No there aren't any simple, or even any constructive, examples of everywhere defined unbounded operators. The only way to obtain such a thing is to use the Hahn-Banach Theorem to extend a densely defined unbounded operator. Densely defined unbounded operators are easy to find.

No there aren't any simple, or even any constructive, examples of everywhere defined unbounded operators. The only way to obtain such a thing is to use Zorn's Lemma to extend a densely defined unbounded operator. Densely defined unbounded operators are easy to find.

Zorn's lemma is applied as follows. Let $A$ be an operator on a domain $\mathcal D$. Consider the set $E$ of all extensions of $A$, that is the collection of operators $A'$ on domains $\mathcal D' \supset \mathcal D$ that agree with $A$ when restricted to $\mathcal D$. Then $E$ is partially ordered by inclusion on domains. Furthermore, any linear chain has an upper bound, by taking unions of domains. So there is a maximal element by Zorn. Finally, suppose the maximal element $A$ is defined on a domain $\mathcal D'$ that is not all of $\ell^2$. Let $v$ be any vector in the complement of $\mathcal D'$. Define an extension of $A'$ on $\mathcal D'+\{a v\}$ by, say, mapping $v$ to zero. This contradicts maximality, so any maximal element is globally defined.