Sequences of Functions
Introduction
Imagine you have a list of different functions, where each function in the list is linked to a number $n = 1 , 2 , 3 \ldots \in \mathbb{N}$. 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 $( f_{n} ) = ( f_{1} , f_{2} , f_{3} , \ldots )$ 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 $n = 0 , 1 , 2 , 3 , \ldots$, we have:
\[f_{0} ( x ) & = \frac{x}{0 + 1} = x \\ f_{1} ( x ) & = \frac{x}{1 + 1} = \frac{x}{2} \\ f_{2} ( x ) & = \frac{x}{2 + 1} = \frac{x}{3} \\ f_{3} ( x ) & = \frac{x}{3 + 1} = \frac{x}{4} \\ \vdots\]Thus, the sequence of functions is:
\[( f_{n} ) = ( x , \frac{x}{2} , \frac{x}{3} , \frac{x}{4} , \ldots )\]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 ${ f_{n} ( x ) }$ be a sequence of functions defined on a common domain $A \subseteq \mathbb{R}$, with $n \in \mathbb{N}$. We say that the sequence ${ f_{n} ( x ) }$ converges pointwise on a set $C \subseteq A$ if, for every $x \in C$, the numerical sequence ${ f_{n} ( x ) }$ converges. In this case, the limit function $f ( x )$ is defined as:
\[f ( x ) = \underset{n \rightarrow + \infty}{lim} f_{n} ( x ) \forall x \in C\]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 $f : A \rightarrow \mathbb{R}$ if and only if $\forall x \in A$ and $\forall \epsilon > 0$ exists $K \in \mathbb{N}$ such that:
\[| f_{n} ( x ) - f ( x ) | < \epsilon \forall n \geq K\]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 $\epsilon$.
Let us consider the sequence of functions as previously discussed:
\[f_{n} ( x ) = \frac{x}{n + 1} , x \in \mathbb{R} , n \in \mathbb{N}_{0}\]Let us examine what happens for each $x$ as $n \rightarrow \infty$. If we fix a generic $x$, for example $x = 2$, the associated sequence is:
\[f_{0} ( 2 ) & = 2 \\ f_{1} ( 2 ) & = 1 \\ f_{2} ( 2 ) & = \frac{2}{3} \\ f_{3} ( 2 ) & = \frac{2}{4} \\ & \vdots\]This sequence of numbers tends to zero as $n \rightarrow \infty$. In general, for every $x \in \mathbb{R}$:
\[\underset{n \rightarrow \infty}{lim} f_{n} ( x ) = \underset{n \rightarrow \infty}{lim} \frac{x}{n + 1} = 0\]Therefore, the limit function is:
\[f ( x ) = 0 \forall x \in \mathbb{R}\]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: |
| 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.
\[f ( x ) = 0 \forall x \in ( - 1 , 1 )\]Consequences of pointwise convergence
Let $f_{n}$ be a sequence of functions $f_{n} : A \rightarrow \mathbb{R}$ that converges pointwise to a function $f : A \rightarrow \mathbb{R}$. 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 $f : A \rightarrow \mathbb{R}$ if, for any $\epsilon > 0$, there exists a natural number $K$ such that, for all $n \geq K$ and all $x \in A$, the following inequality holds:
\[| f_{n} ( x ) - f ( x ) | < \epsilon\]If $( f_{n} )$ converges uniformly to $f$, then $( f_{n} )$ also converges pointwise to $f .$
Let us consider the sequence of functions:
\[f_{n} ( x ) = \frac{x}{n} , x \in [ 0 , 1 ] , n \in \mathbb{N} .\]For each fixed $x$ in the interval $[ 0 , 1 ]$, we have:
\[\underset{n \rightarrow \infty}{lim} f_{n} ( x ) = 0\]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 $f ( x )$:
\[| f_{n} ( x ) - f ( x ) | = | \frac{x}{n} - 0 | = \frac{x}{n}\]The maximum value of this difference on the interval $[ 0 , 1 ]$ is:
\[\underset{x \in [ 0 , 1 ]}{sup} | f_{n} ( x ) - f ( x ) | = \frac{1}{n}\]Given any $\epsilon > 0$, we can choose $N$ such that:
\[\frac{1}{N} < \epsilon\]Therefore, for all $n \geq N$ and for all $x \in [ 0 , 1 ]$, we have:
\[| f_{n} ( x ) - f ( x ) | < \epsilon\]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 $[ 0 , 1 ]$.