$f_n(x):= \frac{nx^3}{1+nx^2}$. Show that the sequence $(f_n)$ converges uniformly on $\mathbb{R}$

Question

$f_n:\mathbb{R} \rightarrow \mathbb{R}$

A bit stuck on this revision question. I first determine the pointwise limit easy enough:

$$\lim_{n\rightarrow \infty} \frac{nx^3}{1+nx^2} = x$$

To show that the sequence $(f_n)$ converges uniformly on $\mathbb{R}$, I need to show that $||f_n-f||_{\infty} \rightarrow 0 $, so I need to find a comparable function $g_n$ (independent of $x$) > $\frac{nx^3}{1+nx^2}$, that I can use to get rid of the $x$.

$$\bigg|\bigg|\frac{nx^3}{1+nx^2} - x\bigg|\bigg|$$

I understand this conceptually, but how do I go about deciding what $g_n$ to try?

Any help would be greatly appreciated.

You could rewrite $f_n$ as $$f_n(x) = x \left( \frac{1}{\frac{1}{nx^2}+1} \right)$$ then show that the term in parenthesis goes to 1 uniformly outside of a neighborhood of zero. Then handle the neighborhood containing zero in another fashion. — Joel, May 27 '14 at 14:33
Do you mean that $$ ||x(\frac{1}{\frac{1}{nx^2}+1}) - x|| = 1$$ Because I tried that and I still retain the $x$... — JJJ, May 27 '14 at 14:44

score 3 · Answer 1 · answered May 27 '14 at 14:45

3

$$\left| \frac{n x^3}{1+nx^2} -x\right| =\left| \frac{ x}{1+nx^2}\right|$$

Now $$\left| \frac{ x}{1+nx^2}\right| \leq \frac{1}{2\sqrt{n}}$$ is equivalent to

$$2\sqrt{n}|x| \leq nx^2+1$$

which follows from $$0\leq (\sqrt{n}x -1)^2$$

answered May 27 '14 at 14:45

Rene Schipperus

39,526

That's great. How were you able to come up with your last line derive that? – JJJ May 27 '14 at 14:59
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$2xy \leq x^2+y^2$ is a well known inequality. – Rene Schipperus May 27 '14 at 15:27

score 2 · Answer 2 · answered May 27 '14 at 14:54

A different approach, which is useful in many other similar problems: compute the maximum value of $$ d_n(x)=|f_n(x)-f(x)|=\Bigl|\frac{n\,x^3}{1+n\,x^2}-x\Bigr|=\frac{|x|}{1+n\,x^2}. $$ It is enough to consider $x\ge0$. Then $$ d_n'(x)=\frac{1-n\,x^2}{(1+n\,x^2)^2}=0\implies x=\frac{1}{\sqrt n}. $$ It is easy to see that this corresponds in fact to a maximum, from where we get $$ 0\le d_n(x)\le d_n\Bigl(\frac{1}{\sqrt n}\Bigr)=\frac{1}{2\sqrt n}. $$

Oh cool. Very nice to know. Thanks a lot. – JJJ May 27 '14 at 15:00 — JJJ, May 27 '14 at 15:00

score 0 · Answer 3 · answered May 27 '14 at 14:39

0

Show that the sequence $f_n$ is uniformly Cauchy, that is, find a bound on $f_n(x) - f_m(x)$ that does not depend on $x$. This will probably involve considering cases $|x| > 1$, $|x| \le 1$.

answered May 27 '14 at 14:39

PA6OTA

1,972
10
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$f_n(x):= \frac{nx^3}{1+nx^2}$. Show that the sequence $(f_n)$ converges uniformly on $\mathbb{R}$

3 Answers3