If $f_n\to f$ uniformly and $f$ bounded, then $f_n$ is uniformly bounded proof method question

Question

Let $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded. Show that $f_n$ is uniformly bounded.

This statement is clearly false. But, to provide a counter example to this statement, I passed by a quite strange way using contraposition (because didn't find a counter example for the original statement :( ).

By contraposition, if $f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded $\implies$ $f_n$ is uniformly bounded, then the following must be true as well: not ($f_n$ is uniformly bounded)$\implies$ not$(f_n\to f$ uniformly as $n\to\infty$ and $f$ bounded) or equivalently :

If $f_n$ is not uniformly bounded, then $f_n\to f$ pointwise as $n\to\infty$ or $f$ unbounded.

My question is the following: If I give a counter example to the contraposition statement, does it show as well that the original statement can't hold as well?

If yes, I have the following counter example to the contraposition statement:

Let $f_n(x)=\frac{x}{n}$. Clearly $f_n(x)$ is not uniformly bounded and $f_n\to 0$ pointwise. But, $f$ is bounded. So the contraposition statement is false and the original statement as well.

Answer 1

The contraposition would be, in this case: if $(f_n)_{n\in\Bbb N}$ is not uniformly bounded, then either $(f_n)_{n\in\Bbb N}$ does not converges uniformly to $f$ or $f$ is unbounded. Your counter-exampl is not a counter-example to this statement.

That example at the end of your post is not an example, since the convergence is not uniform. (I am assuming that this is a sequence of functions with domain $\Bbb R$. If you meant something else, please say so.)

A counter example (again, with domain $\Bbb R$) would be$$f_n(x)=\begin{cases}x&\text{ if }n=1\\0&\text{ otherwise.}\end{cases}$$The sequence $(f_n)_{n\in\Bbb N}$ converges uniformly to the null function (which is bounded), but it is not uniformly bounded.

Note that if a sequence $(f_n)_{n\in\Bbb N}$ of functions converges uniformly to a bounded function, then, for some $N\in\Bbb N$, $(f_n)_{n\geqslant N}$ is uniformly bounded.

Answer 2

One key flaw in your question is that the negation of $\{f_n\}$ doesn’t uniformly converge towards $f$ is not that $\{f_n\}$ converges towards $f$ pointwise. Which seems to be what you state in your question.

Also, it should be good that you state the definition set of all your maps.

For example on $I=[0,1]$, the sequence $f_n(x)=n$ doesn’t converge uniformly to the always vanishing map, nor does it pointwise.

Now if a sequence of maps $\{f_n\}$ converges uniformly to $f$ and that $f$ is supposed to be bounded, then $\{f_n\}$ is eventually uniformly bounded.

Note: also the initial statement of the question is true if each of the $f_n$ is supposed to be bounded.