Let $f_n(x):\mathbb R \to \mathbb R$ be defined by $f_n(x)=\frac{x}{1+nx^2}$. Which of the following statements are correct?

Question

For $n$ a positive integer, let $f_n(x):\mathbb R \to \mathbb R$ be defined by $f_n(x)=\frac{x}{1+nx^2}$. Which of the following statements are correct?

(a) The sequence $\{f_n(x)\}$ of functions converges uniformly on $\mathbb R$.

$f_n(x)=\frac{x}{1+nx^2}$ and $f_n'(x)=\frac{1-nx^2}{(1+nx^2)^2}$. The function $f_n(x)$ converges point wise to zero. Now, $f_n'(x)=0 \implies x=\sqrt{\frac{1}{n}}$, this is the point of maxima as the 2nd derivative $f_n''(x)=\frac{2nx(nx^2-3)}{(nx^2+1)^3}$ turns out to be negative. So, $M_n=\sup_{x\in \mathbb R}\Vert f_n(x)-f(x)\Vert=\Vert \frac{1}{2\sqrt{n}}-0\Vert$ and $\lim_{n \to \infty}M_n=0$ and hence the sequence of functions converges uniformly.

(b) The sequence $\{f_n(x)\}$ of functions converges uniformly on $[1,b]$ for any $b>1$.

this is true and directly follows from part (a).

(c) The sequence $\{f_n'(x)\}$ of derivatives converges uniformly on $\mathbb R$.

The function $f_n(x)$ does not converges point wise to a fixed value as for $x=0$ :$f_n'(0)=1$ and for $x=\sqrt{\frac{3}{n}}$ : $f_n'(\sqrt{\frac{3}{n}})=\frac{-1}{8}$.The sequence $f_n'(x)$ is a series of continuous functions but $f(x)$ is not continuous and hence $f_n'(x)$ doesnot converge uniformly.

(d) The sequence $\{f_n'(x)\}$ of derivatives converges uniformly on $[1,b]$ for any $b>1$.

I didn't get how to prove the (d) part. can someone please help me proving part (d) and also make sure that the other options (a), (b) and (c) have the correct explanation. Thanks !

Answer 1

(a) What you did is correct.

(b) Again, it is correct.

(c) The sequence $(f_n')_{n\in\Bbb N}$ converges pointwise to a discontinuous function ($1$ at $0$ and $0$ elsewhere), and therefore the convergence cannot be uniform.

(d) You have $f_n''(x)=0$ if and only if $x=0$ or $x=\pm\sqrt{\frac3n}$. On $[1,\infty)$, $f_n'$ is then negative and strictly increasing, if $n\geqslant3$. So, on an interval $[1,b]$ with $b>1$, the maximum (when $n\geqslant3$) is attained at $1$, and that maximum of $|f_n'|$ is$$|f_n'(1)|=\frac{n-1}{(1+n)^2}.$$Since$$\lim_{n\to\infty}\frac{n-1}{(1+n)^2}=0,$$the convergence is uniform.

Answer 2

For (d), note

$$|f_n'(x)| = \left |\frac{1-nx^2}{(1+nx^2)^2}\right | \le \frac{1+nx^2}{(1+nx^2)^2} =\frac{1}{1+nx^2}.$$

On $[1,\infty),$ the term on the right is bounded above by $$ \frac{1}{1+n\cdot 1^2} = \frac{1}{1+n}\to 0.$$Thus $f_n'\to 0$ uniformly on $[1,\infty),$ hence certainly on $[1,b).$