Why Uniform Convergence Preserves Continuity
A sequence of continuous functions can converge pointwise to a discontinuous function. Uniform convergence rules out this failure: it ensures that one index in the sequence gives a good approximation at every point of the domain at once. That shared index lets us combine a uniform error estimate with continuity of one approximating function.
Let \((X,\mathcal{T})\) be a topological space, let \((Y,d)\) be a metric space, and let \(f_n:X\to Y\) be maps. Continuity of a map \(g:X\to Y\) at \(x_0\in X\) means that for every \(\varepsilon>0\), there is an open neighborhood \(U\) of \(x_0\) such that \(d(g(x),g(x_0))<\varepsilon\) for every \(x\in U\).
By contrast, pointwise convergence allows the index to depend on both \(\varepsilon\) and \(x\). The theorem below uses exactly the extra uniformity: near a chosen point, we will compare the limit to one fixed continuous function \(f_N\), and the approximation errors will be small throughout the whole domain.
The Uniform Limit Theorem
Proof. Fix \(x_0\in X\). We show that \(f\) is continuous at \(x_0\). Let \(\varepsilon>0\). Uniform convergence, applied with \(\varepsilon/3\), gives \(N\in\mathbb{N}\) such that for every \(x\in X\), \(d(f_N(x),f(x))<\varepsilon/3\). Since \(f_N\) is continuous at \(x_0\), there is an open neighborhood \(U\) of \(x_0\) such that \(d(f_N(x),f_N(x_0))<\varepsilon/3\) for every \(x\in U\). For each such \(x\), the triangle inequality gives $$ d(f(x),f(x_0)) \le d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0)) <\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} =\varepsilon. $$ Thus \(U\) is a neighborhood on which \(f(x)\) is within \(\varepsilon\) of \(f(x_0)\). Since \(x_0\) and \(\varepsilon\) were arbitrary, \(f\) is continuous on \(X\). \(\square\)
The proof requires no metric on the domain: the domain needs only a topology so that continuity can be expressed through open neighborhoods. Nor does it require \(X\) to be compact or \(Y\) to be complete. It assumes that the uniform limit \(f\) already exists as a map into \(Y\). Completeness can matter when asking whether a uniformly Cauchy sequence has a limit, but it is not needed to prove continuity once that limit exists.
Proof. Let \(\varepsilon>0\). Since \(a_n\to 0\), there is \(N\in\mathbb{N}\) such that \(a_n<\varepsilon\) whenever \(n\ge N\). For every such \(n\) and every \(x\in X\), the assumed bound gives \(d(f_n(x),f(x))\le a_n<\varepsilon\). This is the definition of uniform convergence. \(\square\)
This proposition is often used by proving a single estimate that holds for all \(x\), then showing that its right-hand side tends to zero. The strict inequality needed in the definition follows because the bound is at most \(a_n\), and \(a_n\) itself is strictly less than \(\varepsilon\) for sufficiently large \(n\).
Worked Applications on the Unit Interval
Worked Example: A Uniform Limit of Linear Functions
Give \([0,1]\) its subspace topology from the standard topology on \(\mathbb{R}\). For \(n\in\mathbb{N}\), define \(f_n:[0,1]\to\mathbb{R}\) by \(f_n(x)=x+x/(n+1)\). Find the limit and verify uniform convergence.
For each fixed \(x\in[0,1]\), the term \(x/(n+1)\) tends to zero, so the pointwise limit is \(f(x)=x\). Each \(f_n\) is a linear function of \(x\), and is therefore continuous on \([0,1]\). For every \(x\in[0,1]\), $$ |f_n(x)-f(x)|=\left|x+\frac{x}{n+1}-x\right|=\frac{x}{n+1}\le\frac{1}{n+1}. $$ Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/(N+1)<\varepsilon\). If \(n\ge N\), then $$ |f_n(x)-f(x)|\le\frac{1}{n+1}\le\frac{1}{N+1}<\varepsilon $$ for every \(x\in[0,1]\). This proves uniform convergence. The error bound is attained at \(x=1\), where \(|f_n(1)-f(1)|=1/(n+1)\), so it also gives the exact largest error. The theorem now confirms that the limit \(f(x)=x\) is continuous.
Check: At \(n=2\) and \(x=1\), \(f_2(1)=1+1/3=4/3\), while \(f(1)=1\). The error is \(4/3-1=1/3\), matching the bound \(1/(2+1)=1/3\).
Worked Example: A Geometric Sequence of Polynomials
On \([0,1]\) with its usual subspace topology, define $$ f_n(x)=\sum_{k=0}^{n}\left(\frac{x}{3}\right)^k. $$ Find the limit and prove uniform convergence to it.
