Consider the following map:
$$ l^{\infty} \ni \{x_n\}_{n=1}^{\infty} \longmapsto \left\{x_n - \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right\}_{n=1}^{\infty} \in l^{\infty},$$
where $\underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j$ is a Banach limit. I need to check whether the map is well-defined and continuous. If it is continuous, I have to compute the norm.
My attempt at solution:
The map is well-defined since we only shift the sequence by some finite number from $\mathbb{C}$. Continuity follows from the following inequality:
$$ \left\|\left\{x_n - \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right\}_{n=1}^{\infty}\right\| = \sup\limits_{n\geq 1}\left|\left\{x_n - \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right\}_{n=1}^{\infty}\right| \leq \|\{x_n\}_{n=1}^{\infty}\| + \left|\underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right|. $$
Now if $l^\infty$ is over $\mathbb{R}$ then:
$$ \left|\underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right|\leq \left|\limsup\limits_{j\rightarrow\infty}x_j \right|, \text{if}\ \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j > 0, $$
$$ \left|\underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right|\leq \left|\liminf\limits_{j\rightarrow\infty}x_j \right|, \text{if}\ \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j < 0. $$
Both $\left|\liminf\limits_{j\rightarrow\infty}x_j \right|$ and $\left|\limsup\limits_{j\rightarrow\infty}x_j \right|$ are less than $\|\{x_n\}_{n=1}^{\infty}\|$, so:
$$\left\|\left\{x_n - \underset{j\rightarrow\infty}{\mathrm{Lim}}\ x_j\right\}_{n=1}^{\infty}\right\| \leq 2\|\{x_n\}_{n=1}^{\infty}\|$$.
However, I don't know how to generalize this for a complex case. I would also like to know whether a sequence:
$$ (1,-1,-1,\dots,-1)$$
is a good choice to show that the norm of the map is $2$. The norm of this sequence is $1$. It is convergent to $-1$, so the Banach limit will be equal to this limit and then one gets the desired norm of $2$.
I would appreciate any hints regarding the complex case.