Let $f_n:(0,1)\to\Bbb{R}$ be a sequence of continuous functions such that $\forall$ cauchy sequence $\{x_n\}$ in $(0,1)$, we have $\lim\limits_{n\to\infty}f_n(x_n)= 0$. Then
- Find the pointwise limit of $\{f_n\}$.
- Is the the above convergence uniform?
I'm able to prove the part 1. For each $x\in(0,1)$, the constant sequence $\{x\}_n$ is cauchy, hence by the hypothesis $\lim\limits_{n\to\infty}f_n(x)=0$. So the sequence $\{f_n\}$ converges pointwise to $0$. For part 2, I think there is no reason for this convergence to be uniform. But I'm unable to construct a counter example. Initially, I thought $f_n(x)=x^n$. But it's not working (take $x_n=1-\frac{1}{n}$, then $f_n(x_n)\to 1$).
Can anyone help me complete the proof? Thanks for your help in advance.
With the help of @copper.hat, I'm writing the proof. The part 2 is also true.
Suppose the convergence is not uniform. Then we have $\epsilon>0$ such that $\forall k\in\Bbb{N}$ there exists $n_k>k$ and $x_{n_k}\in(0,1)$ such that $|f_{n_k}(x_{n_k})|\ge\epsilon$. Using the compactness of $[0,1]$, WLOG we can assume $\{x_{n_k}\}_k$ is a cauchy sequence. (As $\{x_{n_k}\}$ is sequence in $(0,1)$, in particular in $[0,1]$, we have a convergence sub sequence of it in $[0,1]$, then we'll work with that sub sequence)
By the given hypothesis we should have $f_{n_k}(x_{n_k})\to 0$ but this is not possible as $|f_{n_k}(x_{n_k})|\ge\epsilon$, contradiction. (Although I didn't need continuity anywhere in this argument)