$(e_n)_{n=1}^{\infty}$ orthonormal in a Hilbert space H, $\overline{\text{Span}(E)}\neq H$, why does there exist an $0\neq f\in H$, $f\in E^{\perp}$ If $E=(e_n)_{n=1}^{\infty}$ is an orthonormal sequence in a Hilbert space H such that $\overline{\text{Span}(E)}\neq H$, why does it follow that there exists an $0\neq f\in H$ with $f\in E^{\perp}$. This is part of a larger question on finding the eigenvalues of the compact self-adjoint operator $$Ax=\sum_{n=1}^{\infty}\lambda_n\langle x,e_n\rangle e_n$$ so the spectral theorem says that each $\lambda_i$ is an eigenvalue, but in the solutions to the problem, they claim the existence of the $f$ above as an eigenvector to the eigenvalue $0$. Is it obvious that it exists? What am I missing?
 A: Take a vector $\xi\in H$ that does not belong to $\overline{\text{span}(E)}$. Set $\eta=\xi-\sum_{e\in E}\langle\xi,e\rangle e$. Note that $\eta\neq0$, otherwise $\xi\in\overline{\text{span}(E)}$. Also note that if $e_0\in E$, then
$$\langle\eta,e_0\rangle=\langle\xi,e_0\rangle-\sum_{e\in E}\langle\xi,e\rangle\cdot\langle e,e_0\rangle=\langle\xi,e_0\rangle-\langle\xi,e_0\rangle=0.$$
Since $e_0\in E$ is arbitrary, this shows that $\eta\bot e$ for all $e\in E$.
Comment: This is trivial if one knows that a Hilbert space is decomposed as $H=K\oplus K^\bot$, for any Hilbert subspace $K$. If $K\neq H$, then $K^\bot$ is non-zero so just pick any vector of $K^\bot$. Actually, the above answer is pretty much this argument with a few more details: we take a vector $\xi$ that does not belong to $K:=\overline{\text{span}(E)}$, so if we decompose$\xi=\xi_K+\xi_{K^\bot}$, then $\xi_{K^\bot}$ is non-zero. The vector $\eta$ is precisely this $\xi_{K^\bot}$ vector, i.e. the orthogonal projection of $\xi$ onto $K^\bot$.
