Prove there are infinitely many extensions of linear functional on $l_1$ Let $X = l_1$ and $S = Span(e_1)$ and define $f$ on $S$ by $f(\alpha e_1) = \alpha$ where $\alpha\in \mathbb{C}.$ Then $f$ is a bounded linear functional on $S$ with $\|f\| = 1.$ Prove that $f$ has infinitely many extensions $g\in (l_1)^{*}$ with $\|g\| = 1.$
If $X = l_2$ and $S, f$ as above. $f$ is again a bounded linear functional on $S$ with $\|f\| = 1$. Prove that $f$ has only one extension $g\in (l_2)^{*}$ with $\|g\| = 1.$
(The lecture was on Hahn-Banach theorem.)
 A: Consider the linear hull $$S:= \operatorname{Span}e_1$$ of the sequence $e_1 \in l^1$, i.e. $$e_1(1) = 1 \text{ and } e_1(n) = 0 \text{ for all } n > 1.\tag{$\ast$}$$
Let $f$ be defined as $f(\alpha e_1) = \alpha$ and let $g$ denote its continuous extension to $l_1$, i.e. $g \in (l^1)^* \simeq l^\infty$. 
The duality gives you some $u \in l^\infty$ such that for all $x \in l^1$ you have 
$$g(x) = \sum_{n = 1}^\infty u(n)\, x(n).\tag{$\ast\ast$}$$
But since $g$ extends $f$, we necessarily have 
$$1 = f(e_1) = g(e_1) \overset{(\ast\ast)}= \sum_{n = 1}^\infty u(n) \,e_1(n) \overset{(\ast)}{=} u(1)\, e_1(1) = u(1).$$
Furthermore, $g$ has norm $1$ and so does $u$. Thus, as long as $u$ is bounded with norm $1$ and fulfills $u(1) = 1$ you have a corresponding functional $g$ in $(l^1)^*$.
On the other side, let $f$ now be defined on the subspace $S$ in $l^2$. Then, $$g(x) = \langle x, e_1\rangle$$ is a valid extension of $f$ with norm $1$. We now want to show it's the only one. Let $h$ be another extension of $f$ with norm $1$. By the duality, there exists some $y \in l^2$ with norm $1$ such that $$h(x) = \langle x, y\rangle$$ for all $x \in l^2$. But $h \neq g$ so there has to exists some $e_k$, $k > 1$ such that 
$$h(e_k) \neq 0 = g(e_k).$$ 
But then, we have $$\|y\|^2 = \sum_{n = 1}^\infty |\langle e_n, y  \rangle|^2 \geq |\langle e_1, y \rangle|^2 + |\langle e_k, y \rangle|^2 = |h(e_1)|^2 + |h(e_k)|^2 > 1,$$ a contradiction.
