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Let's consider functional $$L(x_n) = \sum_{n=1}^\infty \sin(n)(x_n - x_{n-1})$$

I'm looking for such a sequence $x_n$ such that $\sum_{n=1}^\infty |x_n| \le 1$ and $|L(x_n)| = 2\sin(\frac12)$

I was mostly trying to point a sequnece which has only one value at certain point and zeroes elsewhere. (easiest thing to calculate and easiest way to match condition of bounded sum). Apart from that I also tried sequneces with several points but those also bring me to nothing (I was using R software to check a lot of possibilites). Could you please give me a hand with figuring out the accurate example ?

EDIT I got your answers but I don't understand one thing within it:

You are defining sequence $x_n$ in such way that this is $x_n = 1$ for $n = u_k$ and $0$ otherwise, and then you are describing $L(x_n)$.

But why $\sum_{n=1}^\infty |x_n| \le 1$ i.e. $\exists!n_0: x_{n_0} = u_k$ ?

Is this obvious ? I was trying to find argument for that but I wasn't able to

Lucian
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1 Answers1

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Rearranging your series, it becomes $$\sum_{n\ge 1}x_n(\sin(n) - \sin(n+1)).$$ Applying the trigonometric identity, $$\sin(n) - \sin(n+1) = -2\sin(1/2)\cos(n+1/2),$$ we find that you seek for a family $x_n$ for which $$\sum_{n\ge 1}x_n\cos(n+1/2) = -1.$$ Now we see why there isn't any solution. We may bound $$|\sum_{n\ge 1}x_n\cos(n+1/2)| \le \sum_{n\ge 1}|x_n\cos(n+1/2)|,$$ since $|\cos(n+1/2)| < 1$ for all $n\in \mathbb{N}$, (why?), we have $|x_n\cos(n+1/2)| < |x_n|$ for all $n\in \mathbb{N}$, such that $$|\sum_{n\ge 1}x_n\cos(n+1/2)| < \sum_{n\ge 1}|x_n| \le 1.$$ Therefore there aren't any solutions $x_n$ to your equation.

Is this a homework question? If so, please add the tag homework.

kindasorta
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  • FYI, although we showed that $2\sin(1/2)$ is an upper bound on the absolute value of your sum, it should be mentioned that from the equivalent expression $\sum_{n\ge 1}x_n\cos(n+1/2) = -1$ we see that it is in fact a supremum. Since $\pi$ is irrational, you can choose a sequence $n_i$ that approximates $-1/2$ modulo $2\pi$ as close as you'd like. Fixing $x_{n_i} = -1$ would make your sum almost as close to its supremum. – kindasorta Mar 18 '21 at 00:33
  • Could you please expand on fact that such defined $x_{n_i}$ aproximates my sum as close as I want ? ;)) – Lucian Mar 18 '21 at 09:02
  • Once you understand my answer, it should be sufficiently clear that a solution to $\sum_{n\ge 1}\sin(n)(x_n - x_{n-1}) = 2\sin(1/2)$ is a solution to $\sum_{n\ge 1}x_n\cos(n+1/2) = -1$ and vice versa. From this, one sees that $n=-1/2$ and $x_n=-1$ is a solution. Since $n=-1/2$ is not a natural number, this is not a solution, but since $\cos(x)$ is a periodic function, we don't actually need $n=-1/2$ it suffices to have $n=-1/2 + 2\pi k$ for some $k\in \mathbb{Z}$. This would also not be a natural number, but could be sufficiently close to one by choosing your $k$'s in a "smart" way. – kindasorta Mar 18 '21 at 16:12