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?

  • $\begingroup$ The exact condition is that $\sum_n|\alpha_n|^2|\langle x,e_n\rangle|^2 < \infty$ for all $x\in E$. A sufficient (and, I think, also necessary) condition is that $(\alpha_n)_n\in\ell^\infty$. $\endgroup$ – amsmath Oct 16 '17 at 2:24
  • $\begingroup$ Thanks, yes saw a theorem that states that $\sum_{n=1}^\infty\alpha_ne_n<\infty$, $(e_n)$ ON-sequence, if and only if $\sum_{n=1}^\infty|\alpha_n|^2<\infty$. But why does boundedness of the sum ensure that it is in E? Can't E contain unbounded elements? $\endgroup$ – JoeDoe Oct 16 '17 at 2:36
  • $\begingroup$ Is it becuase E is a vector space? And addition and multiplication becomes vaguely defined for infinite elements? But can't there be infinities of different orders, making the operations defined? $\endgroup$ – JoeDoe Oct 16 '17 at 2:39
  • $\begingroup$ How does infinite elements interfere with the definiton of hilbert spaces? $\endgroup$ – JoeDoe Oct 16 '17 at 2:40
  • $\begingroup$ If $E$ is a Hilbert space, then for any $f\in E$ you have $\|f\| < \infty$. That follows immediately from the definition of a norm. $\endgroup$ – amsmath Oct 16 '17 at 2:41

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)".

  • $\begingroup$ Thank you, really cleared stuff up! $\endgroup$ – JoeDoe Oct 16 '17 at 9:50

As the other comments and answers clearly stat, the series only makes sense in the framework of Hilbert space if the coefficients are square-integrable. However, 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.

  • $\begingroup$ Thank you for the help! $\endgroup$ – JoeDoe Oct 16 '17 at 9:52

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