Convergence of $f_n(x)=\frac{x}{1+nx^2}$ and its derivative $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $\mathbb{R}$

Question

Consider the sequence $f_n(x)=\frac{x}{1+nx^2}$ on $x\in \mathbb{R}$. Also consider $f'_n(x)=\frac{1-nx^2}{(1+nx^2)^2}$ on $x \in \mathbb{R}$.

I want to check their uniform convergence.

Convergence of $f_n$:

1. $f_n \to 0$ pointwise on $\mathbb{R}$.

2. $f_n(x) \ge 0$ for all $x \in \mathbb{R}^+$ and $f_n$ is an odd function so obtaining a global maximum in $[0, \infty)$ will provide use with an upperbound of $|f_n(x)|$

3. $f_n'(x)=0 \iff x=0 \ \text{or} \ x=\frac{1}{\sqrt n}$
   At $x=0$ $f_n(x)$ is the global minimum so $f_n(x)$ attains its global maximum at $x=\frac{1}{\sqrt n}$ in $\mathbb{R}$
   $\implies M_n:=\operatorname{lub}\{|f_n(x)-0|:x \in \mathbb{R}\}=\frac{2}{\sqrt n} \to 0$
   $\implies f_n \to 0$ uniformly in $\mathbb{R}$.

Convergence of $f'_n$:

1. $f'_n(x) \to \begin{cases} 0 & x \ne 0 \\ 1 & x=0 \end{cases}$ pointwise in $\mathbb{R}$ which tells $f'_n$ cannot be uniformly convergent in $\mathbb{R}$ as the limit function is not continuous.

2. Critical points of $f'_n(x)$ are $x=0$ and $x=\sqrt \frac{3}{2n} \to 0$
   $\implies f'_n$ is monotone and continuous in [1,b] for any $b>1$ for sufficiently large $n$ and converges pointwise to $0$.

I want to ensure $f'_n$ converges uniformly to $0$ on $[1,b]$. How should I proceed?

Answer 1

You have not obtained the point-wise limit of $f_n'$ correctly. The limit is $1$ for $x=0$ and $0$ for all other $x$. The convergence is uniform on any compact interval not containing $0$: $|f_n'(x)| \leq \frac { 1+nM^{2}} {(1+\epsilon^{2}n)^{2}}$ if $M$ is a bound for $|x|$ on the given interval $I$ and $\epsilon =\inf \{|x|: x \in I\}$.

Answer 2

Note that application of the AM-GM inequality gives immediately that

$$\frac{|x|}{1+nx^2}\le \frac{|x|}{2\sqrt{nx^2}}=\frac1{2\sqrt n}$$

Therefore, $f_n$ converges uniformly to $0$ for $x\in \mathbb{R}$.

Next, we have

$$f_n'(x)=\frac{1-nx^2}{(1+nx^2)^2}$$

Let $0<a<b$ and suppose $x\in [a,b]$. Then for $n\ge 1$

$$|f_n'(x)|= \frac{|nx^2-1|}{(nx^2+1)^2}\le \frac{nb^2+1}{(na^2+1)^2}$$

Hence, $f_n'(x)$ converges uniformly to $0$ for $x\in [a,b]$ for $0<a<b$. And inasmuch as $f_n'$ is an even function, $f_n'(x)$ converges on $(-b,-a)$ also.