I'm trying to solve the following problem:
Let $f_n: \mathbb{R} \to \mathbb{R}$ be the sequence of functions given by: $f_n(x) = \dfrac{nx}{nx^2+1}$. Determine whether the sequence converges pointwise and find all intervals of uniform convergence.
I have shown that $(f_n)$ converges pointwise but I' having trouble proving uniform convergence
My attempt
For each $n \in \mathbb{N}$ we have $f_n(0)=0$. And for $x \neq 0$: $\lim \limits_{n \to \infty} f_n(x) = \lim \limits_{n \to \infty} \dfrac{nx}{nx^2+1} = \dfrac{1}{x}$, so the sequence $(f_n)$ converges pointwise to the function $f:\mathbb{R} \to \mathbb{R}$:
$$ f(x) = \begin{cases} 0, \text{ if } x = 0\\
\dfrac{1}{x}, \text{ if } x \neq 0\end{cases}$$
Since $f$ is not continuous on $\mathbb{R}$ the sequence $(f_n)$ does not converge uniformly on $\mathbb{R}$ to $f$.
I would want to show that $(f_n)$ converges uniformly to $\dfrac{1}{x}$ on $(-\infty,0)$ and $(0, \infty)$. What I've seen so far is that:
$$\left \vert f_n(x) – \dfrac{1}{x} \right \vert = \left \vert \dfrac{nx}{nx^2+1}-\dfrac{1}{x}\right \vert = \left \vert \dfrac{-1}{x(nx^2+1)}\right \vert$$
I know that if $\sup_{x \in \mathbb{R}- \{ 0\}} \left \vert f_n(x) – \dfrac{1}{x} \right \vert \to 0$ then $(f_n)$ converges uniformly, but $\dfrac{-1}{x(nx^2+1)}$ is not bounded, so how can I find the intervals of uniform convergence?
Best Answer
Your reasoning so far is correct. You are left with showing that $$\sup_{x\in I} \frac{1}{|nx^2 + 1||x|} \to 0$$ for your desired interval $I$.
Now, if you take an interval which is arbitrarily close to $0$, as in for example $I = (0;+\infty)$, you would intuitively guess that you get uniform convergence (since the interval does not contain the problematic point $0$), however, this is not the case. Since $x\in (0;+\infty)$ can approach $0$ arbitrarily closely, you can take e.g. $x=1/\sqrt{n}\in(0;+\infty)$ to get $$\sup_{x\in I} \frac{1}{|nx^2 + 1||x|} \geq \frac{\sqrt{n}}{2}\to+\infty,$$ i.e. we don't have uniform convergence.
If you want to assure uniform convergence, you want to stay away from $0$. Any interval $I = [a;+\infty)$ for $a>0$ (or subsets thereof) will work (analogously $I = (-\infty; -a]$), since now $x\in I$ satisfies $|x|\geq a$, so that $$\sup_{x\in I} \frac{1}{|nx^2 + 1||x|} \leq \frac{1}{(na^2 + 1)(a)} \to 0.$$