Uniform convergence of $f_n(x) = \frac{nx}{1+n^2x^2}$ on $[a,\infty), a>0$

Question

I believe it does converge uniformly to 0. Note that $|f_n(x)| = \frac{nx}{1+n^2x^2}$ and its derivative is $$\frac{n-n^3x^2}{(1+n^2x^2)^2}$$ which is zero at $x = \sqrt{n/n^3}$ where the function has its maximum. So this function has a peak which moves towards $x=0$ as $n \to \infty$, which means that for some $N$, $\forall n \ge N$, the location of the peak has moved to the left of $x=a$, and since the function then decays we have $\forall n \ge N$: $$\sup_{x\in [a,\infty)}|f_n(x)|= \left. \frac{nx}{1+n^2x^2}\right\rvert_{x=a} = \frac{na}{1+n^2a^2} \to 0$$

Is my reasoning correct? Could you make it more rigorous?

Answer 1

You can make it more precise by noting that $\frac t {1+t^{2}}$ is decreasing in $[1,\infty)$ (because its derivative is negative there) so $f_{n} (x)$ is decreasing for $x \geq a$ as long as $na \geq 1$. So you can take $N=[\frac 1 a] +1$.

Answer 2

Since the limit is the all-zero function, we must have$$\forall\epsilon>0\quad,\quad \exists N\quad,\quad n>N\implies |f_n(x)|<\epsilon$$but$$|f_n(x)|<\epsilon\implies \sup_{\Bbb R} |f_n(x)|\le \epsilon$$and since for $x_0={1\over n}$ we have $f_n(x_0)={1\over 2}$ then we deduce $${1\over 2}<\epsilon$$which is a contradiction. Hence the convergence is not uniform.

An intuition

Since a persistent peak of ${1\over 2}$ always remains in $f_n(x)$ and does not tend to zero though its location displaces gradually to zero, the convergence cannot be uniform. Uniform convergence means that we can draw a tape of arbitrarily small width around the limit function within which all the function sequence terms fall for large enough $n$. A peak of $1\over 2$ prevents us from happening so.

Edit

As the uniform convergence of $\{f_n(x)\}$ is questioned over $[a,\infty)$ for some $a>0$ and based on the former implications the ${1\over 2}$ peak becomes located at a very small position near zero, then we can conclude that the convergence is uniform over $[a,\infty)$.

Answer 3

$|f_n(x)| \le \frac{nx}{nx^2}= \frac{1}{x}\frac{1}{n} \le \frac{1}{a}\frac{1}{n}$ for $x \ge a.$

Can you proceed ?