Question

Show that the sequence (fn) sequence where

fn(x)= x/(1+nx^2), x∈[2,∞] is uniformly convergent in [2,∞]

Answer 1

To prove that the given sequence of functions converges uniformly on $\displaystyle [2,\infty)$, it suffices to show that the sequence of functions converges uniformly on $\displaystyle \R$.

Now,

$\displaystyle \left|f_n(x)\right|=\left|\frac{x}{1+nx^2}\right|=\left|\frac{x\sqrt{n}}{(1+nx^2)\sqrt{n}}\right|=\frac{1}{\sqrt{n}}\left(\frac{\sqrt{n}\ |x|}{1+{(\sqrt{n}|x|)}^2}\right)=\frac{1}{\sqrt{n}}\left(\frac{t}{1+t^2}\right)$

where $\displaystyle t=\sqrt{n}\ |x|.\text{ But, }\frac{t}{1+t^2}\leq\frac{1}{2}\ \ \ \forall\ t\in\R\ \text{since } (1−t)^2 ≥ 0,\ \Rightarrow 2t ≤ 1 + t^2.$

Thus, from the above inequality, we have that; $\displaystyle |f_n(x)|=\frac{1}{\sqrt{n}}\left(\frac{t}{1+t^2}\right)≤\frac{1}{\sqrt{n}}\left(\frac{1}{2}\right)\rightarrow0,\ \text{as } n\rightarrow \infty\ \ \ \forall\ \ \ x\in\R.\\ \text{Hence, given }\epsilon>0\ \text{choose } N = \frac{1}{4\epsilon^2}.\ \text{Then } |f_n(x)| < \epsilon\ \ \ \forall\ \ x\in\R\ \text{if}\ n>N.$ Showing that the sequence of functions converges uniformly to $\displaystyle 0 \text{ on }\R.$ And since $\displaystyle [2,\infty)\subset\R,$ then the given sequence of functions converges uniformly to $\displaystyle 0 \text{ on }[2,\infty).$