# Solutions to a parametrized optimization problem

I have the following maximization program

\begin{align} \max\limits_{\{q_i\}}&\quad\sum\limits_{i=1}^nq_i \\ \text{s.t.}&\quad\begin{cases} k_j \geq \sum\limits_{i=1}^n q_i^{1 \over \alpha}x_{ij} & j=\{1,\dots,J\} \\ q_i \geq 0 & i=\{1,\dots,n\} \\ \end{cases} \end{align}

with $$\alpha>0$$, $$x_{ij}\geq 0$$ and $$k_j \geq 0$$ for all indices. I'd like to prove that if $$\mathbf q^*=(q_1^*,\dots,q_n^*)$$ is a solution to this problem for $$\mathbf k^*=(k_1^*,\dots,k_J^*),$$ then $$\delta^{\alpha}\mathbf q^*$$ is a solution for $$\delta \mathbf k^*.$$

Assume that the solution for $$\delta \mathbf k^*$$ was $$\mathbf {\hat q}$$ and such that $$\sum\limits_{i=1}^n \hat q_i>\delta^{\alpha}\sum\limits_{i=1}^n q_i^*$$. Consider $$\mathbf {\bar q}=\dfrac1{\delta^{\alpha}}\mathbf {\hat q}.$$ It is easily seen that $$\mathbf {\bar q}$$ is feasible for $$\mathbf k^*$$. But we have that

$$\sum_{i=1}^n \bar q_i>\sum_{i=1}^n q_i^*,$$

so $$\mathbf q^*$$ cannot be a solution for $$\mathbf k^*$$.

Is this correct? I worry that if the program has multiple solutions (and this can be the case when $$\alpha>1$$) my proof is not general enough or even wrong altogether.

Edit: Perhaps it would be more accurate to say that, letting $$\mathbf S(\mathbf k)$$ denote the set of solutions to the maximization program for a given $$\mathbf k$$, $$\mathbf q^*\in \mathbf S(\mathbf k^*) \iff \delta^{\alpha}\mathbf q^*\in \mathbf S(\delta\mathbf k^*)$$

Your logic seems fine to me. There's no issue with multiple solutions: just make $$\hat{\bf{q}}$$ any optimal solution to the modified problem.