Same as you did. $F*(\Vec_\Pt)=\lim_{x\to\infty}\frac{d}{dt}F(\Pt+t\Vec)$ — DOTDO Oct 10 '16 at 1:42

Any ideas how I could go about finding $\range(T_d)$? — Ben Crossley Oct 27 '16 at 8:20

@BenCrossley-hobbyist To be honest I don't quite really understand the definition of the function very well or its motivation. If nothing else, one could say that: $$\range(T_d) = \left\{ n\in \mathbb{N}: n= \frac{2^{x_1+\dots+x_d}}{3^d}-\frac{1}{3^d}\sum_{a=1}^{d-1} 3^{d-a-1}\times2^{x_0+\dots+x_a} \ \text{for some }(x_1, \dots, x_d)\in \mathbb{N}^d \right\}$$ but my guess is that this obviously isn't very helpful for you. I guess the only comment I can think of is that a value for the left term determines all possible values for the right term, so first determine all possible values for — Chill2Macht Oct 27 '16 at 10:04

Note that we usually denote $I= [0,a_1]\times [0,\infty)\times [0,\infty)\times [0,a_2)$ also the word "the" in the sentence "Now, the function $f$ is $f:I\rightarrow O$" makes no sense because there are infinitely many functions with such domain\range. — Yanko Mar 16 at 20:33

Dear Camilo, There is certainly such a $y$, e.g. $y = 0$, so there is no problem applying your Zorn's lemma argument to get a maximal $N$. The hard part will be to show that $\langle x \range \oplus N$ is actually equal to $M$; this is where you have to use the assumption on the $a_i$. If you have done this (as you say), then you are in good shape! (And there are other ways to prove this result, but your approach is as good as any.) Regards, — Matt E May 14 '13 at 5:14

Alternatively, you can check that $T$ takes a linearly independent set for $\ker (T)^\perp$ to a linearly independent set in $\range (T)$ — TYS Jun 26 '15 at 1:11