Is the solution of a convex combination of the objective in simple problems a convex combination of the solutions of the same problems?

Question

Let $\mathbf{A}=\left(a_{ij}\right)$ be a $n\times J$ matrix with $a_{ij}\geq 0$, $n>J$ and such that no row has all its entries equal to zero, and each column has at most one zero. Let also $\mathbf{q}=\left(q_i\right)$ a $n\times 1$ vector of variables. Abusing notation, I'll write $\mathbf{q}^{\beta}=\left(q_i^{\beta}\right)$ for some $\beta>1$. Finally, let $\mathbf{w}=\left(w_j\right)_{1\leq j\leq J}$ with $w_{j}\geq 0$ and $\sum_{j=1}^{J}w_{j}=1$.

Consider the problem

$$\begin{align} \min\limits_{\{q_i\}}& \quad \left(\mathbf{A}\mathbf{w}\right)^{\top}\mathbf{q}^{\beta}\\ \text{s.t.}&\quad \quad \begin{cases} \sum_{i=1}^n q_i=1\\ \mathbf{q}\geq 0. \end{cases} \end{align}$$

and, in particular, the specific problems that result from choosing $\mathbf{w}=\mathbf{e}_k=\left(e_j\right)_{1\leq j\leq J}$ with $e_{k}=1$ and $e_{j}=0\;\forall j\neq k$ for $k=\{1,\dots,J\}$. Observe that the objetive function in the general problem above is a convex combination of the objective in the $J$ specific problems. Let $\mathbf{q}^k$ denote the minimand for each $\mathbf{e}_k$ and call $\mathbf{z}^k=\mathbf{A}^{\top}\left(\mathbf{q}^k\right)^{\beta}$.

If we call $\mathbf{\hat q}$ the solution to the general problem for $\mathbf{w}\neq\mathbf{e}_k\;\forall k$, I believe that there exists $\mathbf{w}^{\prime}$ (with $w_j^\prime\geq 0$ and $\sum w_j^\prime>0$) such that $\mathbf{A}^{\top}\mathbf{\hat q}^{\beta}=\sum_{k=1}^J\mathbf{z}^kw_k^\prime$. In other words, the solution to the general problem is a convex combination of the solutions to the specific problems. Is this correct? If it is, can you provide a reference for the proof?

Edit:

Following @mtanneau's suggestion, I've obtained the closed form solution for the optimal $\mathbf{q}\left(\mathbf{w}\right)=\left(q_i\left(\mathbf{w}\right)\right)$ given $\mathbf{w}$. To simplify notation, write $\mathbf{\bar a}\left(\mathbf{w}\right)=\left(\bar a_i\left(\mathbf{w}\right)\right)=\mathbf{Aw}$, where $\bar a_i\left(\mathbf{w}\right)$ is just the weighted average of the values on row $i$ (with weights given by $\mathbf{w}$). Two cases:

1. Given $j$, $\exists i^{*}\,|\,a_{i^{*}j}=0$. The assumptions on $\mathbf{A}$ warrant that $i^*$ is unique, and that if $\mathbf{w}$ is such that $\bar a_{i^*}\left(\mathbf{w}\right)=0$, it must be the case that $\bar a_{i}\left(\mathbf{w}\right)>0\;\forall i\neq i^*$. The solution to the specific problem is $q_{i^*}\left(\mathbf{e}_j\right)=1$ and $q_{i}\left(\mathbf{e}_j\right)=0\,\forall i\neq i^*$, and, simmilarly, for those $\mathbf{w}$ such that $\bar a_{i^*}\left(\mathbf{w}\right)=0$, that of the general problem is $q_{i^*}\left(\mathbf{w}\right)=1$ and $q_{i}\left(\mathbf{w}\right)=0\,\forall i\neq i^*$ (if $\bar a_{i}\left(\mathbf{w}\right)>0\,\forall i$, the solution follows the structure in (2) below).

2. Given $j$, $a_{ij}>0\;\forall i$

$$ q_i\left(\mathbf{w}\right)=\frac{1}{\bar a_i\left(\mathbf{w}\right)^\frac{1}{\beta-1}\sum_{i^{*}=1}^n\left(\frac{1}{\bar a_{i^{*}}\left(\mathbf{w}\right)}\right)^\frac{1}{\beta-1}}$$.

When $\mathbf{w}=\mathbf{e}_k$, this simplifies to

$$ q_i\left(\mathbf{e}_k\right)=\frac{1}{a_{ik}^\frac{1}{\beta-1}\sum_{i^{*}=1}^n\left(\frac{1}{a_{i^{*}k}}\right)^\frac{1}{\beta-1}}$$.

I don't see how to proceed from here.

Answer 1

It is not. Not with the same weights.

Regarding general positive coefficients (not necessarily weights) this is a question of existence of a positive solution to the system $Q_{n\times J} w' = q^w_{n\times 1}$, where $Q = [q^1 ... q^J]$ collects the horizontally stacked column solution vectors to pure objectives, and $q^w$ is the column solution vector of the averaged objective. A natural first step towards resolving this problem is Farkas' lemma.

Proof by counterexample:

Just to have the peace of mind, we shall use the minimal example: $n=J=2$, $\beta = 2$ and $w=(\frac{1}{2},\frac{1}{2})$.

Let $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2\end{bmatrix}$ be the matrix with positive entries.

So we have 3 maximization programs with solutions :

\begin{align} x_1 & = \arg \min_{x\in[0,1]} a_1 x^2 + b_1(1-x)^2 = \frac{b_1}{a_1+b_1}\\ x_2 & = \arg \min_{x\in[0,1]} a_2 x^2 + b_2(1-x)^2 = \frac{b_2}{a_2+b_2}\\ x_w & = \arg \min_{x\in[0,1]} \bar a x^2 + \bar b (1-x)^2 = \frac{\bar b}{\bar a+\bar b}, \end{align}

where $\bar a = \frac{a_1+a_2}{2}$ and $\bar b = \frac{b_1+b_2}{2}$.

It is immediate to see that the minimum of the averaged objective is not the average of the minima of the pure objectives:

$$\frac{x_1+x_2}{2} = \frac{1}{2}\left(\frac{b_1}{a_1+b_1}+ \frac{b_2}{a_2+b_2}\right) \neq \frac{1}{2} \frac{b_1 + b_2}{\bar a + \bar b} = x_w$$.