Is $j_n(x)=\frac{x}{1+n^2x^2}$ uniformly convergent on $[0,1]$?

Question

Is $j_n(x)=\frac{x}{1+n^2x^2}$ uniformly continuous on $[0,1]$?

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I have basically been working with the following theorem:

Let E be a non-empty subset of $\mathbb{R}$ Let $f_n, f: E \rightarrow \mathbb{R}$. Then the following are equivalent

i. $f_n\rightarrow f$ uniformly on E

ii. $\exists N$ s.t. $\forall n >N$, $m_n :=sup_{x\in E}|f_n(x)-f(x)|$ exists and $m_n \rightarrow 0$ as $n \rightarrow\infty$

I am not sure whether the given function is increasing or decreasing on the interval since

$j_n(x)=\frac{x}{1+n^2x^2}=\frac{1}{\frac{1}{x}+n^2x}$

The reason that I am concerned with the increasing/decreasing behavior of this function is because I am looking for a way to meaningful describe $m_n$. Another tool I was considering, but am less comfortable with is Cauchy's Criterion.

Answer 1

We want to show that $$ \sup\bigg| \frac{x}{1+n^2 x^2}-0 \bigg| < \epsilon. $$

To do this we find the maximum of the function $\frac{x}{1+n^2 x^2},\, x\in[0,1]$ (you can use derivative test). The max appears as $x=\frac{1}{n}$ which gives

$$ \frac{1}{2n} < \epsilon. $$

Answer 2

Note $j_n(0) = 0$ and for $x \neq 0$, $j_n(x) \le \frac{x}{2nx} = \frac{1}{2n}$. Therefore, $|j_n(x) - 0| = j_n(x) \le \frac{1}{2n}$ for all $x\in [0,1]$. Consequently, $\sup_{x\in [0,1]} |j_n(x) - 0| \le \frac{1}{2n} \to 0$ as $n\to \infty$. This shows that $j_n\to 0$ uniformly on $[0,1]$.