Proof That Uniform Limits Are Continuous

Why Uniform Convergence Is the Key

Earlier in this course, the theorem Uniform Limits of Continuous Functions Are Continuous was stated and used. This tutorial supplies its epsilon-delta proof. The central point is that continuity of one approximating function controls nearby inputs, while uniform convergence makes its error from the limit small at every input at once. Both controls are needed to estimate the change in the limit.

Let $E\subseteq\mathbb{R}$. Continuity on $E$ is understood relative to $E$: a function $f:E\to\mathbb{R}$ is continuous at $a\in E$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that $x\in E$ and $|x-a|<\delta$ imply $|f(x)-f(a)|<\varepsilon$. The domain need not be an interval, and the point $a$ need not be an interior point of $E$.

Definition: A sequence $f_n:E\to\mathbb{R}$ converges uniformly to $f:E\to\mathbb{R}$ if, for every $\varepsilon>0$, there is an integer $N$ such that $n\geq N$ and $x\in E$ imply $|f_n(x)-f(x)|<\varepsilon$. The choice of $N$ may depend on $\varepsilon$, but not on $x$.

The proof will divide a desired error $\varepsilon$ into three parts. Two parts control the errors between the approximating function and the limit, at the two inputs being compared. The remaining part controls the change in the approximating function itself. This is often called the three-term estimate.

The Epsilon-Delta Proof

Theorem (Uniform Limits of Continuous Functions Are Continuous): Let $E\subseteq\mathbb{R}$, and suppose each $f_n:E\to\mathbb{R}$ is continuous on $E$. If $f_n$ converges uniformly on $E$ to $f:E\to\mathbb{R}$, then $f$ is continuous on $E$.

Proof. If $E$ is empty, continuity on $E$ holds vacuously. Otherwise, fix an arbitrary $a\in E$. We will prove that $f$ is continuous at $a$.

Let $\varepsilon>0$. By uniform convergence, choose $N$ such that for every $x\in E$,

|f_N(x)-f(x)|<\frac{\varepsilon}{3}.

The index $N$ works for all $x\in E$; in particular, the displayed estimate holds both at $x$ and at $a$. Since $f_N$ is continuous at $a$, there is a $\delta>0$ such that $x\in E$ and $|x-a|<\delta$ imply

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

For such an $x$, the triangle inequality gives

\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}

Thus, for this $\varepsilon$, the chosen $\delta$ ensures that $x\in E$ and $|x-a|<\delta$ imply $|f(x)-f(a)|<\varepsilon$. This proves continuity of $f$ at $a$. Since $a$ was arbitrary, $f$ is continuous on $E$. $\square$

Notice the order of choices. First choose one approximating function, $f_N$, whose error from $f$ is small everywhere. Then use continuity of that fixed $f_N$ to choose a neighborhood of $a$. Choosing $N$ after choosing $x$ would not provide the uniform control the argument needs.

A Quantitative Three-Term Estimate

The proof gives more than a yes-or-no continuity conclusion. It gives an explicit bound on how much the limit can change near a point, in terms of the approximation error and the change in a chosen approximant.

Proposition (Three-Term Continuity Estimate): Suppose $f_n:E\to\mathbb{R}$ converges uniformly to $f:E\to\mathbb{R}$. Fix $n$, and suppose $|f_n(x)-f(x)|\leq\eta$ for every $x\in E$, where $\eta\geq0$. Then, for any $a,x\in E$, $$ |f(x)-f(a)|\leq 2\eta+|f_n(x)-f_n(a)|. $$

Proof. Insert the two values of $f_n$ between the values of $f$ and apply the triangle inequality:

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

The estimate holds for every $a,x\in E$, including $x=a$, and proves the proposition. $\square$

For continuity, the estimate suggests an error budget: make $2\eta$ smaller than part of the desired tolerance, then use continuity of $f_n$ to control the last term. For example, if $\eta<\varepsilon/4$, then $2\eta<\varepsilon/2$. A neighborhood on which $|f_n(x)-f_n(a)|<\varepsilon/2$ makes the whole bound strictly less than $\varepsilon$. The proposition makes clear why uniform approximation is useful: its error bound applies simultaneously at both points.

Worked Applications

Worked Example: Uniform Approximation of an Absolute-Value Function

On $E=[-2,2]$, define $f_n(x)=\sqrt{x^2+1/n}$. The radicand is positive for every $x$, so each $f_n$ is continuous. We show directly that $f_n$ converges uniformly to $f(x)=|x|$.

