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My Approach. To prove the given statement it is sufficient to show that $$\exists \epsilon > 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n > N $$ $$ \Rightarrow | f_n(x) - f(x)| > \epsilon$$

Let $\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1$. Hence $| f_n(x) - f(x)| = \frac{1}{2} > \epsilon = \frac{1}{10}$. Hence, we have shown for every $N \in \mathbb{N}$, how to choose $n$ such that $| f_n(x) - f(x)| > \epsilon$. Thus, $f_n(x)$ does not converge uniformly on $[0,1]$.

Is there any logical flaw or wrong steps in my proof. Any help would be appreciated.

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    $\begingroup$ This is fine. You've found the maximum point of $|f_n-f|$, as you should have. $\endgroup$ – Ted Shifrin Dec 2 '13 at 2:55
  • $\begingroup$ And those points give the same maximum and converge to 0. $\endgroup$ – ncmathsadist Dec 2 '13 at 3:00
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Dec 2 '13 at 3:11
  • $\begingroup$ @ncmathsadist: Say what? $\endgroup$ – Ted Shifrin Dec 2 '13 at 3:12
  • $\begingroup$ Strange how this exact question due in my advanced calculus class tomorrow morning... coincidence? $\endgroup$ – recursion.ninja Dec 3 '13 at 2:11
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Can you the other characterization of uniform continuity: $(f_n) \to f $ uniformly on $S$ iff

$$ \lim [ \sup\{ |f(x) - f_n(x) : x \in S \} ] = 0 \; \; (proof?)$$

Your problem then is application of this problem. In fact, if $f = 0$, then

$$ |f_n - f | = f_n = \frac{nx}{1 + n^2x^2 } $$

Now,notice

$$ f_n' = \frac{n(1 + n^2x^2) - nx(2xn^2)}{(1 + n^2x^2)^2}$$

and

$$ f'_n(x) = 0 \iff x = \frac{1}{n}, \frac{-1}{n} $$

Hence

$$ f_n( \frac{1}{n} ) = \frac{1}{2} \; \; \text{check this!} $$

$$ \therefore \lim \sup|f_n - f| = \frac{1}{2} \neq 0$$

$$ \therefore f_n \; \; \text{does not converge uniformly to} \; \; f $$

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  • $\begingroup$ Should one check the second (negative) x? Why didn't you do this? $\endgroup$ – Bartłomiej Szałach Apr 7 '16 at 20:47
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I actually have a question that how would test the convergence if the given function is a series instead of a sequence with n varying from 1 to infinity for any value of x.

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Look at $f(x)$ at $x = 1/(n\sqrt{3}).$

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    $\begingroup$ So what is wrong if I consider $x = \frac{1}{n}$. $\endgroup$ – aghost Dec 2 '13 at 2:52
  • $\begingroup$ That works too. $\endgroup$ – ncmathsadist Dec 2 '13 at 11:26
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To phrase things in a different language, what you (correctly) proved is that $$ \|f_n-f\|_\infty\geq\frac12 $$ for all $n$, where $\|\cdot\|_\infty$ is the $\sup$-norm (it is actually an equality, but that is not important for your proof).

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