Prove Uniform Limits Preserve Continuity

Uniform Approximation and Local Continuity

The Weierstrass M-Test supplies a common way to control the tails of a function series. In particular, it can show that partial sums approach their sum uniformly. This raises a natural question: when the approximating functions are continuous, what properties must their uniform limit retain?

The real-domain version of the answer was established earlier in the course as the Uniform Limit of Continuous Functions Is Continuous theorem. Here we examine the proof technique in a metric-domain formulation and make its error control explicit. The essential point is to divide the permitted error into two parts: one for replacing the limit by a single approximating function, and one for using that function's continuity near the point of interest.

Definition: Let $(X,d)$ be a metric space, let $E\subseteq X$, and let $f:E\to\mathbb R$. The function $f$ is continuous at $a\in E$ if, for every $\varepsilon>0$, there is a $\delta>0$ such that $$ d(x,a)<\delta,\quad x\in E \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$ A sequence of functions $f_n:E\to\mathbb R$ converges uniformly to $f:E\to\mathbb R$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $$ |f_n(x)-f(x)|<\varepsilon $$ for every $x\in E$ and every $n\geq N$.

The order of the quantifiers in uniform convergence matters. Once $N$ is chosen, the approximation bound holds at every point of $E$, including the point $a$ and all nearby points $x$. That lets us compare both $f(x)$ and $f(a)$ with values of the same continuous approximant.

A Quantitative Error Estimate

The triangle inequality provides the basic estimate behind the continuity argument. It is useful to first keep the approximation error visible rather than immediately choosing it to be a fraction of $\varepsilon$.

Proof. Insert $g(x)$ and $g(a)$ between $f(x)$ and $f(a)$, then apply the triangle inequality:

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

The last two bounds use the assumed estimate first at $y=x$, then at $y=a$. Both points must be included: controlling the approximation error only at nearby $x$, without controlling it at $a$, would not justify the estimate. $\square$

The factor of two has a simple origin: the value at $x$ and the value at $a$ each contribute one approximation error. If $g$ is continuous at $a$, its remaining change $|g(x)-g(a)|$ can be made small by restricting $x$ to a suitable neighborhood.

Worked Example: Turning a Uniform Error Bound into a Radius

Suppose $f,g:E\to\mathbb R$, $|f(y)-g(y)|\leq 1/20$ for every $y\in E$, and

|g(x)-g(a)|\leq 3d(x,a)

for the points under consideration. The Two-Point Approximation Estimate gives

|f(x)-f(a)|\leq \frac{2}{20}+3d(x,a) =\frac{1}{10}+3d(x,a).

For a target error $\varepsilon=1/2$, it is enough to require $3d(x,a)<2/5$, or $d(x,a)<2/15$. Then

|f(x)-f(a)|<\frac{1}{10}+\frac{2}{5} =\frac{1}{2}.

This example shows how a fixed approximation error consumes part of the error budget. The remaining allowance must be supplied by the local variation of $g$.

Uniform Limits Preserve Continuity

We now prove the metric-domain formulation. This statement includes functions on subsets of $\mathbb R$, but it does not require an order, an interval, or any special feature of the real line. The domain enters only through its metric and the definition of continuity.

Theorem (Continuity of a Uniform Limit on a Metric Domain): Let $(X,d)$ be a metric space, let $E\subseteq X$, and let $f_n:E\to\mathbb R$ be continuous at $a\in E$ for every positive integer $n$. If $f_n$ converges uniformly on $E$ to $f:E\to\mathbb R$, then $f$ is continuous at $a$.

Proof. Let $\varepsilon>0$. Uniform convergence gives a positive integer $N$ such that

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

for every $y\in E$. We use this one function $f_N$ to approximate $f$ at both $x$ and $a$. Since $f_N$ is continuous at $a$, there is a $\delta>0$ such that, for $x\in E$,

d(x,a)<\delta \quad\Longrightarrow\quad |f_N(x)-f_N(a)|<\frac{\varepsilon}{3}.

For any such $x$, the triangle inequality yields

\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 every $\varepsilon>0$, the chosen $\delta$ ensures that $d(x,a)<\delta$ and $x\in E$ imply $|f(x)-f(a)|<\varepsilon$. This is continuity of $f$ at $a$. $\square$

Only continuity at the point $a$ is needed to conclude continuity there. Consequently, if every $f_n$ is continuous at every point of $E$, then the same argument applies separately at each $a\in E$, and $f$ is continuous throughout $E$. The choice of $N$ is uniform in $x$; the continuity radius $\delta$, however, may depend on $a$ and on the selected approximant $f_N$.

Worked Example: A Uniformly Convergent Sequence on the Real Line

For $x\in\mathbb R$, define

f_n(x)=\frac{x}{1+n x^2}.

