Maximizing a hyperplane $\sum\_i a\_ix\_i$ where $a\_i\in\mathbb R$ and each $a\_i$ are fixed and non-negative and $x\_i$ are variables over a standard simplex $\sum\_i x\_i\leq 1$ with $0\leq x\_i$ always produces a vertex point on the simplex and maximization corresponds to $\max\_i a\_i$.



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> 1. In infinite dimensions is such a proof considered constructive or does it hold only in classical logic? It seems that we would have to show that at maximizing point there is an $i\in\mathbb N$ such that $x\_i=1$ holds and for that perhaps is it possible we cannot do this without invoking [LLPO](https://en.wikipedia.org/wiki/Limited_principle_of_omniscience)?
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Suppose we are looking for a $0/1$ vector in integers (not reals as in 1.) on the standard simplex and we know the optimal vector has either sum of even coordinates summing to $1$ or odd coordinates summing to $1$ then in finite dimensions it is a process of enumerating vertices.



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> 2. In infinite dimensions is such a proof considered constructive or does it hold only in classical logic? It seems that we would have to show that at optimizing point there is an $i\in\mathbb N$ such that $x\_i=1$ holds and for that perhaps is it possible we cannot do this without invoking [LLPO](https://en.wikipedia.org/wiki/Limited_principle_of_omniscience)?
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In general is the proofs of optimization over infinite dimensions considered constructive?


3a. How about when each $a\_i$ are fixed and positive?


3b. How about when each $a\_i$ are fixed and distinct and non-negative thus guaranteeing an unique vertex point?


3c. How about when each $a\_i$ are fixed and distinct and positive thus guaranteeing an unique vertex point?



Constructivity of two problems on a standard simplex?
You make an erroneous assumption in your question, as already in dimension 1 you need LLPO to know that the maximum is actually attained at some point.


We work constructively.


**Theorem:** *LLPO is equivalent to the statement that every affine map $[0,1] \to \mathbb{R}$ attains its maximum*


*Proof.* The general form of an affine map on $[0,1]$ is $f\_{a,b}(x) = a \cdot (1 - x) + b \cdot x$. Suppose then that for every such $f\_{a,b}$ there exists $x\_0 \in [0,1]$ such that $f\_{a,b}(x) \leq f\_{a,b}(x\_0)$ for all $x \in [0,1]$.


Let us first show that LLPO implies attainment of maximum. Given any $f\_{a,b}$, by LLPO either $a \leq b$ or $b \leq a$:


* If $a \leq b$ then the maximum of $f\_{a,b}$ is attained at $x\_0 = 1$.
* If $b \leq a$ then the maximum of $f\_{a,b}$ is attained at $x\_0 = 0$.


The converse is more interesting. First note that the following holds: if $f\_{a,b}(0) \leq f\_{a,b}(t)$ for some $t > 0$ then $f\_{a,b}(0) \leq f\_{a,b}(1)$. Similarly, if $f\_{a,b}(t) \geq f\_{a,b}(1)$ for some $t < 1$ then $f\_{a,b}(0) \geq f\_{a,b}(1)$.


Consider any two reals $a, b \in \mathbb{R}$. We shall decide $a \leq b \lor b \leq a$, which implies LLPO. By assumption, the map $f\_{a,b}$ attains its maximum at some $x\_0 \in [0,1]$. Either $x\_0 < 2/3$ or $x\_0 > 1/3$:


* If $x\_0 > 1/3$ then from $f(0) \leq f(x\_0)$ it follows that $a = f(0) \leq f(1) = b$.
* If $x\_0 < 2/3$ then from $f(x\_0) \geq f(1)$ it follows that $a = f(0) \geq f(1) = b$. $\Box$


Of course, since affine maps are very simple, the maximal *value* of $f\_{a,b}$ exists, but the above argument shows it takes LLPO to know *where* it is attained.


