Suppose $T$ is a compact, self-adjoint operator on a Hilbert space $\mathcal{H}$. Then there exists an orthonormal set $\{e_n\}_{n=1}^{\infty}$ of eigenvalues of $T$ such that every $x \in \mathcal{H}$ has a unique decomposition of the form $$ x = \sum_{n=1}^{\infty} c_n e_n + y, $$ where $c_n \in \mathbb{C}$ and $y \in \ker(T)$.

The proof I know of the above theorem boils down to showing that $\mathcal{H} = L \oplus \ker(T)$, where $$L = \overline{\text{span}\{e_n\}_{n=1}^{\infty}}.$$

I'm confused about the structure of $L$. How do we know that an element of $L$ (say $z$) looks like $$z = \sum_{n=1}^{\infty} c_n e_n?$$ I thought that, in general, the closure of the span of an infinite set did not coincide with the set of all infinite sums of elements from that set?

  • 1
    $\begingroup$ I think the set of infinite sums is identical with the span of an orthonormal set $\endgroup$ – Peter Melech Apr 25 '16 at 12:14
  • 1
    $\begingroup$ Sorry. I mean the closure of the span. Orthogonality is crucial for this. $\endgroup$ – Peter Melech Apr 25 '16 at 12:20
  • $\begingroup$ @PeterMelech That sounds useful - do you have any idea what the sketch of the proof would be? $\endgroup$ – MadMonty Apr 25 '16 at 12:57
  • $\begingroup$ May be something like when $\{e_i\}$ ON then $f\in \overline{span\{e_i\}}$ implies $f=\sum_{i=1}^{\infty}(f,e_i)e_i$ just as a vague idea $\endgroup$ – Peter Melech Apr 27 '16 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.