Hahn-Banach Theorem and non bounded linear functionals. I have read the Hahn-Banach theorem, which states de following:
Let $X$  be  a  real  vector space  and $p$ a  sublinear functional on $X$.  Furthermore,
let  $f$  be  a  linear  functional  which  is  defined  on a  subspace  $Z$  of $X$  and
satisfies
$$f(z) \leq p(z), \forall z \in Z.$$
Then, $f$ has a linear extension $\tilde{f}$ from $Z$ to $X$ and it is dominated by $p$ on $X$.
Thus, nothing is said about the boundedness of $\tilde{f}$. I am looking for an explicit example where $f$ is a linear functional that satisfies the hypotesis and, however, $\tilde{f}$ is not bounded.
I was trying with some finite normed spaces but I could not find nothing.
 A: I have solved this problem and I am going to post my answer.
First of all, it's clear that you can't find a non bounded linear functional in a finite-dimensional normed space. Then, I chose $X = \mathcal{C}[a,b]$ as $\mathbb{R}$-space. Thus, we know that every infinite-dimensional space has a non bounded linear functional (it is true because every vector space has a Hamel basis). Let's call this real functional $p$. Then, it's easy to show that $p$ is a sublinear functional on $X$. On the other hand, I will consider $Z = \{0\}$, subspace of $X$ and define $f: Z \to \mathbb{R}$ given by
$$f(0) = 0.$$
Of course, $f$ is a linear functional on $Z$ and, moreover, $f(0) = 0 = p(0)$. For this reason, we are able to apply the Hahn-Banach theorem, which states that $f$ has a linear extension $\tilde{f}$ from $Z$ to $X$ and
$$\tilde{f}(x) \leq p(x), \forall x \in X = \mathcal{C}[a,b].$$
Even more, we are going to prove that $\tilde{f} = p$. Suppose that $\tilde{f} \neq p$, then exists $x_0 \in X$ such that $\tilde{f}(x_0) < p(x_0)$. But, it follows that:
$$\tilde{f}(x_0) < p(x_0) \Longleftrightarrow -\tilde{f}(x_0) > -p(x_0) \Longleftrightarrow \tilde{f}(-x_0) > p(-x_0),$$
because $\tilde{f}$ and $p$ are linear. Contradiction. Then $\tilde{f} = p$ and we have an example where the linear extension $\tilde{f}$ is not bounded.
