$ summand is $a_d d n^{d-1}t_n=a_d d$. For each $j\geq2$, its factor involving $n$ is

n^{d-j}t_n^j=n^{d-j-(d-1)j}=n^{d(1-j)}\longrightarrow0.

For each lower-degree term $k<d$, every summand tends to zero: its power of $n$ is $n^{k-j-(d-1)j}=n^{k-dj}$, and $k-dj\leq k-d<0$. There are only finitely many summands, so $p(y_n)-p(x_n)\to d a_d$. Because $d a_d\ne0$, for all sufficiently large $n$,

|p(y_n)-p(x_n)|>\frac{|d a_d|}{2}>0.

Thus the input distances tend to zero while the output distances stay bounded away from zero. By the Sequential Criterion for Uniform Continuity, $p$ is not uniformly continuous on $\mathbb{R}$. $\square$

The distinction between affine polynomials and polynomials with a nonzero term of degree at least two is important. For the identity polynomial, the difference-of-powers factor for the degree-one term is just $1$; it does not grow with the inputs. By contrast, for a term of degree $k\geq2$, the estimate involves powers of $R$ that can grow as the allowed inputs grow. On bounded sets those factors are finite, which is why the earlier theorem applies.

What the Estimate Does—and Does Not—Say

The coefficient bound supplies a sufficient Lipschitz constant, not necessarily the least one. Choosing $R$ larger than needed or applying the triangle inequality term by term may make the bound conservative. That does not weaken the conclusion: any finite constant satisfying the Lipschitz inequality gives uniform continuity.

A common mistake is to confuse boundedness of the set with boundedness of the polynomial's degree, or to assume the proof needs a closed interval. The degree is fixed and finite, so the coefficient sum contains only finitely many terms. The domain can be any bounded subset, even one with gaps. Another mistake is to conclude that all polynomials fail to be uniformly continuous on $\mathbb{R}$ because the estimate for higher powers grows there. Constant and affine polynomials are globally Lipschitz; the nonuniform-continuity conclusion applies to polynomials with degree at least two.

For a polynomial on a bounded domain, the practical procedure is direct: bound the absolute values of all inputs by some $R\geq1$, calculate $L=\sum_{k=1}^{n}|c_k|kR^{k-1}$, and choose $\delta=\varepsilon/L$ when $L>0$. If $L=0$, the polynomial is constant and any positive $\delta$ works. This turns a pointwise-looking calculation into one estimate valid everywhere on the domain.

Check Your Understanding

Use the difference-of-powers estimate and the results in this tutorial to answer the following questions.

Why does a bounded set allow one value of $R$ to control every factor in the difference-of-powers estimate?
For $p(x)=x^3+2x$ on $[-1,2]$, what Lipschitz constant follows from the coefficient estimate with $R=2$?
Does the proof that polynomials are Lipschitz on bounded sets require the set to contain every point between any two of its elements? Explain.
Why does the global Lipschitz estimate for an affine polynomial not depend on a bound for the inputs?
For a polynomial of degree at least two on $\mathbb{R}$, what do the sequences in the proof show about input and output distances?