# Is sequential criterion of continuity holds for Sequence of continuous functions?

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 $$I$$?

• Does this help: math.stackexchange.com/questions/369469/… Jan 14 at 5:00
• @KaviRamaMurthy , yes sir it does help me partially. But, is that mean, if $I$ is bounded interval then the statement that i had asked, holds true? Jan 14 at 5:11

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 $$[0,\infty)$$.
• 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 $x_n$? Jan 14 at 5:19
• @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 Jan 14 at 5:35
• 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. Jan 14 at 8:28
• @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 $f_n(x_n) = 0$. Jan 14 at 9:58