I have spent a few hours trying to find an example of a sequence of continuous functions $f_n$ $[0,1]\rightarrow \mathbb{R}$ that pointwise converge to a function $f$: $[0,1]\rightarrow \mathbb{R}$ that is unbounded.

Attempt: I have so far only managed to think of an example where $f_n:(0,1]\rightarrow \mathbb{R}$ when we define: $$f_n = \frac{n}{nx+2}$$

This sequence converges to $f=\frac{1}{x}$, which is clearly not bounded on the interval.

Edit thanks to Olivier Moschetta

My issue is when $x=0$ the sequence does not pointwise converge to a function on $[0,1]$.

Can anyone help me fix this example?

  • 1
    $\begingroup$ Aren't the $f_n$ continuous at $0$? Their discontinuity point is $x=-2/n$ which is not in $[0,1]$. The problem is more that $f_n(0)=n/2$ is divergent, so $f_n$ do not converge pointwise to a function defined on $[0,1]$. $\endgroup$ Oct 3, 2020 at 21:21
  • $\begingroup$ @OlivierMoschetta Yes you are correct, I just fixed my wording. Thanks for the help! $\endgroup$
    – Jenny Liu
    Oct 3, 2020 at 21:27

2 Answers 2


It's pretty easy to "fix" your example. Define $f_n(x)=\frac{nx}{nx^2+2}$. At $x=0$ the sequence converges to $0$, at any other point to $\frac{1}{x}$.


You can take, for instance, $f_n\colon[0,1]\longrightarrow\Bbb R$ defined by$$f_n(x)=\begin{cases}2n^2x&\text{ if }x\leqslant\frac1{2n}\\n&\text{ if }\frac1{2n}\leqslant x\leqslant\frac1n\\\frac1x&\text{ otherwise.}\end{cases}$$Each $f_n$ is continuous and the sequence $(f_n)_{n\in\Bbb N}$ converges pointwise to$$\begin{array}{ccc}[0,1]&\longrightarrow&\Bbb R\\x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$

  • $\begingroup$ What a good example ! $\endgroup$ Oct 3, 2020 at 21:35

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