Pointwise Convergence

From Power Series to Sequences of Functions

In Power Series Mastery, convergence was used to represent a function by a power series near a chosen center. A natural next step is to treat the partial sums themselves as functions and ask what it means for a sequence of functions to converge. The central idea is to fix an input $x$, evaluate every function there, and study the resulting sequence of real numbers.

This point-by-point approach is called pointwise convergence. The order of its quantifiers matters: for each fixed point and each desired accuracy, an index can be chosen. That index may depend on the point. It need not work for all points in the domain at once.

The Definition

Definition: Let $E\subseteq\mathbb{R}$, and let $f_n:E\to\mathbb{R}$ for each positive integer $n$. The sequence $(f_n)$ converges pointwise on $E$ to a function $f:E\to\mathbb{R}$ if, for every $x\in E$ and every $\varepsilon>0$, there is a positive integer $N$, which may depend on both $x$ and $\varepsilon$, such that $$ n\geq N \quad\Longrightarrow\quad |f_n(x)-f(x)|<\varepsilon. $$ We write $f_n\to f$ pointwise on $E$.

For a fixed $x$, this is exactly the definition that the numerical sequence $(f_n(x))$ converges to $f(x)$. The definition asks for this convergence at every point of $E$, but it does not require the same $N$ to work at different points.

A useful way to read the definition is to keep $x$ fixed before choosing $N$. Once $x$ is fixed, the functions $f_n$ have become real numbers $f_n(x)$. As $x$ varies, the index needed to meet a given error tolerance can vary as well.

Worked Examples

Worked Example: Powers on the Unit Interval

Define $f_n(x)=x^n$ on $E=[0,1]$. We determine the pointwise limit, treating the endpoint $x=1$ separately from the other points.

At $x=1$, $f_n(1)=1$ for every $n$, so the limit is $1$. At $x=0$, $f_n(0)=0$ for every positive integer $n$, so the limit is $0$. If $0<x<1$, then $x^n\to0$. To see the decay directly, set $r=1/x>1$. The binomial theorem gives

r^n=(1+(r-1))^n\geq 1+n(r-1), \qquad 0<x^n=\frac{1}{r^n}\leq\frac{1}{1+n(r-1)}.

The right-hand side tends to $0$ as $n\to\infty$, so $x^n\to0$. Thus the pointwise limit is

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

In particular, the value at $1$ cannot be inferred by taking the formula for $x<1$ and substituting $x=1$. Pointwise limits are determined separately at each point by the sequence of values there.

Worked Example: A Peak That Moves with the Index

On $E=\mathbb{R}$, define

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

At $x=0$, $f_n(0)=0$ for every $n$. If $x\neq0$, then

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

For this fixed nonzero $x$, the bound tends to $0$. Therefore $f_n(x)\to0$ at every real $x$, and the pointwise limit is the zero function.

However, at the input $x=1/n$, which changes with $n$, direct substitution gives

f_n(1/n) =\frac{n(1/n)}{1+n^2(1/n)^2} =\frac{1}{1+1} =\frac12.

There is no contradiction. Pointwise convergence holds $x$ fixed while $n$ increases; the calculation at $x=1/n$ changes the input as the index changes. This example warns against replacing a fixed-point argument with an estimate at inputs that move with $n$.

Worked Example: Geometric Partial Sums

For $-1<x<1$, define the partial-sum functions

s_n(x)=1+x+x^2+\cdots+x^n.

For each fixed $x\neq1$, the finite geometric-sum identity gives

s_n(x)=\frac{1-x^{n+1}}{1-x}.

The identity can be checked by multiplying the sum by $1-x$: all intermediate powers cancel, leaving $1-x^{n+1}$. Since $|x|<1$, $x^{n+1}\to0$, so

s_n(x)\longrightarrow\frac{1}{1-x}.

At $x=0$, the same formula gives $s_n(0)=1$ and $1/(1-0)=1$, so the conclusion also holds there. Consequently, the partial-sum functions converge pointwise on $(-1,1)$ to $x\mapsto1/(1-x)$. This is a simple example of how convergence of a power series at each point in its interval of convergence gives pointwise convergence of its partial sums.

Uniqueness of the Pointwise Limit

A pointwise limit, if it exists, is determined uniquely. This follows from the uniqueness of limits of real sequences, applied at each input. The short proof also emphasizes why the pointwise definition is a statement about ordinary numerical limits at fixed points.

Theorem (Uniqueness of the Pointwise Limit): Suppose $f_n:E\to\mathbb{R}$ converges pointwise on $E$ to both $f$ and $g$. Then $f(x)=g(x)$ for every $x\in E$.

Proof. Fix $x\in E$. Suppose, for contradiction, that $f(x)\neq g(x)$, and let $d=|f(x)-g(x)|>0$. Since $f_n(x)\to f(x)$, there is an index $N_1$ such that $n\geq N_1$ implies $|f_n(x)-f(x)|<d/3$. Since $f_n(x)\to g(x)$, there is an index $N_2$ such that $n\geq N_2$ implies $|f_n(x)-g(x)|<d/3$. Choose $n\geq\max(N_1,N_2)$. The triangle inequality then gives

d=|f(x)-g(x)| \leq |f(x)-f_n(x)|+|f_n(x)-g(x)| <\frac{2d}{3},

which is impossible because $d>0$. Hence $f(x)=g(x)$. Since $x$ was arbitrary, the functions agree at every point of $E$. $\square$

Algebra of Pointwise Limits

The familiar limit rules for real sequences can be applied separately at each point. The resulting rules let us combine pointwise limits without returning to the definition each time.

