Calculating $\Vert\sum_n^\infty(-1)^nx_n \Vert_\infty$

Question

I want to calculate norm of functional:

$$f:(l_1,\Vert\cdot\Vert_\infty) \ni x_n\rightarrow \sum_n^\infty(-1)^nx_n $$

My work so far

Let's firstly check if this funcional is well defined. To do so let's consider

$$\sum_n^\infty |(-1)^nx_n|= \sum_n^\infty|x_n| <\infty\;\;\; \text{because we are in $l_1$}$$

Let's now consider $|f(x_n)|$ (to calculate norm of functional we want to find $\sup_{\{x_n:\sup_n|x_n| \le 1\}}|f(x_n)|$):

$$|f(x_n)| = |x_1+x_2+x_3+...| \le |x_1| + |x_2|+... \le \Vert x_n \Vert_\infty + \Vert x_n \Vert_\infty +...$$

Now we want to consider such sequences for which $\sum_n^\infty|x_n| < \infty$ and $\sup_n|x_n| \le 1$.

My question is: how can i choose such sequence which satisfies conditions above as well as $|f(x_n)| = \Vert x_n\Vert_\infty + \Vert x_n\Vert_\infty+...$

Would it be sufficient to calculate $\Vert f \Vert$ ?

Answer 1

Call $x^m_n=\begin{cases}0&\text{if }n>m\lor 2\nmid n\\ 1&\text{if }n\le m\land 2\mid n\end{cases}$. Then it is clear that $\lVert x^m\rVert_\infty\le1$ and $f(x^m)=\sum_{n=1}^\infty (-1)^nx_n^m=\left\lfloor \frac m2\right\rfloor$, and therefore $$\sup_{x\in l_1,\ \lVert x\rVert_\infty\le 1}\lvert f(x)\rvert\ge\sup_{m\in\Bbb N}\lvert f(x^m)\rvert=\infty$$

Answer 2

This is not a continuous linear functional, so its norm is $\infty$. If there is a constant $C$ such that $|\sum (-1)^{n}x_n| \leq C\sup |x_n|$ for all $(x_n) \in \ell^{1}$ the we can take $x_n=(-1)^{n}$ for $n \leq N$ and $x_n=0$ for $n >N$ to see that $N\leq C$. This cannot hold for all $N$ so $f$ is not continuous.