If {fn} converges pointwise to {f} and each {fn} is bounded in a closed interval [a,b], is {f} also bounded in [a,b]?

Question

Is it possible to say that if $f_n$ converges pointwise to $f$ and each $f_n$ is bounded in the closed interval $[a,b]$, then $f$ is also bounded in $[a,b]$?

As far as I know, boundness is not preserved in pointwise convergence. This is easy to see with an example such as: $$f_n=\frac{n}{nx+1},x\in(0,1)$$ where $f_n$ is bounded in $(0,1)$ but $f$ isn't.

However, I'm not sure if this conclusion holds with a closed interval.

I was thinking $$f_n(x)=\frac{1}{x-(1+\frac{1}{n})},x\in[0,1]$$ where $f_n$ is bounded in $[0,1]$ but $f$ isn't. Does this example work?

Answer 1

The example does not work because $f_n(1) = -n$, thus $(f_n)$ does not converge pointwise. But you can easily modify your example:

$$f_n(x) = \begin{cases} \frac{1}{x-(1+\frac{1}{n})} & x \le 1 -\frac 1 n \\ -\frac n 2 + \frac{n^2}{2}(x - (1 -\frac 1 n)) & x \ge 1 -\frac 1 n \end{cases} $$ Since both parts of the definition give $f_n(1 - \frac 1 n) = -\frac n 2$, these functions are continuous, thus bounded. If $x < 1$, then for $n \ge \frac{1}{1-x}$ we get $x \le 1 -\frac 1 n$ and therefore $$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{1}{x-(1+\frac{1}{n})} = \frac{1}{x-1} .$$ Moreover, all $f_n(1) = 0$, thus $$\lim_{n \to \infty} f_n(1) = 0 .$$ Thus $(f_n)$ converges pointwise to $$f(x) = \begin{cases} \frac{1}{x-1} & x < 1 \\ 0 & x = 1 \end{cases} $$ This function is unbounded and non-continuous at $1$.

Answer 2

Let $f$ be any unbounded function on $[a,b]$ and define $f_n(x)=f(x) $ if $-n\leq f(x) \leq n$, $f_n(x)=n$ if $f(x) >n$ and $f_n(x)=-n$ if $f(x) <-n$. Then, for each $x$, $f_n(x)=f(x)$ for $n$ sufficiently large so $f_n \to f$ pointwise. Also $|f_n(x)| \leq n$ for all $x,n$.