# Are sequences with Cesaro mean a closed subset of $\ell_\infty$?

How can we show that the bounded sequences which are Cesaro summable, i.e., the sequences such that the limit $$\lim\limits_{n\to\infty} \frac{x_1+\dots+x_n}n$$ exists, form a closed subset of $\ell_\infty$?

As usually, $\ell_\infty$ denotes the space of all bounded sequences with the sup-norm $\|x\|=\sup\limits_{n\in\mathbb N} |x_n|$.

Closedness of this set was brought up in comments to an answer discussing a proof of existence of Banach limit based on Hahnâ€“Banach theorem.

I can think of this quick argument (I hope I did not miss something there):

It's relatively easy to show that the function \begin{align*} T &\colon \ell_\infty \to \ell_\infty\\ T &\colon (x_n) \mapsto \left(\frac{x_1+\dots+x_n}n\right) \end{align*} is continuous simply by noticing that $\|Tx\|\le\|x\|$. And since the set $c$ of all convergent sequences is closed in $\ell_\infty$, so is $T^{-1}(c)$; which is exactly the set of all sequences that have Cesaro mean.

Are there some other proofs how to show the closedness of this set?

• I think by using the sequential characterization of continuity you can derive a result – Marios Gretsas Sep 16 '17 at 12:59

Another way is to show that the complement is open: for this, let $x=(x_n)$ not be Cesaro summable. Then the sequence $$y_n=\frac{x_1+\dots+x_n}{n}$$ is not Cauchy, meaning that there exists some $\varepsilon>0$ such that, for any $N\in\mathbb N$, there exist $n>m\geq N$ such that $|y_n-y_m|\geq\varepsilon$.
Let now $\delta=\varepsilon/3$. Then $B_{\delta}(x)$ is a subset of the sequences that are not Cesaro summable: to show this, let $z=(z_n)\in\ell_{\infty}$ with $\|z\|<\delta$. Then, for any $N\in\mathbb N$, if $n>m$ are the ones that appear above, $$\left|\frac{(x_1+z_1)+\dots+(x_n+z_n)}{n}-\frac{(x_1+z_1)+\dots+(x_m+z_m)}{m}\right|\geq\\\left|\frac{x_1+\dots+x_n}{n}-\frac{x_1+\dots+x_m}{m}\right|-\left|\frac{z_1+\dots+z_n}{n}\right|-\left|\frac{z_1+\dots+z_m}{m}\right|\geq\\|y_n-y_m|-2\delta\geq\varepsilon-2\delta=\frac{\varepsilon}{3},$$ which shows that $x+z$ is not Cesaro summable.
For $n\in \mathbb N$ let $\sigma_n\in (l_{\infty})^*$ where $\sigma_n(z)=n^{-1}\sum_{j=1}^nz_j$ for $z=(z_j)_{j\in \mathbb N} \in l_{\infty}.$
Let $C$ be the set of those $y\in l_{\infty}$ such that $c(y)=\lim_{n\to \infty}\sigma_n(y)$ exists.
Since $C$ is a vector sub-space of $l_{\infty}$ and $\sup \{|c(y)|/\|y\|: 0\ne y\in C\}=1$ we can extend $c$ (uniquely) to a functional $c'\in (\bar C)^*$ with $\|c'\|=1.$
Let $x\in \bar C$ and let $(y^{(i)})_i$ be a sequence in $C$ converging in norm to $x.$ For $r>0$ take $i_r\in \mathbb N$ such that $\forall i\geq i_r\;(\|x-y^{(i)}\|<r).$ Then $$\forall i\geq i_r\;| c'(x)-c(y^{(i)})|=|c'(x-y^{(i)})|<r.$$ Now $\|\sigma_n\|=1$ for all $n.$ From this and the previous line, we have for each $i\geq i_r$ that $$\lim \sup_{n\to \infty} \sigma_n (x)=\lim \sup_{n\to \infty} \sigma_n (y^{(i)}+(x-y^{(i)})=$$ $$=c(y^{(i)})+\lim \sup_{n\to \infty}\sigma_n(x-y^{(i)}) <$$ $$<(c'(x)+r)+\|x-y^{(i)}\|<c'(x)+2r.$$ Similarly for $i>i_r$ we have $$\lim \inf_{n\to \infty}\sigma _n(x)>c'(x)-2r.$$ The positive value $r$ is arbitrary, so the rest is obvious.