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I want to find supremum of |L| where $$L = \sum_{n=1}^\infty sin(n)\cdot (x_n - x_{n+1})$$ where $x_n$ is such a sequence that $\sum_{n=1}^\infty |x_n| \le 1$.

My work so far

Let's expand expression of $|L|$ to get more knowledge about possible candidates for maximum.

$$|L| = |\sum_{n=1}^\infty sin(n)\cdot (x_n - x_{n+1})|\le\sum_{n=1}^\infty |sin(n)\cdot (x_n - x_{n+1})|\le$$ $$\le\sum_{n=1}^\infty |(x_n - x_{n+1})|$$

And here I'm not sure what to do next. I'm sure I have to somehow use that $\sum_{n=1}^\infty|x_n| \le 1$ but I don't see how it can be forced in expression $\sum_{n=1}^\infty|x_n-x_{n+1}|$. Could you give me a hint how to bound this expression to obtain maximum ?

EDIT

Due to the hints mentioned in comment section I got that:

$$|L| \le \sum_{n=1}^\infty|x_n - x_{n+1}| \le \sum_{n=1}^\infty |x_n| + |x_{n+1}| \le 2$$

So the candidate for my supremum is $2$. But to prove it I have to point exactly the sequence $x_n$ such that $\sum_{n=1}^\infty |x_n| \le 1$ and $|L| = 2$. Am I correct ? If so, could you please also give a hint how to choose this sequence ? I tried to choose $\frac{1}{2^n}$ but value of L is far away from $2$

Lucian
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    Can't you just use the triangle inequality on the summands and get

    $$\sum_{n=1}^\infty |x_n - x_{n+1}| \le \sum_{n=1}^\infty |x_n| + |x_{n+1}|$$

    and then continue in the obvious way from there?

     – PrincessEev Mar 16 '21 at 22:45
  • What do you mean exactly? Do you want to calulate $\sup L(x)$ over all possible sequences $x=(x_n)$ with $\sum |x_n|<1$? – Matematleta Mar 17 '21 at 01:33

1 Answers1

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First, we solve the following problem

The supremum of $$P = \left|\sum_{n=1}^{+\infty}a_n(x_n-x_{n+1})\right|$$ given $$\sum_{n=1}^{+\infty}|x_n| \le 1$$ and $|a_n| \le 1 $ for all $n\in \Bbb N^*$

is $$\sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$$

Proof: We have

\begin{align} P &= \left|\sum_{n=1}^{+\infty}a_n(x_n-x_{n+1})\right| \\ &= \left|\sum_{n=1}^{+\infty}a_nx_n-\sum_{n=2}^{+\infty}a_{n-1}x_n\right| \\ &= \left|a_1x_1+\sum_{n=2}^{+\infty}(a_n-a_{n-1})x_n\right| \le |a_1| |x_1|+\sum_{n=2}^{+\infty}(|a_n-a_{n-1}| |x_n|) \end{align} For the sake of simplicity, denote $a_0 = 0$, then

$$P \le \sum_{n=1}^{+\infty}(|a_n-a_{n-1}| |x_n|)$$

By definition of supremum, $|a_n-a_{n-1}| \le \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$ for all $n\in \Bbb N^*$. Then

\begin{align} P \le \sum_{n=1}^{+\infty}(|a_n-a_{n-1}| |x_n|) \le \sum_{n=1}^{+\infty}(\sup_{k\in \Bbb N^*}|a_k-a_{k-1}| |x_n|) &= \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|\sum_{n=1}^{+\infty}|x_n| \\ &\le \sup_{k\in \Bbb N^*}|a_k-a_{k-1}| \end{align}

$$$$

Second, it suffice to find the supremum $|\sin(n) - \sin(n-1)|$ for $n\in \Bbb N^*$

We have $$|\sin(n) - \sin(n-1)| = 2\sin \left( \frac{1}{2} \right) \left|\cos \left( n-\frac{1}{2} \right) \right|$$

We notice that $\sup_{n\in \Bbb N^*} \left|\cos \left( n-\frac{1}{2} \right) \right| = 1$ because there always exists a sequence $(u_n,v_n) \in \Bbb N^* \times \mathbb{N}^*$ such that $$\frac{u_n-\frac{1}{2}}{v_n} \to \pi$$

or $$ \left|\cos \left( u_n-\frac{1}{2} \right) \right| \to 1$$

Then, $$ \sup_{n\in \Bbb N^*}|\sin(n) - \sin(n-1)| \le 2\sin \left( \frac{1}{2} \right)$$

Third, we prove that the supremum of $|L|$ is $2\sin \left( \frac{1}{2} \right)$

For each value $u_k$ of the sequence $\{ u_k\}_{k\in \Bbb N^*}$ (constructed in the second step), let's construct the sequence $\{x_n\}_{n\in \Bbb N^*}$ as follows: $x_{u_n} = 1$ and $x_k = 0$ for $k\ne u_n$. Hence for each construction of the sequence $\{x_n\}_{n\in \Bbb N^*}$, the value of $L$ (we denote $L(u_k)$) is

$$L(u_k) = |\sin(u_k) - \sin (u_k-1) | = 2\sin \left( \frac{1}{2} \right) \left|\cos \left( u_k-\frac{1}{2} \right) \right| \to 2\sin \left( \frac{1}{2} \right) $$

Hence, this supremum can be atteined.

