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 $y$: $$ \sigma_1(y) \geq \sigma_2(y) \geq \dots \geq \sigma_K(y) \geq 0.$$


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

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

  • $\begingroup$ 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. $\endgroup$ – Wolfgang Bangerth Dec 10 '19 at 3:49
  • $\begingroup$ @WolfgangBangerth thanks for the comments. Changes are done. I also realized that there is something fishy in my attempt at the question. $\endgroup$ – hari Dec 10 '19 at 5:35
  • $\begingroup$ In (2), I don't think there should be a $K$ factor on the left. $\endgroup$ – Wolfgang Bangerth Dec 10 '19 at 15:18
  • $\begingroup$ 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 $\sigma_{k}$. $\endgroup$ – Alone Programmer Dec 10 '19 at 17:47
  • $\begingroup$ @AloneProgrammer, you are right. I observed later that my attempt does not say anything about the minimizers. $\endgroup$ – 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: enter image description here 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: enter image description here 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.

  • $\begingroup$ 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: $\sum_{k=1}^{K} \sigma_{k}(y) \neq K$. $\endgroup$ – Alone Programmer Dec 11 '19 at 15:50
  • $\begingroup$ 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. $\endgroup$ – Wolfgang Bangerth Dec 11 '19 at 16:01
  • $\begingroup$ You mean $J(y)$ would be constant? I don't think so... $\endgroup$ – Alone Programmer Dec 11 '19 at 16:03
  • $\begingroup$ But for $q=1$ it is! $\endgroup$ – Wolfgang Bangerth Dec 11 '19 at 17:27
  • $\begingroup$ Yes, I know but I'm interested to see if it is possible to create counter example for $q > 1$ or not. $\endgroup$ – Alone Programmer Dec 11 '19 at 17:52

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