1. ## Uniform Convergence

Let $\displaystyle h_n (x) = sin(nx)$. Is $\displaystyle h_n (x)$ uniformly convergent on the interval [0,1]?

I don't think that it is uniformly convergent but I'm having a hard time proving it. I thought to use the theorem that it will be uniformly convergent iff $\displaystyle \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall n,m>N |f_n(x) - f_m(x)|\ <\epsilon$, but I can't see how to show this. Any help?

2. Originally Posted by worc3247
Let $\displaystyle h_n (x) = sin(nx)$. Is $\displaystyle h_n (x)$ uniformly convergent on the interval [0,1]?

I don't think that it is uniformly convergent but I'm having a hard time proving it. I thought to use the theorem that it will be uniformly convergent iff $\displaystyle \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall n,m>N |f_n(x) - f_m(x)|\ <\epsilon$, but I can't see how to show this. Any help?
I don't think the sequence is even point-wise convergent. for example note that

$\displaystyle \displaystyle \frac{\pi}{4} \in [0,1]$ let $\displaystyle \displaystyle x=\frac{\pi}{4}$

$\displaystyle \displastyle f_{n}\left( \frac{\pi}{4}\right)=\sin\left( \frac{\pi}{4}n\right)$

this will be both $\displaystyle 1,-1$ infinitely often.

So for any $\displaystyle N$ There exsists $\displaystyle m,n > N$ such that

$\displaystyle \displastyle |f_{n}\left( \frac{\pi}{4}\right)-f_{m}\left( \frac{\pi}{4}\right)|> \frac{1}{2}$

3. Originally Posted by TheEmptySet
I don't think the sequence is even point-wise convergent. for example note that

$\displaystyle \displaystyle \frac{\pi}{4} \in [0,1]$ let $\displaystyle \displaystyle x=\frac{\pi}{4}$

$\displaystyle \displastyle f_{n}\left( \frac{\pi}{4}\right)=\sin\left( \frac{\pi}{4}n\right)$

this will be both $\displaystyle 1,-1$ infinitely often.

So for any $\displaystyle N$ There exsists $\displaystyle m,n > N$ such that

$\displaystyle \displastyle |f_{n}\left( \frac{\pi}{4}\right)-f_{m}\left( \frac{\pi}{4}\right)|> \frac{1}{2}$
Moreover, no subsequence of $\displaystyle \sin(nx)$ is uniformly convergent on $\displaystyle [0,1]$. To see this recall that a uniformly convergent sequence of functions on a compact metric space is necessarily equicontinuous. That said, it's evident that $\displaystyle f_n(x)=\sin\left(nx\right)$ is not equicontinuous. Indeed, let $\displaystyle \left\{f_{n_k}\right\}_{k\in\mathbb{N}}$ be any subsequence of $\displaystyle \left\{f_n(x)\right\}$ suppose that there existed $\displaystyle \delta>0$ such that $\displaystyle d(x,y)<\delta\;\; x,y\in[0,1]$ implies $\displaystyle d(f_{n_k}(x),f_{n_k}(y))<1$. Note though that the period of $\displaystyle \sin\left(n_k x\right)$ is $\displaystyle \frac{2\pi}{n_k}$. Thus, since $\displaystyle \lim_{k\to\infty}n_k=\infty$ we may choose $\displaystyle K$ large enough so that $\displaystyle \displaystyle \frac{2\pi}{n_K}<\delta$. Thus, since $\displaystyle \sin\left(n_K x\right)$ runs through a full period on $\displaystyle \displaystyle \left[0,\frac{2\pi}{n_K}\right]\subseteq[0,\delta)$ there exists $\displaystyle \displaystyle x,y\in\left[0,\frac{2\pi}{n_K}\right]$ such that $\displaystyle \sin\left(n_K x\right)=1$ and $\displaystyle \sin\left(n_K y\right)=0$ and so $\displaystyle d\left(\sin\left(n_K x\right),\sin\left(n_K y\right)\right)=1$...but since $\displaystyle \displaystyle x,y\in\left[0\frac{2\pi}{n_K}\right]\subseteq[0,1]$ and $\displaystyle \displaystyle d(x,y)<\delta$ (since they're contained within $\displaystyle [0,\delta)$) this contradicts our choice of $\displaystyle \delta$. Thus, no such $\displaystyle \delta$ exists and so $\displaystyle \left\{f_{n_k}\right\}$ is not equicontinuous on $\displaystyle [0,1]$. It follows then from previous discussion that $\displaystyle \left\{f_{n_k}\right\}$ is not uniformly convergent on $\displaystyle [0,1]$.

Moreover, it might be feasible to prove that no subsequence of $\displaystyle \left{\sin(nx)\right\}_{n\in\mathbb{N}}$ has a pointwise convergent subsequence on $\displaystyle [0,1]$. The proof for why it cannot possess a pointwise convergent subequence on $\displaystyle [0,2\pi]$ is easy...I'll see if I can adapt my proof.