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Let $(f_n)$ be a sequence of continuous functions on $[a,b]$ that converges uniformly to $f$ on $[a,b]$. Show that if $(x_n) \subseteq[a,b]$ and if $x_n \rightarrow x$, then $\lim f_n(x_n) = f(x) $

My solution: Let $\epsilon > 0$. Take $N \in \mathbb{N}$ such that

$$|f_n(x) - f(x)| < \frac{\epsilon}{2}, \forall n>N \text{ and } \forall x \in [a,b].$$

Now, since $(f_n)$ is continuous, take $\delta > 0$ such that:

$$|x_n - x| < \delta \Rightarrow |f_{N+1}(x_n) - f_{N+1}(x)| < \frac{\epsilon}{2}.$$

Now, if $|x_n - x| < \delta$, we must have that:

$$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Is this correct?

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up vote 2 down vote accepted

In the last sentence how do you know that $|x_n-x|<\delta \Rightarrow |f_n(x_n)-f_n(x)|<\frac{\epsilon}{2}?$ Also where did you use uniformly convergence? Note that $x$ is fixed. Your proof is not correct.

For a correct proof use that $f_n \xrightarrow[n\to\infty]{}f$ uniformly, $f$ is continuous and split $|f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$ for all $n\geq N$ where $N$ is such that $$|f_n(y)-f(y)|<\dfrac{\epsilon}{2} \text{ and } |f(x_n)-f(x)|<\dfrac{\epsilon}{2}$$ for all $n\geq N$ and $y \in [a,b].$

The assumption of $(f_n)$ being continuous is necessary as the following (counter) example shows.

Consider the sequence of functions $(f_n(x))_{n\in\mathbb N}$ on $[0,1]$ defined by $$f_n(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\ \dfrac1n \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}.$$ The sequence $f_n$ converges uniformly to the function $$f(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\ 0 \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}$$ and if $x_n=\dfrac1n$ (or any sequence with $x_n\to0$ and $x_n\neq0 , \ \forall n\in\mathbb N$) then $x_n\xrightarrow[n\to\infty]{}0$ but $f_n(x_n)\not\xrightarrow[n\to\infty]{}f(0).$

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I Took $n = N + 1$. I did use uniform convergence in first line – ILoveMath Dec 9 '12 at 13:44
So in the last sentence you proved correctly that $|f_{N+1}(x_{N+1})-f(x)|<\epsilon$. But you must prove that $|f_n(x_n)-f(x)|<\epsilon$ for all $n\geq N+1$. And maybe is better to rewrite your exercise as ... if $x_n\to x_0$ then $f_n(x_n)\to f(x_0)$. – P.. Dec 9 '12 at 14:00
Is the assumption of $f$ being continuous necessary? – user43901 Mar 13 '13 at 0:19
@user43901: Yes! For example the sequence of functions $f_n(x)=x^n$ on $[0,1]$ converges pointwise to the function $f(x)=0 \ , x\in[0,1)$ and $f(1)=1$. The sequence $x_n=\sqrt[n]{0.5}$ converges to $1$ but $f_n(x_n)=0.5\not\to f(1)=1$. – P.. Mar 13 '13 at 6:44
@P..:Thanks for explain that! So how do I construct a counter-example to the claim "if $fn$ converges uniformly to $f$ and $x_n \rightarrow x$, then $f_n(x_n) \rightarrow f(x)$". Given the discussion above, I know the $f$ must be continuous to this statement to be true. But I am having hard time seeing a counter example. Any ideas? – user43901 Mar 14 '13 at 23:10

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