# Prove that the set of maximizers are independent of parameter in the objective function

A maximization problem reads as $$J(y) = \sum_{k=1}^{K} \sigma_k(y)^q \mathop{\rightarrow}^{y} max$$ where $$q \in [1,\infty]$$ is a user-defined parameter and functions $$\sigma_k, k=\{1,\dots,K\}$$ satisfy these two conditions for any given $$y$$: $$\sigma_1(y) \geq \sigma_2(y) \geq \dots \geq \sigma_K(y) \geq 0.$$

and

$$\sum_{k=1}^{K} \sigma_{k}(y) = K$$

I would like to prove that the set of maximizers for $$J(y)$$ is independent of the choice of $$q$$.

• When you write $\sigma_k^q(y)$, do you mean exponentiation by $q$? It would probably be easier to understand if you wrote $\sigma_k(y)^q$ in that case. – Wolfgang Bangerth Dec 10 '19 at 3:49
• @WolfgangBangerth thanks for the comments. Changes are done. I also realized that there is something fishy in my attempt at the question. – hari Dec 10 '19 at 5:35
• In (2), I don't think there should be a $K$ factor on the left. – Wolfgang Bangerth Dec 10 '19 at 15:18
• I don't think you proved that $y$ at the maximum point of $J$ is independent from $q$. You just found an upper bound probably for $\sigma_{k}$. – Alone Programmer Dec 10 '19 at 17:47
• @AloneProgrammer, you are right. I observed later that my attempt does not say anything about the minimizers. – hari Dec 11 '19 at 9:01

Following up on my comment on the original question, I have finally managed to construct a counter example that shows that the statement is not in fact correct. Define the positive part of a function, $$[x]^+ = \begin{cases}x & \text{if x\ge 0} \\ 0 & \text{otherwise.}\end{cases}$$ The let $$\sigma_1(y) = 1+ \left[\tfrac 14 - |y-1|\right]^+$$ and $$\sigma_2(y) = \left[\tfrac 12 - |y|\right]^+.$$ These are both non-negative functions with the requested ordering. They look like this: The point is that the bump of $$\sigma_2$$ is larger than the bump of $$\sigma_1$$, and consequently the maximum of their sum is at $$y=0$$. But $$\sigma_1\ge 1$$, and so if you take a positive power of it, its bump gets larger; on the other hand, $$\sigma_2<1$$ and so its bump gets smaller if you take some power of it. Indeed, plotting both $$\sigma_1(y)+\sigma_2(y)$$ and $$\sigma_1(y)^4+\sigma_2(y)^4$$ shows how this works: In other words, for $$q=1$$ the maximum is at $$y=0$$, whereas for $$q=4$$, it is at $$y=1$$. This contradicts your claimed independence of the location of the maximizer.
• OP added another condition in comments as: $$\sum_{k=1}^{K} \sigma_{k}(y) = K$$ for any value of $y$. In your counter example case: $K=2$, obviously: $\sum_{k=1}^{K} \sigma_{k}(y) \neq K$. – Alone Programmer Dec 11 '19 at 15:50
• Uh, that doesn't make any sense. If $\sum_k \sigma_k(y)=K$, then any $y$ is a maximizer -- the objective function is simply constant. – Wolfgang Bangerth Dec 11 '19 at 16:01
• You mean $J(y)$ would be constant? I don't think so... – Alone Programmer Dec 11 '19 at 16:03
• But for $q=1$ it is! – Wolfgang Bangerth Dec 11 '19 at 17:27
• Yes, I know but I'm interested to see if it is possible to create counter example for $q > 1$ or not. – Alone Programmer Dec 11 '19 at 17:52