Each $f_n$ is continuous on $\mathbb R$, since its denominator $1+nx^2$ is positive for every real $x$. To bound it uniformly, put $t=|x|$. The nonnegative square $(\sqrt n\,t-1)^2\geq0$ implies

n t^2+1\geq2\sqrt n\,t.

Therefore, including $t=0$,

|f_n(x)| =\frac{t}{1+nt^2} \leq\frac{1}{2\sqrt n}.

Given $\varepsilon>0$, choose $N$ so large that $1/(2\sqrt N)<\varepsilon$. For $n\geq N$,

|f_n(x)-0|=|f_n(x)| \leq\frac{1}{2\sqrt n} \leq\frac{1}{2\sqrt N} <\varepsilon

for every $x\in\mathbb R$. Thus $f_n$ converges uniformly to the zero function. The theorem guarantees that the limit is continuous; in this example its continuity is also immediate from its formula.

Using the Estimate to Design a Proof

The $\varepsilon/3$ proof is a reliable default, but the Two-Point Approximation Estimate allows more flexible choices. Suppose an approximant $g=f_N$ satisfies $|f(y)-g(y)|\leq\eta$ on all of $E$, and its continuity gives $|g(x)-g(a)|<\gamma$ near $a$. Then

|f(x)-f(a)|\leq2\eta+|g(x)-g(a)|<2\eta+\gamma.

It is enough to choose $\eta$ and $\gamma$ so that $2\eta+\gamma\leq\varepsilon$, with strictness in the local bound ensuring the final strict inequality. For example, one can take $\eta=\varepsilon/4$ and $\gamma=\varepsilon/2$. The familiar choice $\varepsilon/3$ for all three terms is slightly more restrictive than necessary, but is easy to remember and leaves no ambiguity about strict inequalities.

Worked Example: Lipschitz Approximants Give an Explicit Radius

Suppose that for some $N$, the uniform error satisfies

|f(y)-f_N(y)|\leq\frac{\varepsilon}{8}

for all $y\in E$, and suppose that for some constant $L>0$,

|f_N(x)-f_N(a)|\leq Ld(x,a).

The Two-Point Approximation Estimate now gives

|f(x)-f(a)| \leq\frac{\varepsilon}{4}+Ld(x,a).

Choose $\delta=\varepsilon/(2L)$. If $d(x,a)<\delta$, then $Ld(x,a)<\varepsilon/2$, and hence

|f(x)-f(a)| <\frac{\varepsilon}{4}+\frac{\varepsilon}{2} =\frac{3\varepsilon}{4} <\varepsilon.

This gives an explicit continuity radius once a suitable approximant and uniform error bound have been found. The calculation also shows why the approximant's local regularity and the uniform accuracy are separate inputs.

Why Pointwise Convergence Is Not Enough

Uniformity cannot simply be omitted. On $[0,1]$, let $f_n(x)=x^n$. Each $f_n$ is continuous. For $0\leq x<1$, the powers $x^n$ tend to $0$, while $f_n(1)=1$ for every $n$. The pointwise limit is therefore

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

This limit is not continuous at $1$. In fact, the convergence is not uniform: for each $n$, take $x_n=1-1/n$ when $n\geq2$. Then $x_n<1$, so $f(x_n)=0$, whereas

|f_n(x_n)-f(x_n)| =\left(1-\frac1n\right)^n.

This quantity does not approach zero; for example, the standard limit is $e^{-1}$. Thus no single index makes the pointwise error small over the whole interval. The failure is exactly what the continuity proof cannot tolerate: it needs one approximant that is accurate both at the point and uniformly around it.

When applying the theorem, check the hypotheses separately. The approximating functions must be continuous at the point in question, and convergence must be uniform on the domain under consideration. Pointwise convergence gives an index that may depend on $x$, so it cannot provide the common approximation bound used in the proof. Nor does uniform convergence alone guarantee continuity if the approximants are not continuous.

Fix the target error.
Start with an arbitrary $\varepsilon>0$ from the definition of continuity.

Select one approximant.
Use uniform convergence to choose an index whose error is small at every point of the domain.

Use local continuity.
Apply continuity of that selected approximant at the point to choose a radius.

Add the three changes.
Compare the limit to the approximant at the nearby point, compare the approximant's two values, and compare back to the limit at the fixed point.

Check Your Understanding

Use the estimates and proof strategy above to answer these questions.

Why does the Two-Point Approximation Estimate contain two copies of the uniform error bound?
In the metric-domain proof, which choice is made using uniform convergence, and which choice is made using continuity at $a$?
If the approximation error is at most $\varepsilon/10$ everywhere, how much error remains for controlling the approximant's change if the total target is $\varepsilon$?
Why does pointwise convergence of $x^n$ on $[0,1]$ fail to provide the estimate needed in the continuity proof?
Does continuity of a uniform limit at $a$ require the approximants to be continuous everywhere on $E$, or only at $a$?