Answer to Question #294905 in Real Analysis for Nikhil Singh

Question

Show that the sequence (fn) sequence where

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

Answer 1

ANSWER .

Let $x$ be fixed and $x\in [2,\infty )$ . Then $f(x)=\lim _{ n\rightarrow \infty }{ { f }_{ n } } \left( x \right) =\lim _{ n\rightarrow \infty }{ \frac { x }{ 1+n{ x }^{ 2 } } = } x\lim _{ n\rightarrow \infty }{ \left( \frac { 1 }{ n } \cdot \frac { 1 }{ \frac { 1 }{ n } +{ x }^{ 2 } } \right) = }$ $=x\left( \lim _{ n\rightarrow \infty }{ \left( \frac { 1 }{ n } \ \right) \cdot \frac { 1 }{ { x }^{ 2 }+\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } } } } \right) =0$ . So $f_n$ converges pointwise to $f(x)=0.$

Denote ${ d }_{ n }=\underset { 2\le x<\infty }{ sup } \left| { f }_{ n }\left( x \right) -f\left( x \right) \right|$ . Since ${ f }_{ n }^{ ' }\left( x \right) =\frac { 1+n{ x }^{ 2 }-2n{ x }^{ 2 } }{ { \left( 1+n{ x }^{ 2 } \right) }^{ 2 } } =\frac { 1-n{ x }^{ 2 } }{ { \left( 1+n{ x }^{ 2 } \right) }^{ 2 } } <0^{}$ for all $x\ge 2$ $\left( n\ge 1 \right)$ ,then each function $f_n$ decreases on the interval $[2,\infty )$

Therefore, $\underset { 2\le x<\infty }{ sup } \frac { x }{ 1+n{ x }^{ 2 } } ={ f }_{ n }\left( 2 \right) =\frac { 2 }{ 1+4n }$ . So, ${ d }_{ n }=\underset { 2\le x<\infty }{ sup } \left| \frac { x }{ 1+n{ x }^{ 2 } } -0 \right| =\frac { 2 }{ 1+4n }$ . Since , $\lim _{ n\rightarrow \infty }{ { d }_{ n }= } \lim _{ n\rightarrow \infty }{ \frac { 2 }{ 1+4n } =0 }$, then by definition, we get that the sequence converges uniformly in $[2,\infty )$ .