0

I have a problem that essentially boils down to the above question. Another important fact is that $f$ is continuous. My approach has basically been a proof by contradiction. We assume that the sequence doesn't converge. However, we also know, by assumption, that \begin{equation} \lim_{n\rightarrow\infty}f_n(t_n) = x. \end{equation} Now, since $f$ is continuous, we can take the limit inside and get \begin{equation} f\left(\lim_{n\rightarrow\infty}t_n\right) = x \end{equation} which would contradict the fact that $t_n$ doesn't converge.

I sort of feel uneasy about this approach. Is it correct, and if not, how can I improve it?

Arturo Magidin
  • 398,050
muaddib
  • 71
  • "Now, since $f$ is continuous". There is no $f$ in that limit: there are a bunch of $f_n$; and while $f_n$ are continuous, they also depend on $n$ and so cannot be taken out of the limit. You are not calculating $\lim_{k\to\infty}f_n(t_k)$ (which would be equivalent to $f_n(\lim_{k\to\infty} t_k)$). – Arturo Magidin Apr 21 '21 at 18:18
  • @ArturoMagidin What other restrictions do I have to impose on $f$ to get the limit inside the function? – muaddib Apr 21 '21 at 18:21
  • The point is that there is no $f$ in the limit. You never even said what $f$ was, but I assume it is supposed to be the limit of the $f_n$. – Arturo Magidin Apr 21 '21 at 18:24
  • part1. i think the fallacy here is $$\lim_{n\rightarrow\infty}f_n(t_n) = \lim_{n\rightarrow\infty}f_n(\lim_{n\rightarrow\infty}t_n)$$ @ArturoMagidin ? it seems like OP knows something's wrong but is unable to articulate precisely what is wrong. so yeah. it sounds a little like Freshman's dream... – BCLC Apr 21 '21 at 20:06
  • part2. i'm thinking it's like $$\lim_{n\rightarrow\infty}a_n b_n = \lim_{n\rightarrow\infty}a_n(\lim_{n\rightarrow\infty}b_n)$$ for sequences $a_n$ and $b_n$, although in this case i believe this is right assuming the limits exist (and are finite). as for the case of $$\lim_{n\rightarrow\infty}f_n(t_n) = \lim_{n\rightarrow\infty}f_n(\lim_{n\rightarrow\infty}t_n)$$ assuming the limits exist...i forgot the rules. probably some uniformity thing, but i have a feeling that the limits exist isn't enough – BCLC Apr 21 '21 at 20:08
  • 1
    @BCLC: Well, yes, that first display makes no sense. As to your second one, it is an instance of problems with double-indexed families, which are very tricky. See here. Here it is worse, because you are only "going down the diagonal", so a lot of information gets lost. – Arturo Magidin Apr 21 '21 at 20:47

1 Answers1

1

No. Suppose that each $f_n$ is the null function from $\Bbb R$ into $\Bbb R$ and that $t_n=(-1)^n$. Then $(f_n)_{n\in\Bbb N}$ converges uniformly and $\lim_{n\to\infty}f_n(t_n)=0$, but $(t_n)_{n\in\Bbb N}$ diverges.

José Carlos Santos
  • 427,504
  • Didn't think of that. However, if I also know that $|f[\mathbb{R}]|>1$ how would this change the outcome? – muaddib Apr 21 '21 at 18:23
  • Do you mean that each $f_n$ is non-constant? Not really. Just take $f_n(x)=x^2$ and the same sequence $(t_n)_{n\in\Bbb N}$. – José Carlos Santos Apr 21 '21 at 18:24
  • Ah, yes I see. One final question. Is it always the case that we have to know an explicit formula for each of the $f_n$s and $t_n$s to be able to say something about the limit in question? – muaddib Apr 21 '21 at 18:29
  • Since the statement is false, the natural way of doing things is to provide a counter example. However, in this case, if, for some $c\in\Bbb R$, $f_n(x)=c$, then any divergent $(t_n)_{n\in\Bbb N}$ would have worked. – José Carlos Santos Apr 21 '21 at 18:33