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If $(f_n)$ is a sequence of real functions on $[0, 1]$ converging uniformly to a function $f$ on $[0, 1]$, and if $f_n$ is continuous at $x_n ∈ [0, 1]$ for each $n$ with $x_n\stackrel{n\to\infty}\longrightarrow x$, must $f$ be continuous at $x$?

I believe that this statement is true but I am having trouble proving it. Could someone please help out?

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  • $\begingroup$ Do you really mean that $f_n$ is continuous at a single point $x_n$? $\endgroup$ – Christian Blatter Oct 19 '18 at 18:14
  • $\begingroup$ @ChristianBlatter I believe it is safe to assume that $f_n$ is continuous at $x_n$ for each $n$. $\endgroup$ – Math1000 Oct 19 '18 at 18:17
  • $\begingroup$ that's what i meant! sorry $\endgroup$ – MathematicianP Oct 19 '18 at 18:19
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Consider the function $f(x) = 1$ for $x>0$ and $f(x)=0$ for $x \le 0$. Let $f_n = f$ for all $f$. Certainly $f_n$ converges uniformly to $f$. Now let $x_n = 1/n$. Then for every $n$ we have $f_n$ is continuous at $x_n$. Also $x_n$ converges to $0$, but $f$ is not continuous at $0$.

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  • $\begingroup$ That's so clear! Thank you! $\endgroup$ – MathematicianP Oct 21 '18 at 17:16
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In the case where all the functions $f_n$ are continuous at a point $x_0$ (so $x_0$ is fixed and doesn’t depend on $n$) then this is true.

The sequence $(f_n)$ converges uniformly to the function $f$.

Hence we have :

$$\forall \epsilon > 0, \exists N, \forall n \geq N, \forall x \in [0,1], \mid f_n(x)-f(x) \mid < \epsilon$$

Moreover all the $f_n$ are continuous at $x_0 \in [0,1]$ which means :

$$ \forall \epsilon > 0, \exists \delta, \forall x \in [x_0-\delta,x_0+\delta], \mid f_n(x_0)-f_n(x)\mid < \epsilon$$

Hence :

$$\forall \epsilon > 0, \exists \delta, \forall x \in [x_0-\delta,x_0+\delta], \mid f(x)-f(x_0) \mid = \mid f(x)-f_N(x)+ f_N(x)-f_N(x_0) + f_N(x_0) -f(x_0) \mid \leq \mid f(x)-f_N(x) \mid+ \mid f_N(x)-f_N(x_0) \mid + \mid f_N(x_0) -f(x_0) \mid \leq 3\epsilon $$

So $f$ is continuous at the point $x_0$.

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  • $\begingroup$ The statement is false as evidenced by @GEdgar's answer. $\endgroup$ – Math1000 Oct 19 '18 at 18:23
  • $\begingroup$ Ok, I miss understood the question. I thought it was at a single point $x_0$ that doesn’t depend on $n$. $\endgroup$ – Thinking Oct 19 '18 at 18:24

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