# Sequence of continuous functions on $[0,1]$ pointwise converging to an unbounded function

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?

• 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 $[0,1]$. Oct 3, 2020 at 21:21
• @OlivierMoschetta Yes you are correct, I just fixed my wording. Thanks for the help! Oct 3, 2020 at 21:27

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$$\begin{array}{ccc}[0,1]&\longrightarrow&\Bbb R\\x&\mapsto&\begin{cases}0&\text{ if }x=0\\\frac1x&\text{ otherwise.}\end{cases}\end{array}$$