Answer to Question #291335 in Real Analysis for Nikhil Singh

Question #291335

Show that the sequence (fn) sequence where


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

1
Expert's answer
2022-01-30T13:15:37-0500

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

Now,

fn(x)=x1+nx2=xn(1+nx2)n=1n(n x1+(nx)2)=1n(t1+t2)\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 t=n x. But, t1+t212    tR since (1t)20, 2t1+t2.\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; fn(x)=1n(t1+t2)1n(12)0, as n      xR.Hence, given ϵ>0 choose N=14ϵ2. Then fn(x)<ϵ     xR if n>N.\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 0 on R.\displaystyle 0 \text{ on }\R. And since [2,)R,\displaystyle [2,\infty)\subset\R, then the given sequence of functions converges uniformly to 0 on [2,).\displaystyle 0 \text{ on }[2,\infty).


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