I'm reading about the definition of exponential function on $\Bbb R$ and I came across a definition of pointwise convergence which I don't understand:

We say that a sequence of functions $(f_n)_n$ where $f_n:I\rightarrow \Bbb R, \ I\subseteq \Bbb R$, converges pointwise to function $f:I\rightarrow \Bbb R$ on the interval $I$ if a sequence of numbers $(f_n(x))_n$ converges to $f(x), \forall x\in I.$

Can someone please explain this definition and provide an example?

  • $\begingroup$ $f_n(x)=\frac1x-\frac1n$, for an example. $I=\mathbb R\setminus \{0\}$. $\endgroup$
    – zoli
    Feb 5, 2017 at 13:18
  • $\begingroup$ What is unclear to you? The definition is itself evident. See personal.psu.edu/auw4/M401-notes1.pdf for some examples. $\endgroup$
    – Crostul
    Feb 5, 2017 at 13:22
  • 2
    $\begingroup$ What the definition says is that for each fixed $x \in I$, the sequence $(f_n(x))_n$ converges to the value $f(x)$. Example: let $f_n(x) = \frac{x}{n}$. For a fixed $x \in \Bbb R$, the sequence $(f_n(x))$ converges to $0$ because $\lim_{n \to\infty} f_n(x) = \lim_{n \to \infty} \frac{x}{n} = x \lim_{n \to \infty} \frac1n = x \cdot 0 = 0$. So if we define $f: \Bbb R \to \mathbb R$ by $f: x \mapsto 0$, we get that $(f_n)$ converges to $f$ pointwise on $\mathbb R$. $\endgroup$
    – user384138
    Feb 5, 2017 at 13:23

2 Answers 2


Pointwise convergence of $(f_n)_{n\in\mathbb{N}}$ to $f$ means that for each point $x \in I$, we have $\lim_{n \rightarrow \infty}{f_n(x)}=f(x)$.

Essentially we take a point $x \in I$ and look at $f_n(x)$ as $n \rightarrow \infty$. If this converges to a limit and does so for all $x \in I$, then it makes sense to say $(f_n)_{n\in\mathbb{N}}$ converges to the function $f(x)=\lim_{n\rightarrow\infty}f_n(x)$.

We call this pointwise convergence because it only looks at individual points rather than the functions as a whole, which is in contrast to something stronger like uniform convergence.

An example would be $$f_n(x)=\frac{x^2}{n} (x \in \mathbb{R})$$

For any $x \in \mathbb{R}$ we then have $$ \lim_{n \rightarrow \infty}f_n(x) = \lim_{n \rightarrow \infty}\frac{x^2}{n} = x^2\lim_{n \rightarrow \infty}\frac{1}{n}=x^2 \cdot0 = 0$$

So $(f_n)_{n\in\mathbb{N}}$ converges pointwise to the null function $f(x)=0$


To converge pointwise in this context means that you can take any $x\in I$ and then look at the function $f_n$ evaluated at $x$. This gives you a value in $\mathbb{R}$ and we call it $f_n(x)$. Now if you look at this with $x$ fixed you look at a sequence of numbers! For every $n$ you have a different number in $\mathbb{R}$ and this sequence converges to the number $f(x)$. If the same holds for any $x \in I$, i.e. any sequence $f_n(x)$ (now with $x$ arbitrary, not fixed) converges to its respective limit $f(x)$, we say the the sequence of functions $f_n$ converges to the function $f$.

An example is provided by the function $f\colon (0,1) \rightarrow \mathbb{R}, x \mapsto x^n$, which converges pointwise to the zero function on $(0,1)$. For any $x\in (0,1)$, the sequence $x^n$ converges to $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.