# Is the functional mapping $x$ to the statistical limit of $x$ bounded (continuous)?

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.

• What is a statistically convergent sequence of complex numbers? And continuity for what topology? – Kavi Rama Murthy Jun 6 at 9:32
• @KaviRamaMurthy Sir, 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 $||.||_\infty$. So, the topology on $st_0$ is induced by the normed linear space $(st_0, ||.||_\infty)$. – BijanDatta Jun 6 at 9:52

$$\newcommand{\stlim}{\operatorname{st-lim}}$$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 $$\liminf x_n \le \operatorname{\mathcal I-liminf} x_n \le \operatorname{\mathcal I-limsup} x_n \le \limsup x_n.$$