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Let $(f_n)$ be a sequence of functions defined on $[0,1]$. Show that if $(f_n)$ converges to zero uniformly on $[0,1]$, then for any sequence of points $(x_n)$ with $x_n \in [0,1]$ for every $n$, the sequence $f_n(x_n)$ has limit zero.

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What have you tried so far? Where are you stuck? Kindly show your work, even if only partial. –  cardinal Jan 25 '12 at 2:54
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I think I have the main idea, but I'm not quite sure of the wording. Since (fn) converges uniformly to zero for all n, the limit clearly exists, but I am not sure what reasoning can be used to show the limit is zero. –  phillips0023 Jan 25 '12 at 2:59
    
Assume not. Argue by contradiction. –  mixedmath Jan 25 '12 at 3:00
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For every $n$, $|f_n(x_n)| \le \sup_{x \in [0,1]} |f_n(x)|$. But the right side goes to 0 as $n \to \infty$... –  Nate Eldredge Jan 25 '12 at 3:46
    
You can prove, as a previous exercise, that $f_n\to f$ uniformly in $A$ if and only if the sequence of numbers $\sup_{x\in A} \vert f_n(x)-f(x)\vert$ converges to $0$. Then, look at the Nate's comment. –  leo Jan 25 '12 at 5:42

1 Answer 1

Since $\{f_n\}$ converges uniformly to $0$, given $\epsilon>0$, there is an $N$ so that $|f_m(x)|<\epsilon$ for all $m>N$ and all $x\in[0,1]$.

To show that the sequence $\{f_n(x_n)\}$ has limit $0$, you need to show that for every $\epsilon>0$, there is an $N$ so that $|f_m(x_m)|<\epsilon$ for all $m>N$.

Can you see how to put these together to get what you want?

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But how can the |fn(xm)|< epsilon if the function isn't explicitly defined? –  phillips0023 Jan 25 '12 at 3:04
    
@phillips0023: Because you use the structure available to you. You've been told that you have a sequence of functions $(f_n)$ which converge uniformly to zero on $[0,1]$. That is a rather strong statement to make about the $(f_n)$. Indeed, it is enough to allow you to arrive at the conclusion without having to say anything more explicit about the nature of $(f_n)$. –  cardinal Jan 25 '12 at 3:09
    
@phillips0023 I'm sorry; what function? The limit function is the zero function: $|f_m(x_m)-0|=|f_m(x_m)|$. From the first paragraph, you can make $|f_n(x)|$ as small as you like $for\ all$ numbers $x\in[0,1]$ as long as $n$ is sufficiently large, and that includes the $x$'s in the given sequence. –  David Mitra Jan 25 '12 at 3:13

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