Pointwise Convergence of Sequences of Functions

Pointwise Convergence of Sequences of Functions

Recall from the Sequences of Functions page that an (infinite) sequence of functions $(f_n(x))_{n=1}^{\infty}$ is simply just that - an infinite sequence whose terms are functions.

Like with sequences of real numbers or sequences of elements in metric spaces - we can also define whether the sequence of functions converges (or diverges) to a limit function $f$, however, we must be careful as to what we mean exactly by "convergence". We begin by describing pointwise convergence of sequences of functions which we define below.

Definition: Let $(f_n(x))_{n=1}^{\infty}$ be a sequence of functions with common domain $X$. Then $(f_n)_{n=1}^{\infty}$ is said to be Pointwise Convergent to the the function $f$ written $\lim_{n \to \infty} f_n(x) = f(x)$ if for all $x \in X$ and for all $\epsilon > 0$ there exists a $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid f_n(x) - f(x) \mid < \epsilon$.

For example, consider the following sequence of functions defined on $[0, 1]$:

\begin{align} \quad (f_n(x))_{n=1}^{\infty} = \left ( \frac{1}{n} x \right )_{n=1}^{\infty} = \left (x, \frac{1}{2}x, \frac{1}{3}x, ..., \frac{1}{n}x, ... \right ) \end{align}

We claim that $(f_n(x))_{n=1}^{\infty}$ is pointwise convergent to $f(x) = 0$. The following image shows the first six functions in the sequence given above. It should be intuitively clear that the sequence $(f_n(x))_{n=1}^{\infty}$ converges to the limit function $f(x) = 0$.

To show this, fix $x \in [0, 1]$ and assume that $x \neq 0$ and let $\epsilon > 0$ be given. Then since $n, x > 0$ we have that:

\begin{align} \quad \mid f_n(x) - f(x) \mid = \biggr \lvert \frac{1}{n} x - 0 \biggr \rvert = \biggr \lvert \frac{x}{n} \biggr \rvert = \frac{x}{n} \end{align}

Choose $N \in \mathbb{N}$ such that $N > \frac{x}{\epsilon}$ which can be done by the Archimedean property. Then $\frac{1}{N} < \frac{\epsilon}{x}$ and so for $n \geq N$ we have that:

\begin{align} \quad \frac{x}{n} \leq \frac{x}{N} < \frac{\epsilon x}{x} = \epsilon \end{align}

Therefore $\lim_{n \to \infty} f_n(x) = f(x)$ for $x \in (0, 1]$. Now, for $x = 0$, notice that:

\begin{align} \quad (f_n(0))_{n=1}^{\infty} = (0)_{n=1}^{\infty} = (0, 0, ...) \end{align}

This sequence clearly converges to $f(0) = 0$. So, we conclude that $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x)}$ for all $x \in [0, 1]$. Hence the sequence $(f_n(x))_{n=1}^{\infty}$ is pointwise convergent on all of $[0, 1]$.

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