Maximum of $\sum_{n=1}^\infty \sin(n) \cdot (x_n - x_{n+1})$

Question

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 $2$

Answer 1

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:

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