What is the closure of the following subspace $A$ of $\ell_\infty$ with the standard sup norm of $\ell_\infty$:
$$A=\{(a_n)\in \ell_\infty\mid (A_n)=a_1+a_2+\ldots+a_n\; \text{belongs to} \;\ell_\infty \}$$
Is it true to say that $\bar{A}$ is the space of all $(a_n)\in \ell_\infty$ such that the Cesaro sum $\frac{a_1+a_2 +\ldots +a_n}{n}$ goes to zero?
If it is the case, what is a proof?
Consider the linear map $ T :\ell ^\infty \to \ell ^\infty , $ given by $T ((s_n)_{n\geq 1}) = (a_n)_{n\geq 1}$, where $$ a_n = \left\{\matrix{ s_1, & \text{if } n=1,\hfill\cr s_n-s_{n-1}, & \text{otherwise }, }\right. $$ and observe that $A$ is precisely the range of $T $.
Moreover, observe that $T =I-S$, where $S$ is the right shift on $\ell ^\infty $.
For a continuous linear functional $f$ on $\ell ^\infty $ one then has that $f$ vanishes on $A$ if and only if $f =f\circ S$.
The answer is then that $\bar A$ coincides with the intersection of the kernels of all right-invariant linear functionals on $\ell ^\infty $.