Htykuut

Suppose $X = ell_{infty}$ and $Y = c_0$, considered as a subspace of $X$. How can I show that if $y^{ast}in Y^{ast}$ and $ | y^{ast}| = 1$, then the norm preserving Hahn Banach extension is unique?

It is clear that $y^*$ can be identified with $y in ell^1$ and this generates the functional
$g : x mapsto sum_{i = 1}^infty x_i y_i$
for $x in ell^infty$.

Now, suppose that $f in X^*$ is a different extension with the same norm. Then, $f - g$ vanishes on $c_0$. However, there must be $hat x in ell^infty$ with
$$(f-g)(hat x) = 1.$$

Even more, let $T_n$ be the operator that sets the first elements of a sequence to zero. Then,
$$(f-g)(T_n hat x) = (f-g)(hat x) = 1.$$
However, $g(T_n hat x) to 0$ and, thus, $f(T_n hat x) to 1$.
Finally, we consider for $k in mathbb N$ the sequence $z_k$ with entries
$$
(z_k)_n
=
begin{cases}
K,operatorname{sign}(y_n)& text{for } n < k, \
hat x_n&text{else},
end{cases}
$$
Where $K = |hat x|_{ell^infty}$.
Then,
$$f(z_k) to K , | y |_{ell^1} + 1.$$
However, $|z_k|_{ell^infty} le K$. This contradicts that $|f|_{X^star} = |y|_{ell^1}$.