Let $st_0$ be the set of all bounded statistically convergent complex sequences. Then we can define a linear functional $g:st_0\to \mathbb C$ by $g(x)=st-lim x_n$, $\forall x=(x_n)\in st_0$. [where, $st-lim x_n$ denotes the statistical limit of the sequence $(x_n)$.]
a sequence $(x_n)_n$ in a topological space $X$ is called statistically convergent to $\ell$ if for each open set $U\ni\ell$, $\delta\{n\in\mathbb N:x_n\notin U \}=0$. If we take $X=\mathbb C$ then we get statistical convergence of sequence of complex numbers. Clearly $st_0\subset l^\infty$, and $l^\infty$ has well known sup-norm $\|\cdot\|_\infty$. So, the topology on $st_0$ is induced by the normed linear space $(st_0, \|\cdot\|_\infty)$.
My Question : Is the linear functional $g$ bounded (continuous)? Thanks in advance. Any answer will be appreciated.
$\newcommand{\stlim}{\operatorname{st-lim}}$For any sequence we have that $$\liminf x_n \le \stlim x_n \le \limsup x_n.$$ There are various ways how to see it - you just need to show that the statistical limit is in fact a cluster point of the sequence $(x_n)$. You could use some similar techniques as in the answers to your other question about statistical convergence: Sequence Lemma and statistical convergence.
In particular, this implies that $\inf x_n \le \stlim x_n \le \sup x_n$ and $|\stlim x_n|\le \sup |x_n|$. So you have that $|g(x)|\le \|x\|_\infty$. This shows that $g$ is continuous.
I will mention that the same is true if we use convergence along a non-trivial ideal. In fact, we can also define limit superior and limit inferior w.r.t. an ideal, an we'll get $$\liminf x_n \le \operatorname{\mathcal I-liminf} x_n \le \operatorname{\mathcal I-limsup} x_n \le \limsup x_n.$$