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.
Best Answer
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.