Preservation of Continuity in Sequences of Functions
students, in calculus you learned that a function is continuous when small changes in input cause small changes in output. In Real Analysis, we often study sequences of functions like $f_1, f_2, f_3, \dots$ and ask what happens to the limit function $f$ as $n \to \infty$. A natural question is: if every $f_n$ is continuous, is the limit function $f$ also continuous? 🤔
The answer is: sometimes yes, sometimes no. The key idea is that uniform convergence preserves continuity, while pointwise convergence does not necessarily do so. In this lesson, you will learn what that means, why it matters, and how to use it in Real Analysis.
What Does It Mean to Preserve Continuity?
Suppose we have a sequence of functions $\{f_n\}$ defined on a set $D$, and suppose $f_n(x) \to f(x)$ for each $x \in D$. The function $f$ is called the limit function.
A major theorem says:
- If each $f_n$ is continuous on $D$,
- and $f_n \to f$ uniformly on $D$,
- then $f$ is continuous on $D$.
This is the meaning of preservation of continuity in this topic. The continuity of the functions in the sequence is carried over to the limit function, but only under the stronger condition of uniform convergence. âś…
Why is this important? Because in applied mathematics, physics, and numerical methods, functions are often approximated by simpler ones. If the approximations are continuous, we want the limit of the approximation process to stay continuous too.
For example, imagine a smooth animation made from many frames. If each frame changes smoothly, but the total sequence does not converge in the right way, the final image may still have a jump or a break. Uniform convergence prevents that kind of surprise.
Pointwise Convergence Alone Is Not Enough
To understand why uniform convergence matters, we first review pointwise convergence.
A sequence $\{f_n\}$ converges pointwise to $f$ on $D$ if for every fixed $x \in D$, we have
$$\lim_{n\to\infty} f_n(x) = f(x).$$
This means you choose one point $x$ and watch the values $f_n(x)$ as $n$ grows. However, the speed of convergence may depend on the point $x$.
That flexibility can cause continuity to fail in the limit. Consider the functions
$$f_n(x) = x^n$$
on the interval $[0,1]$.
Each $f_n$ is continuous on $[0,1]$. But the pointwise limit is
$$f(x) = \begin{cases}
0, & 0 \le x < 1, \\
$1, & x = 1.$
$\end{cases}$$$
This limit function is not continuous at $x = 1$. So here we have a sequence of continuous functions whose pointwise limit is discontinuous. âť—
This example shows that continuity is not automatically preserved by pointwise convergence.
Why does this happen? Near $x = 1$, the values $x^n$ stay close to $1$ for a long time, but for any fixed $x < 1$, they eventually go to $0$. The convergence is not evenly controlled across the whole interval.
Uniform Convergence and Why It Matters
Uniform convergence is the stronger idea that fixes this problem.
We say $f_n \to f$ uniformly on $D$ if
$$\lim_{n\to\infty} \sup_{x\in D} |f_n(x)-f(x)| = 0.$$
This means the difference $|f_n(x)-f(x)|$ becomes small for all $x \in D$ at once, not just one point at a time.
Another way to say this is: given $\varepsilon > 0$, there exists $N$ such that for all $n \ge N$ and all $x \in D$,
$$|f_n(x)-f(x)| < \varepsilon.$$
That “for all $x$” part is the key. It gives a uniform control over the whole domain.
Here is the big continuity result:
Theorem. If each $f_n$ is continuous on $D$ and $f_n \to f$ uniformly on $D$, then $f$ is continuous on $D$.
This theorem is one of the most important results in the topic of sequences of functions. It tells us that uniform convergence is strong enough to carry continuity from the approximating functions to the limit function.
Why the Theorem Is True
Let us explain the idea of the proof in simple terms.
Fix a point $a \in D$. We want to show that $f$ is continuous at $a$, meaning
$$\lim_{x\to a} f(x) = f(a).$$
Take any $\varepsilon > 0$. Since $f_n \to f$ uniformly, we can choose $N$ so that
$$|f_N(x)-f(x)| < \frac{\varepsilon}{3}$$
for all $x \in D$.
Because $f_N$ is continuous at $a$, there exists $\delta > 0$ such that whenever $|x-a| < \delta$,
$$|f_N(x)-f_N(a)| < \frac{\varepsilon}{3}.$$
Now use the triangle inequality:
$$|f(x)-f(a)| \le |f(x)-f_N(x)| + |f_N(x)-f_N(a)| + |f_N(a)-f(a)|.$$
Each term on the right is less than $\varepsilon/3$, so
$$|f(x)-f(a)| < \varepsilon.$$
That proves $f$ is continuous at $a$.
