Abount linear functional: If $T(B)$ is bounded, is $T$ bounded? Let $H$ be a infinite dimensional Hilbert space and let $B$ be a basis of $H$ that $H=\overline{span(B)}$
moreover, let $T : H \rightarrow K$ be a linear functional 
If $T(B)$ is bounded, is $T$ bounded?
 A: Let $B = (e_i)$ an (infinite) orthonormal basis of $H$ and $T$ defined by $T(e_1) = 1$ and $T(e_i) = 0$ if $ i \neq 1$.
Then $T(B)$ is bounded but $|T(n e_1)| = n |T(e_1)| = n$ so $T$ is not bounded. 
A: With $T(B)$ being bounded, do you mean it is bounded as a subset of $\mathbb K$ or that $T$ is bounded as a linear operator on the span of $B$? Also with $B$ you do not mean a Hamel basis, but rather a set of linearly independent vectors so that $\overline{\text{span}(B)}=H$.
In the first case, consider $B= \{e_i \mid i \in \mathbb N\}$ in $\mathscr l^2 (\mathbb N)$. Consider a linear map $ T: \text{span}(B) \to \mathbb K$ via $T(\sum_i a_i e_i)=\sum_i a_i$. Since the sum must always be finite (we are in the span), this is a well defined linear map. If you believe the axiom of choice then this map can be extended from $\text{span}(B)$ onto $H$ (see here). Finally, $T(B)=\{1\}$, as $T(e_i)=1$ for all $i$.
But $T$ is not bounded on $\text{span}(B)$, as $T(\sum_{i=1}^N \frac{1}{\sqrt N} e_i)=\frac{N}{\sqrt N}=\sqrt{N}$, whereas $\|\sum_i^N \frac1{\sqrt N} e_i\| = \sqrt{\sum_i^N \frac1{N}}=1$. So any extension of $T$ onto $H$ will not be bounded.
In the case you mean that $T$ is bounded as a linear map on $\text{span}(B)$, then it is extendable onto the completion via a bounded linear map. This is so, since $\text{span}(B)$ is dense in $\overline{\text{span}}(B) = H$, and linear maps are bounded iff they are continuous. Continuous maps on subsets of metric spaces are always (uniquely) extendable to a continuous map on the completion of the the subset.
Thus there exists a unique bounded extension of $T$ onto $H$. See here and here for more information. But this extension does not need to be the original map you had on $H$. There is more than one extension unless $\text{span}(B)=H$ already without taking the completion. But since there is only one continuous extension, every other extension must be discontinuous, ie unbounded.
