Proof Uniform Limits Preserve Continuity

From Uniform Approximation to Continuity

The previous tutorial separated pointwise convergence from uniform convergence and identified the global control uniform convergence provides. To prove that continuity passes to a limit, we use that control at two inputs: the limit is close to an approximating function at each input, and the approximating function changes little between nearby inputs. The resulting estimate gives a direct epsilon-delta proof.

Let \(E\subseteq\mathbb{R}\), and suppose \(f_n:E\to\mathbb{R}\) converges uniformly to \(f:E\to\mathbb{R}\). Fix a point \(a\in E\). For any \(x\in E\), insert the value \(f_n(x)\) and the value \(f_n(a)\) between \(f(x)\) and \(f(a)\). The triangle inequality gives

|f(x)-f(a)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|.

The first and third terms are errors of approximation. Uniform convergence makes both small using the same index \(n\), regardless of which \(x\) is considered. Continuity of \(f_n\) at \(a\) makes the middle term small when \(x\) is sufficiently close to \(a\). It is essential to choose the approximant first and then use its continuity to choose a distance around \(a\).

A Stability Estimate for Continuity

The same idea works for a single pair of functions. It gives a useful estimate even before we consider a whole sequence of approximations.

Theorem: Let \(E\subseteq\mathbb{R}\), let \(a\in E\), and let \(f,g:E\to\mathbb{R}\). Suppose \(|f(x)-g(x)|\leq\eta\) for every \(x\in E\), where \(\eta\geq0\), and suppose \(g\) is continuous at \(a\). Then, for every \(\varepsilon>2\eta\), there exists \(\delta>0\) such that \(x\in E\) and \(|x-a|<\delta\) imply \(|f(x)-f(a)|<\varepsilon\).

Proof. Fix \(\varepsilon>2\eta\). Then \(\varepsilon-2\eta>0\). By continuity of \(g\) at \(a\), there is a \(\delta>0\) such that, for \(x\in E\), \(|x-a|<\delta\) implies \(|g(x)-g(a)|<\varepsilon-2\eta\). For any such \(x\), the triangle inequality and the assumed bounds at both \(x\) and \(a\) give

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

Thus \(f\) satisfies the continuity estimate at \(a\) for every \(\varepsilon>2\eta\). \(\square\)

The error allowance \(2\eta\) accounts for two approximation errors, one at each input. In particular, this estimate does not claim that an arbitrary fixed approximation error can meet every desired continuity tolerance. For a sequence converging uniformly, however, the approximation error can be made as small as needed by choosing a sufficiently late term.

Uniform Limits Preserve Continuity

Theorem (Uniform Limits Preserve Continuity): Let \(E\subseteq\mathbb{R}\), and suppose \(f_n:E\to\mathbb{R}\) converges uniformly on \(E\) to \(f:E\to\mathbb{R}\). If every \(f_n\) is continuous at every point of \(E\), then \(f\) is continuous at every point of \(E\).

Proof. If \(E\) is empty, the conclusion is vacuous. Otherwise, fix an arbitrary \(a\in E\), and let \(\varepsilon>0\). Uniform convergence gives an index \(N\) such that, for every \(n\geq N\) and every \(y\in E\),

|f_n(y)-f(y)|<\frac{\varepsilon}{3}.

Choose one such index \(n\), for example \(n=N\). Since \(f_n\) is continuous at \(a\), there exists \(\delta>0\) such that \(x\in E\) and \(|x-a|<\delta\) imply

|f_n(x)-f_n(a)|<\frac{\varepsilon}{3}.

For every \(x\in E\) with \(|x-a|<\delta\), apply the triangle inequality, using uniform approximation at \(x\) and at \(a\), and continuity of this fixed \(f_n\) between them:

\begin{aligned} |f(x)-f(a)| &\leq |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon. \end{aligned}

This is the definition of continuity of \(f\) at \(a\). Since \(a\) was arbitrary, \(f\) is continuous at every point of \(E\). \(\square\)

The order of choices matters. The index \(n\) is chosen from uniform convergence and does not depend on \(x\). Only after choosing \(n\) do we use continuity of that particular \(f_n\) to choose \(\delta\). The value of \(\delta\) may depend on \(a\), \(\varepsilon\), and the chosen approximant; the proof does not claim one \(\delta\) works at every point of \(E\).

Worked Applications

Worked Example: A Geometric Series on a Closed Interval

For each nonnegative integer \(n\), define the polynomial \(p_n:[0,\tfrac12]\to\mathbb{R}\) by \(p_n(x)=1+x+x^2+\cdots+x^n\). The finite geometric sum identity gives

(1-x)p_n(x)=1-x^{n+1}.

Indeed, multiplying out gives \(1+x+\cdots+x^n-(x+x^2+\cdots+x^{n+1})=1-x^{n+1}\). Since \(x\leq\tfrac12\), the denominator \(1-x\) is positive, so the pointwise limit is \(f(x)=1/(1-x)\). Subtracting the finite sum from this limit gives

|f(x)-p_n(x)| =\frac{x^{n+1}}{1-x} \leq \frac{(1/2)^{n+1}}{1/2} =\frac{1}{2^n} \qquad\left(0\leq x\leq\frac12\right).

