Show that $||f*\mu||_1=||f||_1$ for all $f \in L^{1}(\mathbb R)$ where $\mu$ is a complex Borel measure on $\mathbb R$ implies $\mu$ is degenerate. This is closely related to a recent post where it was shown that $\mu$ cannot be absolutely continuous. In his answer David Ullrich used continuity of translates in $L^{1}$. Since translation of a measure is continuous for weak convergence but not for total variation convergence a new proof is required. Thanks for any help.
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Previous post: https://math.stackexchange.com/questions/2634026 – Kavi Rama Murthy Feb 07 '18 at 05:09
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what does 'degenerate' mean – mathworker21 Oct 25 '19 at 05:31
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Measures of the type $\delta_x$ defined by $\delta_x(E)=1$ if $x \in E$ and $0$ otherwise are called degenerate measures. @mathworker21 – Kavi Rama Murthy Oct 25 '19 at 05:33
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May I help you !!! cause I have seen this problem being discussed in an online maths discussion meet – JAO FELIX Jun 06 '20 at 10:45