Fall 2019 - Problem 6

Solution.

1. Let $M = \set{x \in \ell^\infty\p{\N} \mid \lim_{n\to\infty} x_n \text{ exists}}$. It's clear that $0 \in M$, and by linearity of limits (when they exist), we see that $M$ is closed under addition and scalar multiplication, i.e., $M$ is a subspace of $\ell^\infty\p{\N}$. Notice that $\func{\phi}{M}{\R}$ is well-defined, linear, and

   $$\abs{\phi\p{x}} \leq \sup_{n\geq1} \,\abs{x_n} = \norm{x}_\infty,$$

   i.e., $\phi \in M^*$. Hence, by Hahn-Banach, $\phi$ extends to a bounded linear functional on all of $\ell^\infty\p{\N}$, which was what we wanted to show.

2. Let $y = \set{\p{-1}^n}_{n\geq1}$. Observe that we can extend $\phi$ to $M \oplus \span{\set{y}}$ in two different ways: given $x \in M$,

   $$\phi_1\p{x + \lambda y} = \phi\p{x} \quad\text{and}\quad \phi_2\p{x + \lambda y} = \phi\p{x} + \lambda,$$

   i.e., we can extend by sending $y$ to $0$ or by $1$. These are obviously (sequentially) continuous, so $\phi_1, \phi_2$ are distinct bounded linear functionals which both extend $\phi$. By Hahn-Banach, they extend to (distinct) bounded linear functionals on all of $\ell^\infty\p{\N}$, so $\phi$ is not unique.