• $H$ be a $\mathbb R$-Hilbert space
  • $A$ be a compact and self-adjoint bounded linear operator on $H$
  • $I:=\left\{n\in\mathbb N:n\le\operatorname{rank}A\right\}$

By the Hilbert-Schmidt theorem, there is a $(\lambda_i)_{i\in I}\subseteq\mathbb R\setminus\left\{0\right\}$ with $$Ae_i=\lambda_ie_i\;\;\;\text{for all }i\in I$$ for some "orthonormal basis $(e_i)_{i\in I}$ of $AH$".

Unless $AH$ is closed (and hence $AH$ equipped with the inner product inherited from $H$ is again a $\mathbb R$-Hilbert space), I don't understand the notion of an "orthonormal basis"$^1$ of $AH$. Moreover, unless $H$ is finite-dimensional, the assumptions on $H$ don't ensure the closedness of $AH$. So, what am I missing?

$^1$ By the notion that I know, an orthonormal set $S\subseteq U$ of a $\mathbb R$-Hilbert space $U$ is called an orthonormal basis of $U$ iff $U=\overline{\operatorname{lin}}S$. Since for any $M\subseteq U$ the set $\overline{\operatorname{lin}}M$ is closed by definition, I don't a natural extension to this notion for subspaces that aren't closed.


The range of a compact operator $A$ cannot be closed if it is infinite-dimensional.

As you observed the vectors $(e_i)$ are a basis of $\overline{R(A)}$. That is, every element in $R(A)$ can be written as an infinite sum of the $e_i$'s. However, not every element in the closed linear hull of the $(e_i)$ is in $R(A)$.

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  • $\begingroup$ I thought there would be some meaningful distinction and $(e_i)_{i\in I}$ would not be an orthonormal basis of $\overline{AH}$. But if I got you right, $(e_i)_{i\in I}$ is an orthonormal basis of $\overline{AH}$. That trivially means that each element of $AH$ can be written as an infinite sum of the $e_i$'s. $\endgroup$ – 0xbadf00d May 29 '17 at 19:17

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