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A sequence is called lacunary if, in your terminology, its minimum growth rate is strictly greater than $1$. The following articles prove that every lacunary sequence is remote. If I understand your question correctly, this means that the (only) maximal $M$ you seek is the interval $(1,\infty)$. Pollington, Andrew D., On the density of sequences $\{n_k\xi\}... 3 The definition which you wrote can be translated in words like this. You evaluate the solution at time$t$. Then you fix a small distance$\epsilon>0$. Then you can find a time length$\ell(\epsilon)$(independent of$t$) having the following property: within$\ell$time units from whatever instant$a$, you will find an instant$\tau$such that the ... 2 Let$\mu$be Lebesgue measure on$S^1$, and$\delta_P$be a point-mass at a point$P \in S^1$. Then there is no flow on$S^1$whose time averages lead to$\frac{1}{2}(\mu + \delta_P)$. (Consider the orbit of$P$.) This distribution seems like it should count as "well-behaved." Its support is connected, and both$\mu$and$\delta_P$themselves arise ... 1 I suspect that your confusion stems from misinterpreting the norm on$L^2(X|Y)$in your definition 2. You should note that$\mathbb{E}_Y$takes$L^\infty(X)$to$L^\infty(Y)$and the new norm taken on$L^\infty(X)$is with respect to the$L^\infty$-norm on$L^\infty(Y)$, not the$L^2$-norm. That is$\|f\|:=\|\mathbb{E}_Y(f\cdot\overline{f})\|_\infty^{1/2}\$ ...