1
$\begingroup$

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 $q$.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.