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Math Help - Uniform Convergence

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    Uniform Convergence

    Let h_n (x) = sin(nx). Is 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 \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?
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    Quote Originally Posted by worc3247 View Post
    Let h_n (x) = sin(nx). Is 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 \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 \frac{\pi}{4} \in [0,1] let \displaystyle x=\frac{\pi}{4}

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

    this will be both 1,-1 infinitely often.

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

    \displastyle |f_{n}\left( \frac{\pi}{4}\right)-f_{m}\left( \frac{\pi}{4}\right)|> \frac{1}{2}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I don't think the sequence is even point-wise convergent. for example note that

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

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

    this will be both 1,-1 infinitely often.

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

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



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