For $u,v\geq0$, the inequality $\sqrt{u+v}\leq\sqrt{u}+\sqrt{v}$ follows by squaring the nonnegative right-hand side: $(\sqrt{u}+\sqrt{v})^2=u+v+2\sqrt{uv}\geq u+v$. Applying this with $u=x^2$ and $v=1/n$, and also using $\sqrt{x^2+1/n}\geq |x|$, gives

0\leq \sqrt{x^2+\frac1n}-|x|\leq\frac1{\sqrt n}.

The upper bound is independent of $x$ and tends to zero. Therefore the convergence is uniform on $[-2,2]$. The theorem proves that $f(x)=|x|$ is continuous on this domain. In this example its continuity is also familiar, but the argument illustrates how a uniform error estimate can establish continuity of a limit without analyzing that limit separately at each point.

Worked Example: A Geometric Polynomial Sequence

On $E=[0,1]$, define

P_n(x)=\sum_{k=0}^{n}\left(\frac{x}{3}\right)^k.

Each $P_n$ is a polynomial and hence continuous. The finite geometric-sum identity gives

P_n(x)=\frac{1-(x/3)^{n+1}}{1-x/3}.

The denominator is positive on $[0,1]$, since $1-x/3\geq2/3$. The geometric series has sum $F(x)=1/(1-x/3)=3/(3-x)$ there. Subtracting the finite sum from this expression yields

|F(x)-P_n(x)| =\frac{(x/3)^{n+1}}{1-x/3} \leq \frac{(1/3)^{n+1}}{2/3} =\frac{1}{2\cdot3^n}.

The last bound tends to zero independently of $x\in[0,1]$, so $P_n$ converges uniformly to $F$. The theorem therefore gives continuity of $F$ on $[0,1]$. The estimate also identifies precisely how the polynomial approximations control the limit, including near the endpoint $x=1$.

Worked Example: Pointwise Convergence Does Not Suffice

Define $g_n:[0,1]\to\mathbb{R}$ by $g_n(x)=x^{1/n}$, with $g_n(0)=0$. Each $g_n$ is continuous on $[0,1]$. At $x=0$, every term is zero. For fixed $x>0$, $x^{1/n}$ tends to $1$, so the pointwise limit is

g(x)= \begin{cases} 0,&x=0,\\ 1,&0<x\leq1. \end{cases}

This limit is not continuous at zero: for positive inputs approaching zero, its value remains $1$, whereas $g(0)=0$. The convergence cannot be uniform. Indeed, take the moving input $x_n=2^{-n}$, which belongs to $[0,1]$. Then

g_n(x_n)=(2^{-n})^{1/n}=\frac12, \qquad g(x_n)=1, \qquad |g_n(x_n)-g(x_n)|=\frac12.

Since the error is at least $1/2$ at these inputs for every $n$, its supremum over the domain cannot tend to zero. There is no contradiction with the continuity theorem: its uniform-convergence hypothesis fails. The moving-input calculation exhibits exactly how a pointwise limit can retain a discontinuity even though each function in the sequence is continuous.

How to Apply the Proof

When using the theorem, verify the hypotheses on the domain where the conclusion is wanted. Each approximating function must be continuous there, and the convergence must be uniform there. Uniform convergence on one set does not automatically provide uniform convergence on a larger set. Also, the limit need not be defined by a simple formula; it can be enough to know that the continuous functions converge uniformly to it.

A common mistake is to use pointwise convergence as though it supplied one index that controls the error at every input. Pointwise convergence allows the required index to vary with $x$; continuity of the limit requires controlling all sufficiently nearby inputs at once. The three-term estimate avoids this problem by fixing one approximant with a uniform error bound before invoking its continuity.

Takeaway: To prove continuity of a uniform limit at a point, fix an approximant whose error is small everywhere, then use continuity of that single approximant near the point. The three-term estimate records the resulting error control and shows why pointwise convergence alone is insufficient.

Check Your Understanding

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

Why must the index $N$ controlling $|f_N(x)-f(x)|$ be independent of $x$?
In the epsilon-delta proof, where are the two approximation errors evaluated?
If a chosen approximant differs from the limit by at most $\eta$ everywhere, what bound does the three-term estimate give for $|f(x)-f(a)|$?
What uniform error bound proves that $\sqrt{x^2+1/n}$ converges to $|x|$ on $[-2,2]$?
For $g_n(x)=x^{1/n}$, which moving inputs show that convergence to the pointwise limit is not uniform?