Sequences of Functions

What is a Sequence of Functions?

You already know what it means for a sequence of real numbers $(a_n)$ to converge: for every $\varepsilon > 0$, all terms are eventually within $\varepsilon$ of the limit $L$.

Now suppose instead of a single number at each stage, we have a function $f_n : A \to \mathbb{R}$. For each fixed $x \in A$, evaluating the sequence at $x$ gives a sequence of real numbers: $f_1(x), f_2(x), f_3(x), \ldots, f_n(x), \ldots$

So $(f_n)$ is simply a sequence whose terms happen to be functions rather than numbers. Let us consider some examples.

$f_n(x)$	Domain	First few terms of the sequence $(f_n)$
$\frac{x}{n}$	$\mathbb{R}$
$x^n$	$[0,1]$

Question: What does it mean for the sequence $(f_n)$ to converge, and to what kind of object?

Two Examples

Activity: Finding Pointwise Limits

For each family below, fix an arbitrary $x$ in the domain and ask: what does the sequence of numbers $f_1(x), f_2(x), f_3(x), \ldots$ converge to? Compute at several specific values of $x$, then write a formula for the limit.

Family A: $f_n(x) = \frac{x}{n}$ on $\mathbb{R}$

A1. Fix $x = 3$. Write out $f_1(3), f_2(3), f_5(3), f_{10}(3)$. What do these numbers converge to?

A2. For an arbitrary fixed $x \in \mathbb{R}$, compute $\lim_{n\to\infty} \frac{x}{n}$.

A3. Write the limit function $f(x) = \lim_{n\to\infty} f_n(x)$. Is $f$ continuous?

Family B: $g_n(x) = x^n$ on $[0,1]$

B1. Fix $x = 1/2$. Write out $g_1(1/2), g_2(1/2), g_5(1/2), g_{10}(1/2)$. What do these numbers converge to?

B2. Fix $x = 9/10$. How large must $n$ be before $g_n(9/10) < 0.01$? Compare this with the $n$ needed for $x = 1/2$. What does the difference tell you?

B3. For arbitrary fixed $x \in [0,1)$, compute $\lim_{n\to\infty} x^n$. What about $x = 1$? Write the limit function $g(x)$.

B4. Each $g_n$ is continuous on $[0,1]$. Is $g$ continuous on $[0,1]$?

The Problem: Lost Continuity

What just happened?

In both families above, every $f_n$ was continuous. Yet in each case the pointwise limit failed to be continuous at one point.

The obvious conjecture — the pointwise limit of continuous functions is continuous — is false.

The word "limit" is doing two different jobs simultaneously. Continuity of the limit requires interchanging the order of two limit operations:

$\lim_{x \to c}\Bigl(\lim_{n \to \infty} f_n(x)\Bigr) \overset{?}{=} \lim_{n \to \infty}\Bigl(\lim_{x \to c} f_n(x)\Bigr)$

This interchange is not free. The rest of this handout is about finding the right condition that makes it valid.

Pointwise Convergence

Definition (Pointwise Convergence)

For each $n \in \mathbb{N}$, let $f_n$ be a function defined on a set $A \subseteq \mathbb{R}$. The sequence $(f_n)$ converges pointwise on $A$ to $f$ if, for every $x \in A$, $\lim_{n\to\infty} f_n(x) = f(x)$.

Equivalently: for every $\varepsilon > 0$ and every $x \in A$, there exists $N \in \mathbb{N}$ (which may depend on $x$) such that $n \geq N$ implies $|f_n(x) - f(x)| < \varepsilon$.

Example 1. Let $f_n(x) = \frac{x^2 + nx}{n}$ on $\mathbb{R}$. Show that $f_n \to f$ pointwise and identify $f$.

Example 2. Let $f_n(x) = x^n$ on $[0,1]$. Show that $f_n \to g$ pointwise and identify $g$.

Why Pointwise Convergence Is Not Enough

Suppose $(f_n)$ is a sequence of continuous functions on $A$ converging pointwise to $f$. Attempt to prove $f$ is continuous at $c \in A$.

Fix $c \in A$ and $\varepsilon > 0$. By the triangle inequality:

$|f(x) - f(c)| \leq$ + +

(a) Term (III): Since $f_n(c) \to f(c)$, we can choose $N$ so that term (III) $< \varepsilon/3$.

