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We have to show that $f_n(x)=\frac{nx}{1+n^{2}x^{2}}$ is uniform convergent on $[a,\infty),a \gt 0$ but not on $[0,\infty)$

I am trying to prove uniform covegence on $[a, \infty]$ by using the result a sequence $(f_n)$ of bounded functions on $A \subset \Bbb R$ converges uniformly on $A$ iff $ \Vert f_n-f \Vert _A \to 0$.

Here I got $f(x)=0$ and for proving sequence $(f_n)$ is sequence of bounded functions i tried to prove that $f_n(x)$ is a decreasing function and have maxima at $x=a$. For this I differentiated $f_n(x)$ and got $f_n^{'}(x)=n(1-n^{2}x^{2})/(1+n^{2}x^{2})^{2}$ but don't know how to move further.

Can anyone tell is this the right direction if yes then how to proceed further?

TonyK
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sajan
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  • Then what would you get if you let $f_n'(x)=0$? At that point, is it the maximum or the minimum? Then what possible bound would you get for the $|f_n - f|$? Also what does $f$ looks like? – xbh Sep 19 '18 at 11:18
  • i got $x=1/n$ and this point of maximum and for this $\Vert f_n-f \Vert _A \le 1/2$ but from result stated above $\Vert f_n-f \Vert _A$ should approach $0$. – sajan Sep 19 '18 at 11:24
  • Now does this maximum always attained on the interval? If does, how would that influence the expression $|f_n -f|$, i.e. would this be able to converge to 0? If not, then what is the actual maximum? Would this make $|f_n -f|$ tend to 0? – xbh Sep 19 '18 at 11:28
  • There are already quite a few questions about this : https://approach0.xyz/search/?q=%24%5Cfrac%7Bnx%7D%7B1%2Bn%5E2x%5E2%7D%24%2C%20uniform%20convergence&p=1 – Arnaud D. Sep 19 '18 at 12:09
  • The most closely related seems to be this : https://math.stackexchange.com/questions/2820601/uniform-convergence-on-two-separate-intervals – Arnaud D. Sep 19 '18 at 12:12

1 Answers1

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Let $n \in \mathbb{N}$ be large enough so that $\frac1n < a$. Then for any $x \in [a, +\infty)$ we have $x > \frac1n$ or $n^2x^2 > 1$.

Therefore

$$\|f_n\|_{[a, +\infty)} = \sup_{x \in [a, +\infty)} \frac{nx}{1+n^2x^2} \le \sup_{x \in [a, +\infty)} \frac{nx}{2n^2x^2} = \sup_{x \in [a, +\infty)} \frac{1}{2nx} \le \frac1{2na} \xrightarrow{n\to\infty} 0$$

so $f_n \to 0$ uniformly on $[a, +\infty)$.

On the other hand, for $[0, +\infty)$ we have

$$\|f_n\|_{[0, +\infty)} = \sup_{x \in [0, +\infty)} f_n(x) \ge f_n\left(\frac1n\right) =\frac12 \not\to 0$$

so $f_n \not\to 0$ uniformly on $[0, +\infty)$.

mechanodroid
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