Can one define a functional on a Hilbert space based on its action on a Hilbert basis? I know that the actions of a functional on a vector space can be uniquely described by the value the functional takes on each element of a (Hamel) basis.   My question is,  in a Hilbert space,  would this work as well for a Hilbert basis?  
For example,  if the functional is on $l^2$,   can we define a functional T by how it acts on $x_i=(x_n)$ where $x_n=1$ for $i=n$ and $0$ otherwise?
 A: As KCd said in a comment, a continuous linear functional on a Hilbert space $H$ is uniquely determined by its action on a given orthonormal basis. This follows since, if $M$ is an orthonormal basis of $H$, then for any $x\in H$ we have
$$f(x)=\sum_{m\in M}\langle x,m\rangle f(m)$$
which follows from continuity. If $\dim H=\infty$ and we do not require $f$ to be continuous then the functional is no longer determined along the orthonormal basis. Indeed, if $B$ is a Hamel basis of $H$ such that $M\subset B$, then we may freely choose $f(x)\in\mathbb K$ for $x\in B\setminus M$, so clearly there is no unique choice.
A: Let $ \mathcal{H} $ be a Hilbert space and $ (\mathbf{e}_{i})_{i \in I} $ a Hilbert basis of $ \mathcal{H} $. If $ f $ is a linear mapping from $ \mathcal{H} $ to $ \mathbb{C} $, then $ f $ is continuous if and only if
$$
(\spadesuit) \qquad (f(\mathbf{e}_{i}))_{i \in I} \in {\ell^{2}}(I).
$$
Hence, although a linear functional on $ \mathcal{H} $ is uniquely determined by how it acts on members of a Hilbert basis of $ \mathcal{H} $, not every assignment of scalars to members of the basis will result in a linear functional. The square-summability condition $ (\spadesuit) $ must also be satisfied.
Note: By a Hilbert basis, I mean an orthonormal one.