The proof works because uniform convergence controls the error between $f_n$ and $f$ everywhere, allowing the continuity of one continuous function $f_N$ to transfer to the limit $f$. 🔍
A Helpful Example: Polynomials Approximating a Function
Polynomials are continuous everywhere, so they are a good source of examples.
Suppose a sequence of polynomials $p_n$ converges uniformly to a function $f$ on a closed interval $[a,b]$. Since every polynomial is continuous, the theorem tells us that $f$ must also be continuous on $[a,b]$.
This is useful in approximation theory. For example, the Weierstrass approximation theorem says every continuous function on $[a,b]$ can be approximated uniformly by polynomials. The theorem about preservation of continuity guarantees that the uniform limit of those polynomials stays continuous.
Here is a concrete example:
$$p_n(x) = \frac{x}{1+n x^2}$$
on $[0,1]$.
Each $p_n$ is continuous. For each fixed $x \in [0,1]$,
$$\lim_{n\to\infty} p_n(x) = 0.$$
In fact, the convergence is uniform because
$$|p_n(x)| = \frac{x}{1+n x^2} \le \frac{1}{2\sqrt{n}}$$
for $x \in [0,1]$, so the maximum error goes to $0$. Therefore the limit function $f(x)=0$ is continuous, as expected.
What Can Go Wrong Without Uniform Convergence?
If the convergence is only pointwise, continuity can fail badly.
Recall the example $f_n(x)=x^n$ on $[0,1]$. Each $f_n$ is continuous, but the limit function jumps at $x=1$.
This is a classic warning sign in Real Analysis: even if every function in the sequence is nice, the limit may be less nice unless convergence is strong enough. Pointwise convergence asks about each point separately, while continuity is a local but still “whole-neighborhood” idea. Uniform convergence matches that neighborhood control much better.
Another useful fact is that the interval matters. On $[0,1)$, the same sequence $x^n$ converges pointwise to the zero function, which is continuous on $[0,1)$. But on the closed interval $[0,1]$, the discontinuity at $1$ appears. This shows how careful we must be about the domain.
Connection to the Bigger Picture in Sequences of Functions
Preservation of continuity is one piece of a larger story about limits of functions.
In the broader topic of sequences of functions, we study:
- pointwise convergence,
- uniform convergence,
- preservation of continuity,
- and later, preservation of integration and differentiation under stronger conditions.
These ideas are connected because they describe which properties survive when taking limits.
For continuity, the rule is clear:
- Pointwise convergence alone is not enough.
- Uniform convergence is enough.
This is a model example of a general principle in Real Analysis: the stronger the mode of convergence, the more properties are preserved.
You can think of it like building a bridge. If each section is well made but the connections between sections are weak, the whole bridge may fail. Uniform convergence is the strong connection that keeps the structure intact. 🧱
Conclusion
students, the main lesson is that continuity can pass from a sequence of functions to its limit, but only when the convergence is uniform. If $f_n$ are continuous and $f_n \to f$ uniformly, then $f$ is continuous. If the convergence is only pointwise, continuity may be lost, as shown by the example $f_n(x)=x^n$ on $[0,1]$.
This result is central in Real Analysis because it explains when approximation by continuous functions is safe. It also prepares you for deeper theorems about exchanging limits with other operations such as integration and differentiation.
Study Notes
- A sequence of functions $\{f_n\}$ converges pointwise to $f$ if for each fixed $x$, $\lim_{n\to\infty} f_n(x)=f(x)$.
- A sequence of functions $\{f_n\}$ converges uniformly to $f$ if $\lim_{n\to\infty} \sup_{x\in D}|f_n(x)-f(x)|=0$.
- Continuity is preserved under uniform convergence: if each $f_n$ is continuous on $D$ and $f_n \to f$ uniformly on $D$, then $f$ is continuous on $D$.
- Pointwise convergence does not preserve continuity in general.
- Example of failure: $f_n(x)=x^n$ on $[0,1]$ converges pointwise to a discontinuous function.
- The proof of preservation of continuity uses the triangle inequality and the continuity of one fixed function $f_N$.
- Uniform convergence is stronger than pointwise convergence because it controls the error everywhere at once.
- This topic is a key part of the larger study of sequences of functions in Real Analysis.