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Suppose that {${f_n}$} is a sequence of functions on $[a,b]$ that has the property that, for every convergent sequence {$x_n$} of numbers in $[a,b]$,$$\lim\limits_{n \to \infty} f_n(x_n)=0$$ Show that {$f_n$} converges uniformly to the zero function on $[a,b]$.

My attempt:

Assume toward a contradiction that {$f_n$} does not converge uniformly to the zero function on $[a,b]$. Then, there exists some $\epsilon$ $>0$ and some $N$ such that if $n>N$, then $|f_n(x)-0|\geq\epsilon$. Since$$\lim\limits_{n \to \infty} f_n(x_n)=0$$ we know that $|f_n(x)-0|<\epsilon$. This is a contradiction, and thus {$f_n$} must converge uniformly to the zero function on $[a,b]$.

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    $\begingroup$ Your "Then, there exists ..." claim is incorrect. What you can say is that $\sup_{[a,b]}|f_n| \not \to 0.$ It then follows that there is some $\epsilon > 0$ such that $\sup_{[a,b]}|f_n| \ge \epsilon $ for infinitely many $n.$ $\endgroup$ – zhw. Dec 13 '16 at 19:54
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    $\begingroup$ You almost got it! What you can conclude however is that for some $\varepsilon>0$ and all $N$ there exists $n > N$ such that for some $$x = x_{(N,n)}$$ then $|f_n(x)| > \varepsilon$. Try to construct a sequence $(x_{n'})$ such that for an appropriate subsequence $f_{n'},$ you can get $|f_{n'}(x_{n'})|>\varepsilon$ always. $\endgroup$ – Will M. Dec 13 '16 at 19:56

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