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I know that, If $(f_n)$ is Sequence of continuous functions on interval $I$ that converges uniformly on $I$ to function $f$ and If $(x_n)\subset I$ converges to $x_0\in I$ then,

$$lim(f_n(x_n))=f(x_0)$$

My question: Is the other direction holds? I mean, is the following is true?

Let $(f_n)$ be Sequence of continuous functions on $I$ and If for $c\in I$ we have, for every Sequence $(x_n)$ in $I$ that conveges to $c$, $lim(f_n(x_n))=f(c)$ then, Sequence $(f_n)$ converges uniformly to $f$ on $I$?

Please help...

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Not true on the interval $(-\infty,\infty)$: let $$f_0(x) = \begin{cases} 0 & x<0, \\ x & 0\le x<1,\\ 1 & x\ge 1\end{cases}$$ And then set $f_n(x) := f_0(x-n)$. We have $f_n \to 0=:f$ pointwise, but not uniformly. Yet if $x_n\to c$, then $x_n$ is bounded, so there is some $M$ where $x_n < M$ for all $n$. Then for all $n>M$, $f_n(x_n)=0= f(c)$.

The same example works on $[0,\infty)$.

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  • $\begingroup$ Sir, could please add details. For me i am not able to prove pointwise convergence of $f_n$ to $0$ (as the definition of Sequence $f_n$ is not clear to me). And how $f_n(x_n)=0$ as there is no information about $x_n$? $\endgroup$ Jan 14 at 5:19
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    $\begingroup$ @AkashPatalwanshi $f_n(x) = f_0(x-n) $ is zero if $x<n$. Taking $n\to \infty$ covers every possible $x$. The information I used about $x_n$ is that it is convergent and therefore bounded $\endgroup$ Jan 14 at 5:35
  • $\begingroup$ Sir, just one last difficulty. what i am getting by your definition is $f_n(x_n) = \begin{cases} 0 & x_n-n<0, \\ x_{n}-n & 0≤x_n-n<1,\\ 1 & x_n-n\ge 1\end{cases}$ but from this how we get $f_n(x_n)=0$ when $n>max(N,1+|c|)$. please help. $\endgroup$ Jan 14 at 8:28
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    $\begingroup$ @AkashPatalwanshi this is simpler - $x_n$ is convergent and therefore bounded, so for some $M$, $x_n<M$ for all $n$. Therefore $x_n ≤ M < n $ for all $n>M+1$ and therefore $f_n(x_n) = 0$. $\endgroup$ Jan 14 at 9:58

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