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Sep 12, 2022 at 15:39 comment added Yemon Choi @LSpice Banach algebraists often call this "unitization" (it does what it says on the tin)
Sep 12, 2022 at 15:39 comment added Yemon Choi As @QiaochuYuan points out below, this is nothng to do with "completeness" in the sense of metric spaces. The Banach space $L^1({\bf R})$ is complete in its usual norm and is a Banach algebra when equipped with the convolution product. However it does not have an identity element for convolution (this would be the Dirac delta, as you observe)
Aug 29, 2022 at 18:09 history edited YCor
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Aug 29, 2022 at 1:52 answer added Qiaochu Yuan timeline score: 7
Aug 28, 2022 at 23:52 comment added LSpice This is an example (though I'd never thought of it as such, and it's more concrete than the usual abstract such process) of adjoining a unit to an algebra that doesn't have one, but I don't know any other name for that than "adjoining a unit".
Aug 28, 2022 at 23:45 history edited askquestions2 CC BY-SA 4.0
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Aug 28, 2022 at 23:34 history asked askquestions2 CC BY-SA 4.0