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Oct 23 at 1:25 comment added Iosif Pinelis @Kaira : Yes, the result holds for $p=\infty$ as well. I have added a remark on this.
Oct 23 at 1:24 history edited Iosif Pinelis CC BY-SA 4.0
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Oct 23 at 1:12 comment added Kaira If we wanted to treat the $p=\infty$ case, I guess using Markov and using the monotonicity of $L^p$ norm to replace $L^1$ norm with $L^{\infty}$ works, right?
Apr 29 at 18:34 vote accept Kaira
Apr 29 at 17:51 history answered Iosif Pinelis CC BY-SA 4.0