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Timeline for Question about Dirac operator

Current License: CC BY-SA 4.0

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Apr 11, 2022 at 9:00 comment added Bernd Ammann With the original $\leq$ and $\geq$ the statement is certanly false. Take as counterexample the disjoint union of countably many S^2 of different radius. It is also not really a question about Dirac operators, but about operators in general. Thus the question is not really well-posed.
Apr 10, 2022 at 15:41 comment added user473423 How can c be the minimum of the $\lambda$ which satisfy $|\lambda|\le c$?
Apr 10, 2022 at 15:31 comment added Radeha Longa These c's are the same one. And I was confused about if the minimum of $\lambda_i$s' is a constant, not a function. @Echo
Apr 10, 2022 at 15:01 comment added user473423 There seem to be two different c's here.
Apr 10, 2022 at 14:21 comment added Radeha Longa I'm sure that the inequality is $\geq$. And I guess we can take the $c=\min\{\lambda_i\}$ to make our claim hold. Does it right?@Echo
Apr 10, 2022 at 14:05 comment added user473423 Oh sorry, I read the inequality the other way round. But see, only this way your claim is true, otherwise it is false. Or did you mean $\le$ in your claim?
Apr 10, 2022 at 11:55 comment added Radeha Longa Excuse me, I checked your calculation and I think it should be $|\lambda_i|^2\leq c^2$ for $\sigma\in H_+$. @Echo
Apr 10, 2022 at 11:08 comment added user473423 If $\sigma=\sum_i\mu_if_i$ with $Df_i=\lambda_if_i$, then $\parallel D\sigma\parallel^2=\sum_i|\mu_i|^2|\lambda_i|^2\ge\sum_ic^2|\mu_i|^2=c^2\parallel \sigma\parallel^2$
Apr 10, 2022 at 7:15 comment added Radeha Longa Could you please write down your idea with some details? I can't follow you. Thanks a lot.@Echo
Apr 10, 2022 at 7:13 comment added user473423 Express $\sigma$ in an orthonormal basis consisting of eigenvectors.
Apr 10, 2022 at 4:10 history asked Radeha Longa CC BY-SA 4.0