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Feb 7, 2020 at 18:05 history edited YCor CC BY-SA 4.0
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Mar 12, 2019 at 22:14 answer added Jochen Glueck timeline score: 1
Mar 7, 2019 at 10:47 comment added Jochen Glueck @AndrásBatkai: This was my first thought, too; but after all, I think it's a bit more subtle: the resolvent set of $A^n$ can indeed by empty (for instance, if the spectrum of $A$ is the left halfplane). So with this argument you can only show that $(A-\lambda)^n$ is closed for some scalar $\lambda$ - but then it is not obvious why this should imply that $A^n$ is closed.
Mar 7, 2019 at 8:39 comment added András Bátkai This is more an exercise and there are many arguments. For example you could show that the resolvent set of $A^n$ is not empty.
Mar 7, 2019 at 8:12 history edited Ma Pa CC BY-SA 4.0
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Mar 7, 2019 at 7:12 comment added Jochen Glueck Cross-posted from math.stackexchange.com.
Mar 7, 2019 at 5:23 history edited Ma Pa CC BY-SA 4.0
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Mar 7, 2019 at 5:11 history asked Ma Pa CC BY-SA 4.0