(b) Term (II): Once $N$ is fixed, continuity of $f_N$ provides $\delta > 0$ so term (II) $< \varepsilon/3$ whenever $|x - c| < \delta$.

(c) Why can't we control term (I) for all $x$ with $|x-c| < \delta$ simultaneously?

Uniform Convergence

Definition (Uniform Convergence)

The sequence $(f_n)$ converges uniformly on $A$ to $f$ if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ (independent of $x$) such that $n \geq N$ implies $|f_n(x) - f(x)| < \varepsilon$ for all $x \in A$.

The Quantifier Picture

Type	Quantifier Order
Pointwise	$\forall \varepsilon > 0, \forall x \in A, \exists N \in \mathbb{N}$ (depending on $x$): $n \geq N \Rightarrow \|f_n(x)-f(x)\| < \varepsilon$
Uniform	$\forall \varepsilon > 0, \exists N \in \mathbb{N}$ (depending only on $\varepsilon$), $\forall x \in A$: $n \geq N \Rightarrow \|f_n(x)-f(x)\| < \varepsilon$

Geometric Picture: Uniform convergence of $f_n \to f$ on $A$ means: for every $\varepsilon$-band around the graph of $f$, there exists $N$ such that the entire graph of $f_n$ lies inside the band for all $n \geq N$.

Example 3. Let $g_n(x) = \frac{1}{n(1+x^2)}$ on $\mathbb{R}$. Find the pointwise limit. Is convergence uniform?

Scratch work:

Formal proof:

Example 4. Let $f_n(x) = \frac{x^2 + nx}{n}$ on $\mathbb{R}$. Find the pointwise limit. Show convergence is not uniform on $\mathbb{R}$ but is uniform on every bounded interval $[-b,b]$.

Scratch work:

Formal proof:

Cauchy Criterion for Uniform Convergence

Theorem

A sequence of functions $(f_n)$ defined on $A \subseteq \mathbb{R}$ converges uniformly on $A$ if and only if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|f_n(x) - f_m(x)| < \varepsilon$ whenever $m,n \geq N$ and $x \in A$.

An important advantage of the Cauchy criterion is that it lets us verify uniform convergence without knowing the limit function in advance.

Proof sketch. ($\Rightarrow$) Suppose $f_n \to f$ uniformly.

— Write down the $\varepsilon$-$N$ definition of uniform convergence for $f_n \to f$

— Use triangle inequality to show that for $m, n \geq N$ and $x\in A$: $|f_n(x)-f_m(x)|<\varepsilon.$

($\Leftarrow$) Assume the Cauchy condition holds.

— Explain why for each fixed $x \in A$, the sequence $(f_n(x))$ is Cauchy in $\mathbb{R}$, and converges.

— Define $f(x) = \lim_{n\to\infty} f_n(x)$. Now show $f_n \to f$ uniformly:

Uniform Convergence and Continuity

Theorem

Let $(f_n)$ be a sequence of functions defined on $A \subseteq \mathbb{R}$ converging uniformly on $A$ to $f$. If each $f_n$ is continuous at $c \in A$, then $f$ is continuous at $c$.

Proof sketch: Fix $c \in A$ and $\varepsilon > 0$.

Step 1. Write the $\varepsilon$-$N$ definition of $f_n \to f$ uniformly.

Step 2. Now $N \in \mathbb{N}$ is fixed. Write down the $\varepsilon$-$\delta$ definition of continuity for $f_N$ at $c$.

Step 3. For $|x - c| < \delta$, apply the triangle inequality:

$|f(x) - f(c)| \leq$ + +
$< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$

Since $\varepsilon>0$ and $c \in A$ was arbitrary, $f$ is continuous at $c$. $\square$

Activity

Problem 1. Find the pointwise limit and determine uniform convergence

(a) $f_n(x) = \frac{nx}{1+nx^2}$ on $(0,\infty)$.

(b) $f_n(x) = \frac{x}{1+nx^2}$ on $\mathbb{R}$.

Problem 2. True or False (justify each)

(a) If $f_n \to f$ uniformly on $A$ and each $f_n$ is bounded, then $f$ is bounded.

(b) If $f_n \to f$ uniformly on $A$ and each $f_n$ is uniformly continuous on $A$, then $f$ is uniformly continuous on $A$.