Sequence of functions $\sin^{n} (\pi x)$ cauchy in C[0,1]

Question

Equipping C[0,1] with its usual (supremum) norm, for each $n \in \mathbb{N}$, let $g_n(x)=\sin^{n} (\pi x)$ for $x \in [0,1]$. I'm trying to prove (or disprove) this sequence is cauchy.

Intuitively, I would think it is cauchy.

From the definition, I'm trying to show that for any $\epsilon >0$, sup$|\sin^{n} (\pi x)-\sin^{m} (\pi x)| < \epsilon$ whenever m,n>N.

I'm looking for a hint on how to go about showing this.

Answer 1

Each $g_m$ is continuous with $g_m(0)=0$ and $g_m(\frac{1}{2})=1$ so let $x_m\in (0,\frac {1}{2})$ with $g_m(x_m)=\frac {1}{2}.$ We have $$\sup_{n>0}\|g_m-g_{m+n}\|\geq \sup_{n>0}|g_m(x_m)-g_{m+n}(x_m)|= \lim_{n\to \infty}\frac {1}{2}-g_{m+n}(x_m)=\frac {1}{2}.$$ So the Cauchy criterion is not satisfied.

Answer 2

A "trick" I use is to bound the supremum of a function using it's derivative. You can look at $f_{nm}(x)=\sin ^n (\pi x) - \sin ^m (\pi x) $ and search for its maxima in $[0,1]$.

Edited: Another way is to recall (or prove) that a sequence of functions convergeces uniformally iff it is uniformally cauchy sequence (or that $C_\sup[0,1]$ is a Banach space).