I am struggling to understand the solution of an item of a question in my exercise sheet,I think it is a standard application of Hahn Banach theorem and found this question An application of Hahn-Banach theorem on the left-shift operator on $l^\infty$, but kinda not still understanding, and it is not exactly in the same way of the question in my problem sheet.
let $l^\infty$ be the space of bounded sequences over the real numbers and $T:l^\infty\rightarrow l^\infty$ be defined as the shift operator, i.e T sends $(x_1,x_2,x_3 \ldots)$ to $(x_2,x_3,x_4 \ldots)$ and let M be the subspace that is $Im(I-T)$ applied in all $x \in l^\infty$. If $c$ is the space of convergent sequences, define the bounded linear functional in this space $f_0(x) = \lim_{n \to \infty} x_n$ for all $x \in c$. By Hahn Banach theorem (applicable since that the absolute value of $f_0(x)$ is bounded by the supremum of x that is a sublinear function), there exists a linear functional in the entire $l^\infty$ that preserves the operator norm (and thus is continuous) and coincides with $f_0$ in the space of convergent sequences, lets call it $f$.
I want to prove that f is invariant under T, i.e $f(T(x))=f(x) \forall x \in l^\infty$ I can see that this is true for x in the space of convergent sequences, since f will preserve the limit, but how to prove that the same is valid for all x in the space of bounded real sequences? It is really not guaranteed by Hahn Banach theorem. Can someone help me? Thanks.