Each \(f_n\) is a polynomial, hence continuous. Since \(0\le x/3\le 1/3\), the geometric series gives the candidate limit $$ f(x)=\sum_{k=0}^{\infty}\left(\frac{x}{3}\right)^k=\frac{1}{1-x/3}. $$ The denominator is positive on the whole interval: \(1-x/3\ge 2/3>0\). Thus \(f\) is continuous there as a quotient of continuous functions with nonzero denominator. The finite geometric-sum formula gives, for every \(x\in[0,1]\), $$ f(x)-f_n(x)=\frac{(x/3)^{n+1}}{1-x/3}. $$ Both numerator and denominator are nonnegative, and the denominator is at least \(2/3\). Therefore $$ 0\le f(x)-f_n(x)\le \frac{(1/3)^{n+1}}{2/3}=\frac{1}{2\cdot 3^n}. $$ For \(n\ge N\), the right-hand side is at most \(1/(2\cdot3^N)\), which tends to zero as \(N\to\infty\). The proposition on uniform error bounds proves that \(f_n\) converges uniformly to \(f\).
Check: At \(n=2\) and \(x=1\), \(f_2(1)=1+1/3+1/9=13/9\), and \(f(1)=1/(1-1/3)=3/2=27/18\). Since \(13/9=26/18\), the error is \(1/18\), exactly \(1/(2\cdot3^2)\). The bound is attained at the endpoint \(x=1\).
Worked Example: Pointwise Convergence Can Lose Continuity
On \([0,1]\), again with the usual subspace topology, consider \(g_n(x)=x^n\). Find its pointwise limit and check whether convergence is uniform.
If \(0\le x<1\), then \(x^n\to0\); at \(x=1\), \(1^n=1\) for every \(n\). Hence the pointwise limit is $$ g(x)= \begin{cases} 0,&0\le x<1,\\ 1,&x=1. \end{cases} $$ This limit is discontinuous at \(1\). Indeed, the points \(x_n=2^{-1/n}\) satisfy \(0\le x_n<1\), \(x_n\to1\), and \(g(x_n)=0\), whereas \(g(1)=1\). If \(g\) were continuous at \(1\), values \(g(x_n)\) would tend to \(g(1)\), but the values are all zero.
The convergence is not uniform. For every \(n\), \(x_n^n=(2^{-1/n})^n=1/2\), while \(g(x_n)=0\). Thus $$ |g_n(x_n)-g(x_n)|=\left|\frac12-0\right|=\frac12. $$ The uniform convergence requirement with \(\varepsilon=1/2\) would require every point to have error strictly less than \(1/2\) for all sufficiently large \(n\). The displayed equality violates that requirement for every \(n\). This example does not contradict the Uniform Limit Theorem: its convergence is pointwise, not uniform.
Check: At the endpoint \(x=1\), the error is \(|g_n(1)-g(1)|=|1-1|=0\). The failure of uniform convergence comes from points \(x_n<1\) approaching the endpoint, not from the endpoint itself.
The Role of the Shared Index
A common mistake is to use pointwise convergence as if it supplied one index that works throughout a neighborhood. For pointwise convergence, after choosing \(\varepsilon\), the index may vary with \(x\). Continuity of a limit requires control at all points in a neighborhood, so those point-dependent indices do not provide the uniform estimate used in the theorem’s proof.
For the sequence \(g_n(x)=x^n\), a mistaken argument might say: “For every \(x<1\), choose \(N\) large enough that \(x^N<\varepsilon\); therefore \(x^n\) is small near \(1\), so the limit is continuous at \(1\).” The first choice depends on \(x\), and it says nothing about the value at \(x=1\), where \(g_n(1)=1\) for every \(n\). More concretely, for each \(n\), the point \(x_n=2^{-1/n}<1\) has \(g_n(x_n)=1/2\), even though \(x_n\) approaches \(1\). There is no single index that makes the error from the limit small at all points.
For real-valued functions, absolute value is the metric on \(\mathbb{R}\), so the uniform error condition is \(|f_n(x)-f(x)|<\varepsilon\) for all \(x\) once \(n\) is sufficiently large. A bound such as \(|f_n(x)-f(x)|\le a_n\), independent of \(x\), is especially useful. The theorem then converts that quantitative estimate into a qualitative conclusion: the limit function is continuous.
Check Your Understanding
In each question, distinguish carefully between pointwise and uniform convergence.
- On \([0,1]\), define \(h_n(x)=x/(n+2)\). Find the pointwise limit and give a uniform error bound.
- In the proof of the Uniform Limit Theorem, why is the same \(N\) valid for every \(x\in X\)?
- For \(g_n(x)=x^n\) on \([0,1]\), verify directly that the pointwise limit is discontinuous at \(1\).
- Suppose \(f_n:X\to Y\) are continuous and \(d(f_n(x),f(x))\le 1/(n+1)\) for all \(x\). Explain why \(f\) is continuous.
- Which hypotheses of the Uniform Limit Theorem concern the domain, and which concern the target?