Theorem (Algebraic Limit Laws for Pointwise Convergence): Suppose $f_n\to f$ and $g_n\to g$ pointwise on $E$. For any real constant $c$, the sequences $f_n+g_n$, $cf_n$, and $f_ng_n$ converge pointwise on $E$ to $f+g$, $cf$, and $fg$, respectively.

Proof. Fix $x\in E$. For sums, let $\varepsilon>0$. Pointwise convergence provides indices after which $|f_n(x)-f(x)|<\varepsilon/2$ and $|g_n(x)-g(x)|<\varepsilon/2$. For every $n$ beyond both indices,

|(f_n+g_n)(x)-(f+g)(x)| \leq |f_n(x)-f(x)|+|g_n(x)-g(x)| <\varepsilon.

For scalar multiples, if $c=0$ the assertion is immediate. If $c\neq0$, choose an index after which $|f_n(x)-f(x)|<\varepsilon/|c|$. Then

|cf_n(x)-cf(x)|=|c|\,|f_n(x)-f(x)|<\varepsilon.

For products, write $a=f(x)$ and $b=g(x)$. Because $f_n(x)\to a$, there is an index after which $|f_n(x)-a|<1$; for those indices, $|f_n(x)|\leq|a|+1$. Given $\varepsilon>0$, pointwise convergence also provides indices after which

|g_n(x)-b|<\frac{\varepsilon}{2(|a|+1)} \quad\text{and}\quad |f_n(x)-a|<\frac{\varepsilon}{2(|b|+1)}.

Take $n$ large enough that all three bounds hold. Adding and subtracting $f_n(x)b$, then using the triangle inequality, yields

|f_n(x)g_n(x)-ab| \leq |f_n(x)|\,|g_n(x)-b|+|b|\,|f_n(x)-a| <\frac{\varepsilon}{2} +\frac{|b|\varepsilon}{2(|b|+1)} <\varepsilon.

Thus the products converge to $ab=f(x)g(x)$ at the fixed point $x$. Since $x$ was arbitrary, all three conclusions hold pointwise on $E$. $\square$

Worked Example: Combining Pointwise Limits

On $(-1,1)$, let $u_n(x)=x^n$ and $v_n(x)=1-x^n$, with $n$ a positive integer. For every fixed $x$ in this interval, $x^n\to0$, including $x=0$. Hence $u_n\to0$ and $v_n\to1$ pointwise there. The algebraic limit laws imply that

u_n(x)+v_n(x)=1 \quad\text{and}\quad u_n(x)v_n(x)=x^n(1-x^n)

converge pointwise to $1$ and $0$, respectively. The product conclusion can also be checked directly: since $|1-x^n|\leq2$ for $|x|<1$,

|x^n(1-x^n)|\leq2|x|^n\longrightarrow0.

This illustrates how the pointwise limit laws reduce function-sequence calculations to numerical limits at each fixed input.

What Pointwise Convergence Does Not Preserve

Pointwise convergence is a local condition in the sense that it tests one input at a time. It does not, by itself, control what happens across the whole domain with one shared index. The moving-peak example showed how an error can remain large at inputs that change with $n$, even though the sequence converges at every fixed point.

Another important consequence is that pointwise limits of continuous functions need not be continuous. In the first worked example, every $f_n(x)=x^n$ is a polynomial and therefore continuous on $[0,1]$. Its pointwise limit is $0$ for $x<1$ and $1$ at $x=1$. This limit is discontinuous at $1$: for any $\delta>0$, choose $x=1-\min(\delta/2,1/2)$. Then $x\in[0,1)$, $|x-1|<\delta$, but $|f(x)-f(1)|=1$. Thus continuity of every function in the sequence does not guarantee continuity of the pointwise limit.

The key distinction is between an index chosen after fixing $x$ and an index that works across the domain. Pointwise convergence only asserts the first. The stronger form of convergence that supplies a single index for all inputs will be the subject of the next tutorial.

Takeaway: Pointwise convergence means that $f_n(x)$ converges as a real sequence for every fixed $x$. The limit is unique, and sums, scalar multiples, and products behave as expected. But the required index may vary with $x$, so pointwise convergence alone does not preserve continuity.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

In the definition of pointwise convergence, which quantities may the index $N$ depend on?
Find the pointwise limit of $f_n(x)=x^n$ on $[0,1]$, including its values at both endpoints.
For the moving-peak example, why does evaluating at $x=1/n$ not contradict pointwise convergence to zero?
If $f_n\to f$ and $g_n\to g$ pointwise, what is the pointwise limit of $f_ng_n$?
What feature of the limit of $x^n$ on $[0,1]$ shows that pointwise convergence need not preserve continuity?

What You'll Learn