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Sep 5, 2017 at 17:36 history edited Gerald Edgar CC BY-SA 3.0
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Sep 5, 2017 at 17:34 vote accept CommunityBot
Sep 5, 2017 at 17:34 comment added user113782 Oh bum, you are right, of course. Thank you!
Sep 5, 2017 at 15:46 comment added user113782 I've had a mistake in (N2) and (N3), now they are fixed. From (N2) $\rho(x)=\rho(\lvert x\rvert)$ follows. But I think your first line cannot hold as stated, just take $x=y=1/2$ to falsify it.
Sep 5, 2017 at 13:37 comment added user113782 You must be pretty close :-). In Musielak-Orlicz definition there is symmetry: $\rho(x)=\rho(-x)$. However, I do not think symmetry implies equality with $\rho(\lvert x \rvert)$ unless the modular is so-called orthogonally additive, i.e. unless one has $\rho(x+y)=\rho(x)+\rho(y)$ when $x,y$ are disjoint.
Sep 5, 2017 at 13:29 history edited Gerald Edgar CC BY-SA 3.0
added 101 characters in body
Sep 5, 2017 at 13:11 history answered Gerald Edgar CC BY-SA 3.0