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Aug 11, 2017 at 12:03 comment added Piero D'Ancona The same argument works near any isolated point of the spectrum of an operator
Aug 11, 2017 at 11:01 history edited CooLee CC BY-SA 3.0
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Aug 11, 2017 at 11:00 comment added CooLee Here I assume $f$ is a fixed $L^2(\Omega)$-function.
Jul 21, 2017 at 2:26 history edited CooLee
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Jul 21, 2017 at 1:46 comment added CooLee Can the same argument work for the operator $-\Delta + \vec B\cdot \nabla$?
Jul 20, 2017 at 20:24 comment added Christian Remling Since $u=(-\Delta-\lambda)^{-1}f$, the bound with a uniform $C$ and for all $f\in L^2$ would give the same uniform bound on $\|(-\Delta-\lambda)^{-1}\|$, which is impossible because the operator norm of the resolvent diverges when $\lambda$ approaches the spectrum (easy to show, from the spectral theorem). (The bound with a uniform $C$ for only specific $f$'s might conceivably hold.)
Jul 20, 2017 at 15:32 comment added Andrew Wouldn't it lead to an a priory estimate in $W_2^2$ from which by the method of continuity would follow that the problem is solvable for $\lambda=\lambda_1$ for any $f\in L_2$?
Jul 20, 2017 at 11:28 history edited CooLee
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Jul 20, 2017 at 11:20 history edited CooLee CC BY-SA 3.0
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Jul 20, 2017 at 9:32 history edited CooLee CC BY-SA 3.0
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Jul 18, 2017 at 8:22 history edited CooLee CC BY-SA 3.0
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Jul 17, 2017 at 7:03 history edited CooLee CC BY-SA 3.0
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Jul 17, 2017 at 6:58 history edited CooLee CC BY-SA 3.0
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Jul 16, 2017 at 12:46 review Close votes
Jul 16, 2017 at 15:47
Jul 16, 2017 at 10:07 history edited CooLee CC BY-SA 3.0
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Jul 16, 2017 at 8:57 history edited CooLee CC BY-SA 3.0
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Jul 16, 2017 at 8:46 history asked CooLee CC BY-SA 3.0