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I'm very bad in coming up with examples and I was asked if there can be sequences that are bounded, but converge pointwise to an unbounded sequence and vice versa.

For the first one I thought that if I just let $f_n(x)=g(x) \chi_{[-n,n]}(x)$, then shouldn't this work for any unbounded function $g: \Bbb R \to \Bbb R$? Are there any other simple examples that I could demonstrate without the usage of the indicator function?

For the other part I couldn't figure out an example so I would be happy to see one!

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  • $\begingroup$ It sounds like you are asking about sequences of bounded functions that converge pointwise to an unbounded function, correct? The title and the first sentence of the post are somewhat confusingly written. $\endgroup$
    – Alex Ortiz
    Jul 24, 2022 at 18:46
  • $\begingroup$ Also, for the example in your post, you need to truncate the vertical height of the function, as opposed to its horizontal width to get a sequence of bounded functions. Something like $f_n = \min(n,|g|)$. When $|g|>n$, $f_n = n$ and when $|g| \le n$, $f_n = g$. This ensures each $f_n$ is bounded by $n$, and the limit is $|g|$, which is unbounded by assumption. Unless your question is about functions with bounded support. An edit to the title and body of the question to make things clearer would be appreciated. $\endgroup$
    – Alex Ortiz
    Jul 24, 2022 at 18:48

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For the first part consider $\{f_n\}$ where each $f_n\colon (0,\infty)\to\mathbb{R}$ is given by $$ f_n(x)= \begin{cases} n & 0<x\leq\frac{1}{n}\\ 1/x & \frac{1}{n}<x \end{cases} $$ We have that $f_n\to 1/x$ pointwise. Clearly every $f_n$ is bounded, but the limit function is not.

Now, if for the other part you mean a sequence of unbounded functions that converges pointwise to a bounded function consider $\{f_n \}$ given by $f_n(x)=x/n$ which converges pointwise to $f(x)=0$.

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