Pointwise/uniform convergence of $ f(x) = \begin{cases} +\infty, & \text{if }0<x<0 \text{ (???)}\\ 1, & \text{if }1/(2n)≤x<1 \end{cases} $

Question

$$ f_n(x) = \begin{cases} -n^3x+2n^2, & \text{if }0<x<1/(2n) \\ n/(n+1), & \text{if }1/(2n)≤x<1 \end{cases} $$ $ n\in \mathbb{N}$

How can I prove that $f_n$ is pointwise convergent/uniform convergent? What I have tried so far: I think the limiting function $f(x)$ is: $$ f(x) = \begin{cases} +\infty, & \text{if }0<x<0 \text{ (???)}\\ 1, & \text{if }1/(2n)≤x<1 \end{cases} $$ But I'm not sure about that and as $n/(n+1)$ converges to 1 for n to $\infty$ I thought that $f_n$ is pointwise convergent, because $f(1)=1$ as well.

Answer 1

It's not that simple. You can't write $0<x<0$ or even $=+\infty$, it's not a number !

In fact, the thing that disturbs you ( and that you noticed ) is that $$ \frac{1}{2n} \underset{n \rightarrow +\infty}{\rightarrow }0 $$ Hence for all $x>0$ it exists $N$ so that if $\ n > N$ then $\displaystyle \frac{1}{2n}<x$. And then, still for $n>N$ $$ f_{n}\left(x\right)=\frac{n}{n+1} $$ Then if you consider the subsequence $\displaystyle \left(f_n\right)_{n \geq N}$, it converges to the function $\displaystyle x \mapsto 1$.

What can you conclude ?

Answer 2

\begin{equation} f(x) = \begin{cases} +\infty, & \text{if }0<x<0 (???)\\ 1, & \text{if }1/(2n)≤x<1 \end{cases} \end{equation}

Note that your limiting function $f$ depends on $n$, which does not make sense.

Limit of $f_n$ is $f(x)$ = 1, a constant function.

Take any point $x\in(0,1)$, and show that $|f_n(x) - f(x)| = |f_n(x) - 1| \to 0$.

Answer 3

Clearly $f_n:(0,1)\to\mathbb{R}$. You have written $0<x<0 $ ??? Indeed: what should be on the right side? Nothing. Since $1/(2n)$ can be aribtrary close to $0$, it holds that for every $x\in (0,1)$ there is some $N$ such that for all $n>N$ we have: $f_n(x)=\frac{n}{n+1}$. And that implies $f(x)=\lim_{n\to\infty}f_n(x)=1$ for all $x\in (0,1)$.

That was pointwise convergence. What about uniform? It does not converge uniformly, take for example $x_n=\frac{1}{4n}<\frac{1}{2n}$. Now we have: \begin{align} \sup_{x\in (0,1)}|f_n(x)-1|\geq |f_n(x_n)-1|=|-n^2/4+2n^2-1|\to \infty \end{align} as $n\to\infty$ hence the convergence cannot be uniform.