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(Note : I'm not sure about the tags, please re-tag this if you think you have the right tag).

I am optimising a certain function over a certain space (that i will describe), and to use non-constraint optimisations routines, i wanted to know if there is a mapping from $\mathbb{R}^{p}$, where $p$ is a constant to be determined, to the following space $S$:

Define first the index set $$\mathbb{N}_{n}^{d} = \left\{\mathbf k \in \mathbb{N}^d\, : \; k_i \le n \;\forall i \in 1,...,d\right\}$$

Define now the space $S$ as being the following restriction of $\mathbb{R}_{+}^{\mathbb{N}_{n}^{d}}$ :

$$y \in S \iff y \in \mathbb{R}_{+}^{\mathbb{N}_{n}^{d}} \text{ and }\forall i \in 1,...,n,\, \forall j \in 1,...,d,\, \sum_{\mathbf k \in \mathbb{N}_{n}^{d}} y_{\mathbf k} \mathbf 1_{k_j = i} =\text c_{i,j}$$

For some constants $\left(c_{i,j}\right)_{i \in 1,...,n,\, j \in 1,...,d}$ such that $\sum\limits_{i=1}^{n} c_{i,j} = C$ for all $j \in 1,...,d$

Note that this implies that $\sum\limits_{\mathbf k \in \mathbb{N}_{n}^{d}} y_{\mathbf k} = C$.

The insights is that, if $C = 1$, y are probabilities of some multivariate discrete distributions with marginal probabilities for the $j^{\text{th}}$ marginal $\left(c_{i,j}\right)_{i}$. $C$ is not one, so this measure is not a probability, but you could set $C = 1$ without any problem.

So the question is : Is there some way to map some power of $\mathbb{R}$ to this space $S$, that is surjective ? I dont care about injectivity by the way.

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    $\begingroup$ What type of mapping do you want? Otherwise $p = 1$ does the job ($S$ and $\mathbb{R}$ have the same cardinality. $\endgroup$ Commented Nov 19, 2020 at 17:03
  • $\begingroup$ Probably p=1 does the job, but i would prefer something like $p = n^d - nd - 1$ which is the number of degree of freedom in my equations. If you have a surjective mapping for p=1, please tel me anyway. $\endgroup$
    – lrnv
    Commented Nov 19, 2020 at 19:07
  • $\begingroup$ I think @DieterKadelka's point is that $p = 1$ not just probably works, but proveably works (it's just testing cardinalities), if you just ask for a surjective mapping of sets. $\endgroup$
    – LSpice
    Commented Nov 19, 2020 at 20:58
  • $\begingroup$ It's not clear what the quantifiers are in the definition of $S$, but I guess you first pick $C$ and then define $S_C$. Then $S_C$ is compact and convex, and you can just project from the ambient Euclidean space on it; this gives a continuous map (from a higher-dimensional space). $\endgroup$
    – LSpice
    Commented Nov 19, 2020 at 20:59
  • $\begingroup$ @LSpice Yeah a simple projection on the space works, but, and this is my mistake for not specifying it before, i'd like something that is smooth, and not only continuous. Like if you optimize on the positive reals, then optimizing on the log makes your problem unconstraint. $\endgroup$
    – lrnv
    Commented Nov 23, 2020 at 11:33

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