I ran into this question and im not sure how to solve it:
Check uniform convergence of:
$$f_{n}(x)=\sin\left ( {\frac{1}{n^{3}x}} \right )\quad \;x\in (0,1]$$
I tried finding the supremum of the function and then take the limit as $n \to\infty$. I got the answer $1$ which means that it is not uniform convergence, but there is no use of the specific bounding of $x\in (0,1]$ in this way, which means there is no uniform convergence for any $x$.
on the other hand, I found that when $x\in [1,\infty)$ there is uniform converegnce.
Please help me understand where is the use of the specific boundry of $x\in (0,1]$.
Thanks in advance.
The pointwise limit of $f_n(x)$ is $f(x) = 0$. For $N \in \mathbb N$, let $x_N = 1/N^3$. This is always possible since $x \in (0, 1]$. We have $f_N(x_N) = \sin(1)$. By letting $\epsilon < \sin(1)$ we see that $f_n(x)$ cannot converge uniformly on $x \in (0, 1]$.
On the other hand, if $x \in [1, \infty)$, we have $0 \le \dfrac{1}{n^3 x} \le \dfrac{1}{n^3} \le 1$. Hence: $$ \left|f_n(x)\right| = \left|\sin\left(\frac{1}{n^3 x}\right)\right| \le \left|\sin\left(\frac{1}{n^3}\right)\right| $$
For any $\epsilon > 0$, we can make $n$ large enough so that: $$ \left|\sin\left(\frac{1}{n^3}\right)\right| < \epsilon $$
Thus, $f_n(x)$ converges uniformly on $x \in [1, \infty)$.