Conclusion:

$$ \sup |L| = 2\sin \left( \frac{1}{2} \right)$$

NN2
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  • In the first part, how do you show that $\sup_{n\in\Bbb{N}} \lvert a_n-a_{n+1}\rvert$ is the least upper bound and not just an upper bound? Also I think you might be writing $a_{n+1}$ when you mean $a_{n-1}$ (although that doesn’t affect the result in this case) – iosce Mar 17 '21 at 00:18
  • The first part of what I’m saying is that you show that $P\le \sup_{n\in\Bbb{N}}\lvert a_k-a_{k-1}\rvert$ for all $x_n$, but don’t show that there is no $s< \sup_{n\in\Bbb{N}}\lvert a_k-a_{k-1}\rvert$ such that $P\le s$ for all $x_n$ – iosce Mar 17 '21 at 00:44
  • @iosce No, I just showed that $P \le sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$. What is $s$ or $x_n$ here? – NN2 Mar 17 '21 at 00:49
  • What I’m getting at is that you proved $\sup\lvert L\rvert\le 2\sin(1/2)$, not that $\sup\lvert L\rvert =2\sin(1/2)$. I was using the same $x_n$ as in op and your post, $s$ is a hypothetical upper bound less than yours – iosce Mar 17 '21 at 00:53
  • @iosce OK, I see. In the part 2, we all agree that $\sup_{n\in \Bbb N^}|\sin(n) - \sin(n-1)|$ because there exists a sequence $u_n$ such that $\left|\cos \left( u_n-\frac{1}{2} \right) \right| \to +\infty$ (in fact, a rigorous proof for this result is long, I just show the idea). Construct the sequence ${x_k}_{k \in \Bbb N^}$ for each $u_N$ such that $x_{u_N} = 1$ and $x_k = 0 $ if $k \ne u_n$. Then if we denote $L(u_n)$ the sum from this construction then $L(u_N) = |\sin (u_N) - \sin(u_N-1)|$ and then $L(n_N)| \to 2 \sin(1/2)$ when $n \to +\infty$ – NN2 Mar 17 '21 at 01:06
  • $x_k$ needs to be a series such that $\sum_{k=1}^\infty \lvert x_k\rvert\le 1$, which isn’t true of your construction. I wasn’t disputing that $2\sin(1/2)$ is the supremum of $\lvert \sin(n)-\sin(n-1)\rvert$ – iosce Mar 17 '21 at 01:19
  • @iosce The sequence $x_{k} = 1$ if $k=u_n$ and $x_k = 0$ for $k\ne u_n$ satisfies indeed the condition $\sum_{k=1}^{+\infty} x_k = 1$ ! – NN2 Mar 17 '21 at 01:29
  • You probably want the condition $|a_n| \le 1$ in the first part, instead of $a_n \le 1$. – Martin R Mar 17 '21 at 05:34
  • Hey! thank you very much for your answer. I understand almost everything except two things - could you please expand on fact that $\sum_{k=1}^\infty|x_k| \le 1$ ? And also why you are constructing sequence $(x_k)$ and then you are deriving $L(u_k)$ ? – Lucian Mar 17 '21 at 17:08
  • @Lucian The fact $\sum_{k=1}^{+\infty}|x_k| \le 1$ is used at the last step of the last inequality of first part. $$\sup_{k\in \Bbb N^}|a_k-a_{k-1}|\sum_{n=1}^{+\infty}|x_n| \le \sup_{k\in \Bbb N^}|a_k-a_{k-1}|$$ – NN2 Mar 17 '21 at 17:15
  • @Lucian The sequence $x_k$ in the third part is constructed because in the second part, we proved only that $$\sup |L| \le \sup_{k\in \Bbb N^}|a_k-a_{k-1}|$$ (and not $\sup |L| = \sup_{k\in \Bbb N^}|a_k-a_{k-1}|$). So, it we prove (and we really did in the third part) that there exists a construction that $L(u_k) \to \sup_{k\in \Bbb N^}|a_k-a_{k-1}|$ then we can have that $$\sup |L| \ge \sup_{k\in \Bbb N^}|a_k-a_{k-1}|$$ From the result of the 2nd and 3rd parts, we can conclude that $$\sup |L| = \sup_{k\in \Bbb N^*}|a_k-a_{k-1}|$$ – NN2 Mar 17 '21 at 17:18
  • @Lucian About $L(u_n) = |\sin(u_k) - \sin (u_k-1) |$, it suffices to notice that $$L =\sum_{k=1}^{+\infty}(a_k-a_{k-1})x_k $$ (obtaining in the 1st part) $$$$ Replace $x_{k} = 1$ if $k = u_n$ and $0$ if $k \ne u_n$, you have $L(u_n) = |\sin(u_k) - \sin (u_k-1) |$ – NN2 Mar 17 '21 at 17:21