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Given the function $f_n:[0,1] \to R$ $$f_n(x)=\frac{nx}{(n^2x^2+1)^2}$$

I can show the pointwise convergence $\forall x \in [0,1]: \lim_{n \to \infty} f_n(x) = f(x) = 0$.

To show uniform convergence seems to be much harder to me. I show you what I did below, but it might not be correct (or at least not formal enough). Any help to get a more formal solution would be really appreciate.

$$\sup_{x \in [0,1]} \left| f_n(x) - f(x) \right| = \sup_{x \in [0,1]} \left| f_n(x)\right| = \sup_{x \in [0,1]} \left| \frac{nx}{(n^2x^2+1)^2} \right| = \sup_{x \in [0,1]} \frac{nx}{(n^2x^2+1)^2} = \sup_{x \in [0,1]} \frac{nx}{(nx)^4+2(nx)^2+1} = \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)+\frac{1}{nx}} = \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)+\frac{1}{nx}} \le \sup_{x \in [0,1]} \frac{1}{(nx)^3+2(nx)} \le \frac{1}{(n)^3+2(n)} \underrightarrow{n \to \infty} 0$$

leo
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    It does not converge uniformly: $f_n(1/n)=1/4$ for every $n$. Your error is in dropping the $x$'s in the last inequality. – David Mitra Jan 14 '13 at 21:01
  • How did you arrive at the bound in the last line, $\frac{1}{(n x)^3+2nx} \leq \frac{1}{n^3 +2n}$? If you took $x\leq 1$, it should be $\geq$ – Alex Jan 14 '13 at 21:13
  • You are right @Alex! I did see that it's not correct. But I as I was confused I thought I just write down what I had on paper. – leo Jan 14 '13 at 21:19
  • @leo: No worries. As others have already noted, this series does not converge uniformly for some $x=\varphi(n)$ – Alex Jan 14 '13 at 21:21

4 Answers4

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The convergence $f_n\to0$ is not uniform since $f_n(1/n)=1/4$ for every $n$.

Did
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  • I see! Thank you very much! @did could you please write shortly how I could see that (come up with this solution). To avoid such a mistake in the future. – leo Jan 14 '13 at 21:05
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    Since $f_n(x)=f_1(nx)$, for any $t$ such that $f_1(t)\ne0$, the sequence $x_n=t/n$ is such that $f_n(x_n)$ does not converge to zero. – Did Jan 14 '13 at 21:09
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To find

$$ \sup_{x \in [0,1]} \frac{nx}{(n^2x^2+1)^2}, $$

you need to find the maximum of the above function $$ \frac{nx}{(n^2x^2+1)^2} $$

for $x\in[0,1]$. You can use applications of the derivative to find the max and min of a function.

Mhenni Benghorbal
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Before deciding if a given sequence is uniformly convergent or not, it's usually a good idea to look at the graphs of the $f_n$.

Below are the graphs of $\color{darkblue}{f_1}$, $\color{maroon}{f_2}$, $\color{darkgreen}{f_5}$, $\color{pink}{f_8}$, $\color{cyan}{f_{10}}$, and $\color{yellow}{f_{40}}$ (sorry if the colors are too vibrant).

enter image description here

From this you might surmise that the convergence isn't uniform, as each $f_n$ seems to "peak" with value $1/4$, and since the pointwise limit is the zero function.

And indeed the convergence of $(f_n)$ to the zero function isn't uniform: you could show that each $f_n$ assumes the value $1/4$ (and even that this is the maximum value) using the methods of the other answers.

David Mitra
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You can find maximum of $ \; \dfrac{u}{u^4+2u^2+1}$

M. Strochyk
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