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Jun 9, 2023 at 21:15 vote accept Curious
Jun 5, 2023 at 17:07 comment added Christian Remling Maybe a slightly better way of saying the same thing is as follows: Lemma: Let $H$ be a Hilbert space with ONB $\{e_n\}$. Then there is a dense subspace $D\subseteq H$ with $e_n\notin D$ for all $n$. Proof: Define $D$ as above. $\square$ To answer the actual question, apply this to the Hilbert space $(D(A),\|\cdot\|_A)$, with $e_n$ chosen as the eigenvectors of $A$.
Jun 3, 2023 at 20:12 history answered Christian Remling CC BY-SA 4.0