5

Let $\{f_{n}\}_{n}$ be a sequence of functions from $[0,1] \to \mathbb{R}$. Let $f:[0,1] \to \mathbb{R}$ such that for any $x\in [0,1]$ and any sequence $\{x_{n}\}$ of elements in [0,1], if $x_{n}\to x$, then $f_{n}(x_{n}) \to f(x)$. Prove that f is continuous.

My attempt: Let $\epsilon>0$. Let $\delta_{x}-ball$ (and $\delta_{y}-ball$) be the ball such that if $x_{n}$ lies in the $\delta_{x}-ball$ (respectively $y_{n}$ and $\delta_{y}-ball$) then $f_{n}(x_{n})$ lies in $\frac{\epsilon}{3}-ball$ of $f(x)$ (same for $f_{n}(y_{n})$ and $\frac{\epsilon}{3}-ball$ of $f(y)$). Now let $\delta$ =min($\delta_{x},\delta_{y})$. So we have $|x-y|<\delta \implies |f(x)-f(y)|=|f(x)-f_{n}(x_{n})+f_{n}(x_{n})-f_{n}(y_{n})+f_{n}(y_{n})-f(y)|<\frac{2}{3}\epsilon+|f_{n}(x_{n})-f_{n}(y_{n})|$. This unfortunately seems like a dead end as we know nothing about $f_{n}$.

Could you provide some hint on how to proceed or use a different method altogether? Any help will be appreciated, thanks.

unoriginalname
  • 103

1 Answers1

2

Here are some hints to show that $f$ is continuous by showing that $f(x_n) \to f(x)$ for any convergent sequence $x_n \to x$ in $[0, 1]$.

Firstly, we know that $\{f_n\}$ converges pointwise to $f$ as any constant sequence is convergent.

Secondly, you should show that for any increasing sequence $\{k_n\} \subset \mathbb{N}$ and convergent sequence $x_n \to x$ in $[0, 1]$ we have $f_{k_n}(x_n) \to f(x)$.

Hopefully this will help you solve the problem, but don't hesitate to ask more questions if something is unclear.

Edit: Here is a proof of the second statement.

Given the increasing sequence $\{k_n\} \subset \mathbb{N}$ and the convergent sequence $x_n \to x$ in $[0, 1]$ let us define a new sequence $\{y_k\} \subset [0, 1]$ by $y_k = x_n$ for $k_{n-1} < k \leq k_n$. Clearly $y_k \to x$, and so by assumption $f_k(y_k) \to f(x)$. It then follows that the subsequence $\{f_{k_n}(y_{k_n})\}$ also converges to $f(x)$. By definition $y_{k_n} = x_n$ and thus we have shown that $f_{k_n}(x_n) \to f(x)$.

user920957
  • 630
  • 3
  • 14
  • Alright so basically we are using the first statement to say $f_{n}(x_{n})$ converges to $f(x_{n})$ right? But we know that $f_{n}(x_{n})$ converges to f(x) so won't this be enough to prove continuity? I don't understand the use of second statement, or maybe I'm skipping some steps in my argument? Thanks alot for the answer, by the way, it makes sense. – unoriginalname Dec 07 '23 at 09:33
  • 1
    Yes the first statement shows that $f_k(x_n) \to f(x_n)$ as $k \to \infty$, but $k$ might need to be much larger than $n$ for $f_k(x_n)$ to be close to $f(x_n)$. Hence you need the second statement as well, as without it you only have $f_n(x_n) \to f(x)$ as $n \to \infty$ but you don't know if $f_n(x_n)$ is close to $f(x_n)$. – user920957 Dec 07 '23 at 09:37
  • 1
    Ohhh, that makes alot of sense, thank you very much! – unoriginalname Dec 07 '23 at 09:42
  • so in order to prove the second statement, we start by taking $|x_{n}-x|< \delta$. Then we have $|f_{k_{n}}(x_{n})-f(x)|=|f_{k_{n}}(x_{n})-f_{n}(x_{n})+f_{n}(x_{n}) -f(x)| <|f_{k_{n}}(x_{n})-f_{n}(x_{n})|+\frac{\epsilon}{2}$. Now due to pointwise convergence we can say that the first term is also less than $\frac{\epsilon}{2}$, does this argument make sense? – unoriginalname Dec 07 '23 at 09:54
  • I don't think that works unfortunately. If you want I can edit my answer with the proof I had in mind for the second claim. – user920957 Dec 07 '23 at 10:07
  • Damn, it doesn't :(. I wanted to get it but I think you can edit it as I can't think of a better way right now. Where do you think the flaw lies, by the way? Isn't like $f_{k_{n}}$ midway between f(x) and $f_{n}$ so it automatically gives you that $< \epsilon/2$?? – unoriginalname Dec 07 '23 at 10:14