The norm of a continuous linear operator

Question

What is the norm of this operator? $S{u} = \sum_{n=1}^{\infty} \frac{(-1)^{n} U_{n}}{n} $ where $U \in \ell^{1}$, where $\ell^{1}=\{ U=(U_{n})_{n \in \mathbb{N}} \subset \mathbb{R}~ such ~ that ~ \sum_{n=1}^{\infty} | U_{n} | < \infty \}$. I don't know if it is right but all I did is $$ |S u | = \left|\sum_{n=1}^{\infty} \frac{(-1)^{n} U_{n}}{n}\right| \le \left| \sum_{n=1}^{\infty}{U_{n}} \sum_{n=1}^{\infty}{\frac{(-1)^n}{n}}\right| \le \ln(2) \sum_{n=1}^{\infty}|U_{n}| = \ln(2)\|U\|_{\ell^{1}} $$

Answer 1

The first inequality in your estimate is wrong.

In fact, $\|S\|_{\mathcal L(l^1,\mathbb R)} = 1 > \ln 2$.

First, let $U = (1,0,0,\dots)$, i.e., $U$ is the first unit vector. Then $\|u\|_{l^1}=|Su| = 1$. So $\|S\|_{\mathcal L(l^1,\mathbb)} \ge 1$. Second, let $u$ be arbitrary. Then $$ |Su| = \left| \sum_n \frac{(-1)^n}n u_n\right| \le \sum_n \frac1n |u_n| \le \sum_n |u_n|=\|u\|_{l^1}. $$ So $\|S\|\le1$.