Let $X=\ell^{\infty}$. Then there exists $f\in X'$ such that $\|f\|= 1$ and $f(x)=\lim_{n\to\infty} x_n$ for $x\in c$ and $f(Sx)=f(x)$ for $x\in X$ where $S$ is the left shift operator: $S(x_1,x_2,\cdots)=(x_2,x_3,\cdots)$
I think explicit construction of such $f$ is out of question.
I first thought this was about Hahh-Banach problem, but then the requirement $f(Sx)=f(x)$ seems very restrictive because the range of $S$ is in fact $X$.
What I ended up doing was:
define $f$ on the dense subspace $c+Y$ ,$Y=\text{span}\{e_n:n\ge 1\}$(where $e_n$ are elementary sequences which has $1$ as $n^{\text{th}}$) by
$$ f(x)=\lim_{n\to\infty} x_n $$
It's easy to see that $f$ is bounded with respect to $\infty$ norm and $\|f\|=1$ in fact. Since the associated field is a Banach space, we may extend $f$ to whole of $X$, and this gives desired $f$ because
for any $x=\lim_n x_n$ (this limit is limit of sequences in dense subset $Y$)
by continuity
$$ f(Sx)=\lim_{n\to\infty} f(Sx_n)=0=\lim f(x_n) = f(x) $$
because $x_n\in Y$ implies limit of sequence $x_n$ is $0$ (as it should eventually be $0$)
But something seems wrong, because this means $f(x)=0$ outside of $c$, so how is this $f$ actually continuous?
If something's wrong with my method, what would be correct approach?