I guess the set $\{f = \cos(nx) : n=1, ... ,\infty\}$ is closed in $\ C([0,1])$ under the sup norm. Therefore, I am trying to show its limit function is of the form $\cos(nx)$. However, I have difficulty proving that for $\cos(n_ix)$ to converge, $\ n_i$ must be constant after some term. At first, I tried to prove c $\leq$ $\mid cos(nx)-cos(mx)\mid$ for some positive c and distinct m and n. Is my guess wrong?
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Yes. This same problem was asked recently. Maybe the other version already has answers. – conditionalMethod Dec 09 '19 at 00:32
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6Does this answer your question? Prove that ${\cos(nx) : n=1, ... ,\infty }$ is a bounded subset – conditionalMethod Dec 09 '19 at 00:34
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@conditionalMethod That question is asking whether the family of functions $x\mapsto \cos nx$ is uniformly bounded, not whether it is closed. – Math1000 Dec 09 '19 at 01:22
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@Math1000 One follows from the other. – conditionalMethod Dec 09 '19 at 01:23
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I don't see how. It was shown that these functions are uniformly bounded, but I don't see how they can be closed considering there is no limit function. – Math1000 Dec 09 '19 at 01:25
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@Math1000 That's not the only thing proven there, nor what was about most the the argument about. ... ${1,2,3,...}$ is closed and there is no limit besides the eventually constant sequences. – conditionalMethod Dec 09 '19 at 01:39
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Ah, good point. I suppose it does not make sense to argue that a set is not closed if does not contain its limit, if it does not have any limit points to begin with. I concede. – Math1000 Dec 09 '19 at 02:07
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1Does the domain [0,1] have no effect on the result? It seems that under the domain, they are not orthogonal. – Steven Chen Dec 09 '19 at 03:28
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The set is closed. If not we would come to contradiction in the following way:
Denote $f_n(x)=cos(nx)$ and assume that some subsequence of $f_n$ converges to $f$ in sense of supremum norm and $f$ is not from the set. Let denote this subsequence as $f_k$.
Let $t\in [0,1]$. we see that $$|\int_0^tf_k-f\;dx|\leq \|f_{k}-f\|_\infty\to 0\mbox{ as }k\to\infty.$$
We see that $\int_0^tf_k\;dx=-sin(kt)/k\to 0$. Hence $\int_0^tf\;dx=0$ for every $t$ and so $f=0$. But $1=\|f_{k}\|_\infty\to \|f\|_\infty$ and so $\|f\|_\infty=1$ what is a contradiction.
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The method is great. I didn't notice we can construct a sequence with increasing k such that fk goes to f if f fails to fall into the set. – Steven Chen Dec 09 '19 at 16:00
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2This is the method of computing weak limits in $L^p$ spaces. It is usefull until you are able to compute the integral on interval $[0,t]$. – elliptic Dec 09 '19 at 16:05