Given \(\varepsilon>0\), choose \(N\) such that \(1/2^N<\varepsilon\). For \(n\geq N\), the error is at most \(1/2^n\leq1/2^N<\varepsilon\) everywhere on the interval. Thus \(p_n\) converges uniformly to \(f\). Each \(p_n\) is continuous, so the theorem proves that \(f(x)=1/(1-x)\) is continuous on \([0,\tfrac12]\).

Worked Example: Using the Stability Estimate

Let \(g(x)=x^2\) and define, for each positive integer \(n\),

f_n(x)=x^2+\frac{x}{n(1+x^2)} \qquad(x\in\mathbb{R}).

The denominator is positive for every real \(x\). Also, \(1+x^2\geq2|x|\), because \((|x|-1)^2\geq0\). Hence

|f_n(x)-g(x)| =\frac{|x|}{n(1+x^2)} \leq\frac{1}{2n} \qquad(x\in\mathbb{R}).

Fix \(n\) and \(a\in\mathbb{R}\). The stability estimate applies with \(\eta=1/(2n)\), so for every \(\varepsilon>1/n\), continuity of \(g\) at \(a\) gives a \(\delta>0\) such that \(|x-a|<\delta\) implies \(|f_n(x)-f_n(a)|<\varepsilon\). This shows precisely how the two endpoint errors limit the tolerance when the approximant is fixed. Each \(f_n\) is continuous at \(a\), since it is the sum of the continuous function \(g(x)=x^2\) and the continuous function \(x\mapsto x/[n(1+x^2)]\), whose denominator is positive for every real \(x\). The example illustrates the estimate directly: closeness to a continuous function controls changes in the nearby function, provided the desired tolerance exceeds twice the uniform error.

Worked Example: Pointwise Convergence Does Not Suffice

For positive integers \(n\), define \(h_n:[0,1]\to\mathbb{R}\) by

h_n(x)=\frac{nx}{1+nx}.

The denominator is positive on \([0,1]\), so each \(h_n\) is continuous there. At \(x=0\), \(h_n(0)=0\) for every \(n\). For each fixed \(x>0\), rewrite the expression as \(h_n(x)=1-1/(1+nx)\); as \(n\) increases, \(1/(1+nx)\) tends to zero. The pointwise limit is therefore \(h(0)=0\) and \(h(x)=1\) for \(x>0\). This limit is not continuous at \(0\): for example, \(x_k=1/k\) tends to \(0\), while \(h(x_k)=1\) for every positive integer \(k\), not to \(h(0)=0\).

The convergence is not uniform. At \(x=1/n\), the approximant has value

h_n(1/n)=\frac{n(1/n)}{1+n(1/n)}=\frac{1}{2}, \qquad h(1/n)=1, \qquad |h_n(1/n)-h(1/n)|=\frac12.

Thus for every \(n\), the error is at least \(1/2\) at some point of the domain. There can be no single index after which the error is uniformly smaller than, for example, \(1/4\). This example shows why the uniform hypothesis in the theorem cannot be replaced by pointwise convergence.

What the Theorem Does—and Does Not—Say

The theorem is useful because it lets a difficult function inherit continuity from simpler approximants, once a uniform error estimate has been established. The estimate need not be expressed using a supremum norm: it is enough to show directly that for every \(\varepsilon>0\), one index works at every point of the domain. The geometric-series example used exactly such an input-independent bound.

A common pitfall is to select an index that works only at the point \(a\), or separately at each \(x\). Either approach gives pointwise control, not the two simultaneous error bounds needed in the proof. Another is to infer uniform continuity of \(f\) from this theorem. The conclusion is continuity at each point, and the corresponding \(\delta\) may depend on that point. Uniform continuity requires a single distance choice to work across the whole domain, which is a stronger conclusion and needs an additional argument or hypothesis.

The proof’s reusable pattern is to compare the limit at two inputs through one fixed approximant. Uniform convergence controls the two outer differences; continuity controls the middle difference. The three terms can then be assigned portions of the desired error, such as \(\varepsilon/3\), to obtain the required estimate.

Check Your Understanding

Use the estimates and proof structure in this tutorial to answer the following questions.

In the proof that uniform limits preserve continuity, why must the index chosen from uniform convergence be independent of \(x\)?
Where do the three terms in the comparison of \(|f(x)-f(a)|\) come from?
In the stability estimate, why must the requested tolerance satisfy \(\varepsilon>2\eta\)?
What uniform error bound proves convergence of the geometric partial sums on \([0,\tfrac12]\)?
For the functions \(h_n(x)=nx/(1+nx)\), what value of \(x\) demonstrates that convergence is not uniform, and what is the error there?
Does continuity of the uniform limit imply uniform continuity on an arbitrary domain? Explain briefly.