Does $\sin ^n x$ converge uniformly on $[0,\frac{\pi}{2})$?

Question

Does $f_n(x)=\sin ^n x$ converge uniformly on $[0,\frac{\pi}{2})$ ?

I know $f_n(x) \rightarrow 0$ pointwise, since $\vert \sin ^n x \vert< 1$. How about the uniform convergence?

Any hint?

Answer 1

Let $x_n=\sin ^{-1} (1-\frac 1 n)$. Then $\sin^{n} (x_n))=(1-\frac 1 n )^{n} \to 1/e$. So the convergence is not uniform.

Answer 2

Uniform convergence of $(f_n)$ to $f(x) = 0$ would require that $$ M_n = \sup \{ |f_n(x) - f(x) | : 0 \le x < \frac \pi 2 \} $$ converges to zero. However, for each $n$ $$ M_n \ge \lim_{x \to \pi /2} f_n(x) = 1 \, . $$