Convergence of $x_n(t)=\frac{2nt}{1+n^2t^2}$ in $C_{[0,1]}$ and $C_{[1,\infty]}$

Question

Ex: Study the convergence in $C_{[0,1]}$ and $C_{[1,\infty]}$ of the following sequence of functions:$$x_n(t)=\frac{2nt}{1+n^2t^2}$$

For $C_{[0,1]}$

$x_n(t)\to t\\ n\to\infty$

So I want to evaluate if:

$\max_{0\leqslant t\leqslant 1}|x_n(t)-t|\to 0\\n\to\infty$

So I decided to start up with:

$|x_n(t)-t|=|\frac{2nt}{1+n^2t^2}-t|\leqslant|\frac{2nt}{2nt}-1|=|t-1|$

Using $(a-b)^2\geqslant 0\iff a^2+b^2\geqslant 2ab$.

So according to this majoring. $x_n(t)\to t$ when $t\to 1$.

So the sequence would converge in both $C_{[0,1]}$ and $C_{[1,\infty]}$ because $t=1$ is in the domain of the functions of both spaces.

Questions:

1) Is these approach right?

2) How could I have solved the question? As it seems to me there could be another way to major $|\frac{2nt}{1+n^2t^2}-t|$.

Thanks in advance!

Answer 1

Note that $$\lim_{n\to \infty}\frac{2nt}{1+n^2t^2}=0$$

for all $t$. To examine whether the convergence is uniform, we simply note from the AM-GM inequality that $1+n^2t^2\ge \frac1{2|nt|}$, with equality if and only if $t=1/n$.

Hence $$\sup_{t} \frac{2nt}{1+n^2t^2}= 1$$

and the convergence is not uniform for $t\in [0,1]$.

The convergence is obviously uniform on all compact subsets of $[\delta,\infty)$ for $\delta>0$.

Answer 2

1. The inequality $|x_n(t)-t| \leq |t-1|$ is not sufficient to conclude because as you have written you need to prove that $\max_{t \in [0,1]} |x_n(t)-t | \rightarrow 0$ and from this inequality only provides: $$ \max_{t \in [0,1]} |x_n(t)-t | \leq \max_{t \in [0,1]} |t-1 |=1$$

2. As pointed out in the comments you need first to compute the pointwise convergence (as uniform convergence implies pointwise convergence) as for $t \neq 0$: $$ \lim_{n \rightarrow + \infty} x_n(t) \neq t$$