Let $l^2$ be a Hilbert space of infinite sequences $(z_0, z_1, \cdots)$ with finite $\sum_{i=0}^{\infty} z_i^2$.
Are there any simple example of unbounded linear opearator $T: l^2 \to l^2$ with $D(T)=l^2$?
Let $l^2$ be a Hilbert space of infinite sequences $(z_0, z_1, \cdots)$ with finite $\sum_{i=0}^{\infty} z_i^2$. Are there any simple example of unbounded linear opearator $T: l^2 \to l^2$ with $D(T)=l^2$? 


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. 

