Talk:Spectrum of Self-Adjoint Bounded Linear Operator is Real and Closed/Proof 2

This seems insufficient. Basically you've shown that eigenvalues of $\R$ are real. I am not sure how to then get that $A - \lambda I$ is surjective. That it's injective and has bounded inverse (defined on the image of $A - \lambda I$) is obvious, but I am not sure how useful this inequality is to get surjectivity or denseness of the image. (which would imply surjectivity) Caliburn (talk) 21:38, 25 February 2023 (UTC)


 * I added a bit more details. Yes, this was an absolutely non-trivial point but it tuns out that the inequality is useful for that. Sorry that I forgot to complete this proof. I will do it soon. --Usagiop (talk) 23:54, 25 February 2023 (UTC)