I think this is a fairly standard problem and has been asked about before. However I am mostly interested in getting help with how to close the argument that I have attempted below.
Task:
The idea is to start with a bounded linear functional on a subspace of $Z \subset l^\infty$ and then invoke Hahn-Banach. The difficult part is choosing the appropriate subspace and functional.
Let $T:l^\infty \to l^\infty$ map $(x_1, x_2, \ldots)$ to $(x_2, x_3, \ldots)$.
Write $Y = \{ x - T(x) : x \in l^\infty \} $
Denote $\boldsymbol{a} := (a,a, \ldots)$ for $a \in \mathbb{C}$.
Show there is a bounded linear functional $f$ from $l^\infty$ to $\mathbb{C}$ such that $f \big( \boldsymbol{1} \big) = 1$, $f(y) = 0$ for each $y \in Y$, and $||f||_{op} = 1$.
$||\cdot ||_{op} $ is the operator norm $\sup \{ |f(x)|/||x||_\infty : x \neq 0 \}$
My attempt:
I have shown $\boldsymbol{1} = (1,1,\ldots)$ is not in the closure of $Y$.
Take the subspace $Z := Y + \text{span} (\boldsymbol{1} ) = \{ y + \lambda \cdot \boldsymbol{1} : y \in Y, \lambda \in \mathbb{C} \}$.
If $z \in Z$ then $z = y + \lambda \boldsymbol{1}$ for some $y \in Y$ and $\lambda \in \mathbb{C}$.
Define a functional $g$ on $Z$ such that $g(z) = g(y + \lambda \boldsymbol{1} ) = \lambda$.
It is immediate that $g \big( \boldsymbol{1}\big) = 1$, $g(y) = 0$ for each $y \in Y$, and $1 \leq ||g||$.
What I am struggling with is showing $||g|| \leq 1$.
I can get this if I can show that for any $z \in Z$, with $z = y + \lambda \boldsymbol{1}$, we must have $||z||_\infty \geq |\lambda|$.
Since $\boldsymbol{1}$ is not in the closure of $Y$, neither is $\lambda \boldsymbol{1}$.
I have shown separately that any ball of radius $0 < \epsilon < |\lambda|$ centered at $\lambda\boldsymbol{1}$ does not intersect the closure of $Y$.
This means that the $y$ component in $z = y + \lambda \boldsymbol{1}$ satisfies $ ||y - \alpha \boldsymbol{1} || \geq \epsilon > 0 $.
If the $y$-component of $z$ is zero then we trivially have $||z||= |\lambda|$. But I am failing to show $||z|| \geq |\lambda|$ when the $y$-component of $z$ is non-zero.
I have tried exploiting $ ||y - \alpha \boldsymbol{1} || \geq \epsilon > 0 $ together with the triangle inequality in an intelligent manner but have had no success at all:
$$ ||z|| = ||z - \lambda \boldsymbol{1} + \lambda \boldsymbol{1} || = || y + \lambda \boldsymbol{1} || \ldots \text{dead end} $$
I have a suspicion the closing argument ought to be rather simple, but I am not seeing it. Does anyone else?
Proof that for any positive $\epsilon < 1$ and any $y \in Y$, we have $||y - \boldsymbol{1} || \geq \epsilon$:
Suppose not.
So say $||y-\boldsymbol{1}|| < \epsilon $. Then $|x_n - x_{n+1} - 1|_{\mathbb{C}} < 1$ for each $n$.
Writing $x_n = a_n + i b_n$, we have
$$ |a_n - a_{n+1} - 1 | < \epsilon $$
Hence $$1-\epsilon + a_{n+1}< a_n$$
This recurrence relation implies for every $n$ we have $$ n (1- \epsilon) + a_{n+1} < a_1 $$
But $\{a_n\}$ is a bounded sequence of reals, so $-M \leq a_n \leq M$. This implies
$$n (1-\epsilon) < 2M $$
but the left-hand side can be made arbitrarily large provided $n$ is large enough, owing to $\epsilon < 1$. This is our contradiction.
Thus for any $y \in Y$ and any positive $\epsilon < 1$ we have $||y-\boldsymbol{1}||\geq \epsilon$.
