Sequences of Functions

Copy Markdown View Source

Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/sequences-of-functions/ Fetched from algebrica.org post 17568; source modified 2026-04-12T20:24:14.

Introduction

Imagine you have a list of different functions, where each function in the list is linked to a number $n = 1, 2, 3… \in \mathbb{N}$. So, for each $n$, you get a different function, and that ordered list of functions is essentially what a sequence of functions is. In formal terms, let $A \subseteq \mathbb{R}$ be a non-empty subset and suppose that for each $n \in \mathbb{N}$ we have a function $f_n: A \rightarrow \mathbb{R}$. We then say that $(f_n) = (f_1, f_2, f_3, \dots)$ is a sequence of functions on $A.$

Let’s consider a simple practical example. Let $(f_n)$ be a sequence of functions, with $n \in \mathbb{N}_0$ and $x \in \mathbb{R}$, defined by:

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

This is a family of functions where each function is linear, and the slope decreases as $n$ increases. For $n = 0, 1, 2, 3, …$, we have:

\begin{aligned} f_0(x) &= \frac{x}{0+1} = x \\[0.5em] f_1(x) &= \frac{x}{1+1} = \frac{x}{2} \\[0.5em] f_2(x) &= \frac{x}{2+1} = \frac{x}{3} \\[0.5em] f_3(x) &= \frac{x}{3+1} = \frac{x}{4} \\[0.5em] \vdots \end{aligned}

Thus, the sequence of functions is:

(f_n) = \left( x, \frac{x}{2}, \frac{x}{3}, \frac{x}{4}, \dots \right)

Graphically, this situation can be observed as follows:

The graph shows how, as the index $n$ increases, the slope of the line $f_n(x)$ progressively decreases. This reflects the fact that the function flattens toward the zero function $f(x) = 0$ for every $x$. In other words, the sequence of functions $f_n(x)$ converges pointwise to the zero function as $n$ approaches infinity.

Pointwise convergence

Let $\lbrace f_n(x) \rbrace$ be a sequence of functions defined on a common domain $A \subseteq \mathbb{R}$, with $n \in \mathbb{N}$. We say that the sequence $\lbrace f_n(x) \rbrace$ converges pointwise on a set $C \subseteq A$ if, for every $x \in C$, the numerical sequence $\lbrace f_n(x) \rbrace$ converges. In this case, the limit function $f(x)$ is defined as:

f(x) = \lim_{n \to +\infty} f_n(x) \quad \forall \, x \in C

The set $C$ is called the pointwise convergence set of the sequence ${f_n(x)}$.

Pointwise convergence can also be expressed as follows. Let $(f_n)$ be a sequence of functions defined on a set $A$. Then $(f_n)$ converges pointwise to $f : A \to \mathbb{R}$ if and only if $\forall \, x \in A$ and $\forall \varepsilon > 0$ exists $K \in \mathbb{N}$ such that:

|f_n(x) - f(x)| < \varepsilon \quad \forall \; n \geq K
In other words, for each fixed point $x$, we can make $f_n(x)$ as close as we like to $f(x)$ by choosing $n$ large enough. The index $K$ required to achieve the desired accuracy may vary depending on $x$ and $\varepsilon$.

Let us consider the sequence of functions as previously discussed:

f_n(x) = \frac{x}{n+1}, \quad x \in \mathbb{R}, \quad n \in \mathbb{N}_0

Let us examine what happens for each $x$ as $n \to \infty$. If we fix a generic $x$, for example $x = 2$, the associated sequence is:

\begin{aligned} f_0(2) &= 2 \\[0.5em] f_1(2) &= 1 \\[0.5em] f_2(2) &= \frac{2}{3} \\[0.5em] f_3(2) &= \frac{2}{4} \\[0.5em] &\vdots \end{aligned}

This sequence of numbers tends to zero as $n \to \infty$. In general, for every $x \in \mathbb{R}$:

\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{x}{n+1} = 0

Therefore, the limit function is:

f(x) = 0 \quad \forall \, x \in \mathbb{R}

By the uniqueness of limits of sequences of real numbers, the pointwise limit of a sequence of functions $(f_n)$ is unique.

Example

Let’s study the behavior of the following sequence of functions in the interval $-1 < x < 1$:

f_n(x) = x^n

For a fixed value of $x$ in this interval, we know that the absolute value of $x$ is less than $1$, that is, $|x| < 1$. This means we are considering powers of a number smaller than $1$ in absolute value. By properties of exponents, when the base has an absolute value less than $1$, the sequence $x^n$ tends to zero as $n$ tends to infinity:

\lim_{n \to \infty} x^n = 0

The sequence of powers of a real number $x$ with $|x| < 1$ converges to zero.

Therefore, for every $x$ in the interval $-1 < x < 1$, the sequence of functions $f_n(x) = x^n$ converges pointwise to the zero function.

f(x) = 0 \quad \forall \, x \in (-1,1)

Consequences of pointwise convergence

Let ${f_n}$ be a sequence of functions $f_n : A \to \mathbb{R}$ that converges pointwise to a function $f : A \to \mathbb{R}$. The following properties hold:

  • If each $f_n(x) \geq 0$ for all $x \in A$, then $f(x) \geq 0$ for all $x \in A$. In practice, if each function $f_n(x)$ is non-negative on $A$, then the limit function $f(x)$ will also be non-negative on $A$. This reflects the fact that the limit of a sequence of non-negative real numbers cannot be negative.

  • If each $f_n$ is non-decreasing on $A$, then $f$ is non-decreasing on $A$. Consequently, if each function $f_n$ is non-decreasing on $A$, then the limit function $f$ will also be non-decreasing on $A$. In other words, the property of monotonicity is preserved under pointwise convergence.

Uniform convergence

Let $(f_n)$ be a sequence of functions defined on a set $A \subseteq \mathbb{R}$. We say that $(f_n)$ converges uniformly on $A$ to the function $f : A \to \mathbb{R}$ if, for any $\varepsilon > 0$, there exists a natural number $K$ such that, for all $n \geq K$ and all $x \in A$, the following inequality holds:

|f_n(x) - f(x)| < \varepsilon

If $(f_n)$ converges uniformly to $f$, then $(f_n)$ also converges pointwise to $f.$

Let us consider the sequence of functions:

f_n(x) = \frac{x}{n}, \quad x \in [0,1], \quad n \in \mathbb{N}.

For each fixed $x$ in the interval $[0,1]$, we have:

\lim_{n \to \infty} f_n(x) = 0

This means that the sequence $f_n(x)$ converges pointwise to the function (f(x) = 0). Let us now check whether the convergence is uniform on $[0,1]$. We compute the difference between $f_n(x)$ and the limit function $f(x)$:

|f_n(x) - f(x)| = \left| \frac{x}{n} - 0 \right| = \frac{x}{n}

The maximum value of this difference on the interval $[0,1]$ is:

\sup_{x \in [0,1]} |f_n(x) - f(x)| = \frac{1}{n}

Given any $\varepsilon > 0$, we can choose $N$ such that:

\frac{1}{N} < \varepsilon

Therefore, for all $n \geq N$ and for all $x \in [0,1]$, we have:

|f_n(x) - f(x)| < \varepsilon

This confirms that the sequence of functions $f_n(x) = \frac{x}{n}$ converges uniformly to the limit function $f(x) = 0$ on the interval $[0,1]$.