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In Nowak (1989), a modular $\rho$ on a vector lattice is defined by the following properties

(N1) $\rho(x)=0\implies x=0$;

(N2) $\lvert x\rvert \le \lvert y\rvert\implies \rho(x) \le \rho(y)$;

(N3) $\rho(x\vee y )\le \rho(x)+\rho(y)$ for all $x\ge 0, y\ge 0$;

(N4) $\rho(\lambda x)\to 0$ if $\lambda \to 0$

This supposedly implies $$ \rho(\alpha x+\beta y)\le \rho( x)+\rho(y) \text{ for all } \alpha,\beta\ge0,\alpha+\beta=1,$$ which features in the definition of modular in Musielak and Orlicz (1959).

I don't see how this follows. NB: convexity is not assumed.

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  • $\begingroup$ Perhaps you need to prove $$\alpha x+\beta y\le x+y \text{ for all } \alpha,\beta\ge0,\alpha+\beta=1,$$ maybe just for positive $x,y$. Is that true in a vector lattice? $\endgroup$ – Gerald Edgar Sep 5 '17 at 12:24
  • $\begingroup$ @GeraldEdgar For positive $x$ and $y$ it is certainly true. But what then? $\endgroup$ – encore Sep 5 '17 at 12:55
  • $\begingroup$ Ok, I can see how this will work for positive and disjoint elements. If $\lvert x\rvert \wedge \lvert y\rvert = 0$ then $\lvert x\rvert \vee \lvert y\rvert = \lvert x\rvert + \lvert y\rvert$ and the result follows from (N3) because $\rho(\alpha x)$ is known to be non-decreasing function of $\alpha$ on $[0,\infty)$ for any fixed $x$. $\endgroup$ – encore Sep 5 '17 at 13:09
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How about this?
Let $\alpha, \beta \ge 0$, $\alpha+\beta = 1$.
Then $$ |\alpha x + \beta y| \le \alpha |x| + \beta |y| \le |x| \vee |y| $$ so $$ \rho(\alpha x + \beta y) = \rho\big(|\alpha x + \beta y|\big) \le \rho\big(|x| \vee |y|\big) \le \rho\big(|x|\big) +\rho\big(|y|\big) = \rho(x)+\rho(y) $$

Note: $\rho(x) = \rho\big(|x|\big)$ from N2.

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  • $\begingroup$ 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. $\endgroup$ – encore Sep 5 '17 at 13:37
  • $\begingroup$ 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. $\endgroup$ – encore Sep 5 '17 at 15:46
  • $\begingroup$ Oh bum, you are right, of course. Thank you! $\endgroup$ – encore Sep 5 '17 at 17:34

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