Extending a linear operator satisfying an order condition

Question

Let $\ell^\infty$ be the usual space of bounded sequences, and consider the subspace $V_1 ⊂ \ell^\infty$ consisting of vectors with finite $1$-norm. (That is, $V_1$ contains those $x ∈ \ell^\infty$ such that $\sum_i |x_i|$ is finite. I'm not calling this space $\ell^1$ because it inherits the $\infty$-norm instead. I don't know what terminology would be standard.) Let $\ell^\infty$ be ordered such that $x \ge y$ iff $x_i \ge y_i$ for each $i$. Let $$ P = \{ x ∈ \ell^\infty ∣ x ≥ y \text{ for some } y ∈ V_1 \} $$

Does there exist a linear operator $f : \ell^\infty \to V_1$ that extends the identity map on $V_1$ with the property that $x \ge f(x)$ for all $x ∈ P$?

This seems like it should follow from some version of the Hahn-Banach theorem, maybe?

Answer 1

No, this is not possible. For, if it were, then $P$ would be a continuous linear projection from $l^{\infty}$ to $V_1$, and so the latter must be complemented in $l^{\infty}$. Since $l^{\infty}$ is prime, we must then have $V_1\cong l^{\infty}$. But the latter is false; $V_1$ is (unlike $l^{\infty}$) separable.