Currently I am trying to solve essentially the same problem as presented in uniform convergence of n sin(x/n) with the exception that the interval is restricted to [0, 1]. I have not made any progress beyond $\left|n\sin(x/n) - x\right| \leq \max_{x \in [0, 1]}\left|n\sin(x/n) - x\right|$ as the derivative $\frac{n\sin(x/n) - x}{\left|n\sin(x/n) - x\right|}$ does not provide anyting useful and I do not know any upperbound for the expression inside the absolute value. Also, I did not get much out of the two answers in the aforementioned link as (to me) it seems that one answers shows that the function does not converge uniformly and one shows that it does.
How should I proceed?
Recall the basic inequalities $y- \frac{y^3}{6} \leqslant \sin y \leqslant y$ for $y>0$. If you are not familar with the LHS inequality, it is easily derived from the RHS inequality by integrating twice.
We have
$$x- x \sin \frac{x}{n} = \begin{cases}x \left(1 - \frac{\sin \frac{x}{n}}{\frac{x}{n}}\right), & 0 < x \leqslant 1 \\ 0 , & x = 0\end{cases}$$
Using the basic inequalities, we obtain
$$\frac{x}{n} \geqslant \sin \frac{x}{n} \geqslant \frac{x}{n} - \frac{x^3}{6n^3},$$
and by rearranging, we get for $0 < x \leqslant 1$,
$$0 \leqslant x \left(1 - \frac{\sin \frac{x}{n}}{\frac{x}{n}}\right) \leqslant \frac{x^3}{6n^2}\leqslant \frac{1}{6n^2}$$
Thus, for all $x \in [0,1]$,
$$\left|x \sin \frac{x}{n} - x \right| \leqslant \frac{1}{6n^2},$$
and it follows that $n \sin \frac{x}{n} \to x$ uniformly on $[0,1]$.