Check if map maps to Hilbert space I am supposed to find a condition s.th. $A(x)=\sum_{n=1}^\infty\alpha_n\langle x,e_n\rangle e_n$ maps from E to E. Where E is an Hilbert space with a complete ON-sequence $(e_n)^\infty_{n=0}$, $(\alpha_n)_{n=1}^\infty\in\mathbb{C}$.
What are the requirements for an element to belong to a general Hilbert space?
Can a Hilbert space contain an element with infinite norm?
And if the series does not converge, does that mean that it is undefined and hence cant be an element?
Would it be a problem if the series does not converge, and if so why?
 A: As the other comments and answers clearly state, the series only makes sense if the coefficients are square-integrable, but this is only in the framework of Hilbert space. There are more general frameworks, because there are cases in which it is necessary to consider generalized vectors, with infinite norm. The concept of rigged Hilbert spaces has been developed to deal with this kind of objects, especially in quantum mechanics and spectral theory.
The basic example is the plane wave function
$$e_\xi(x)=\exp(ix\xi),\qquad \xi, x\in \mathbb R,$$
in the Hilbert space $L^2(\mathbb R)$. It holds that $\|e_\xi\|_{L^2(\mathbb R)}=\infty$, but $e_\xi$ is a valid element of the rigging $(L^2(\mathbb R), C^\infty_c(\mathbb R))$ ($C^\infty_c$ denotes smooth functions with compact support), because the pairing
$$
\int_{\mathbb R} e_{\xi}(x) \phi(x)\, dx 
$$
makes sense for all $\phi\in C^\infty_c(\mathbb R)$.
The rigging of a Hilbert space also introduces a weaker mode of convergence, so in this sense one may also have convergent orthonormal series and integrals even without square integrable coefficients. In the above example, this corresponds to convergence in the sense of distributions.
A: By a general theorem in Hilbert space, your given E is isometrically isomorphic to $l^2 (\mathbb{C})$, so that the first question: "What are the requirements for an element to belong to a general Hilbert space?" with the obvious answer: "An element belongs to a Hilbert space (or more generally to a Banach space) iff its norm is finite" has the tailored answer provided by @amsmath in the comments section:
[quote] The exact condition is that: 
$$\sum_n |\alpha_n|^2 |\langle x, e_n\rangle|^2 <\infty, \forall x\in E\ (1) $$   [/quote] and applies to the coefficients in the given basis expansion.
As for the other questions: "Can a Hilbert space contain an element with infinite norm?", the answer is "NO"; "And if the series does not converge, does that mean that it is undefined and hence cant be an element?", the answer is: "YES, for a series expansion of a vector with respect to an infinite basis, the convergence of this series in the norm is required by the finite-norm condition of the expanded vector and places boundedness conditions on the coefficients of this expansion, such as (1)".  
