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 $q$.

Improve this question
$\endgroup$
9
  • $\begingroup$ 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
    Commented Dec 10, 2019 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
    Commented Dec 10, 2019 at 5:35
  • $\begingroup$ In (2), I don't think there should be a $K$ factor on the left. $\endgroup$
    – Wolfgang Bangerth
    Commented Dec 10, 2019 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 $\sigma_{k}$. $\endgroup$
    – Mithridates the Great
    Commented Dec 10, 2019 at 17:47
  • $\begingroup$ @AloneProgrammer, you are right. I observed later that my attempt does not say anything about the minimizers. $\endgroup$
    – hari
    Commented Dec 11, 2019 at 9:01

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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: $\sum_{k=1}^{K} \sigma_{k}(y) \neq K$. $\endgroup$
    – Mithridates the Great
    Commented Dec 11, 2019 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
    Commented Dec 11, 2019 at 16:01
  • $\begingroup$ You mean $J(y)$ would be constant? I don't think so... $\endgroup$
    – Mithridates the Great
    Commented Dec 11, 2019 at 16:03
  • $\begingroup$ But for $q=1$ it is! $\endgroup$
    – Wolfgang Bangerth
    Commented Dec 11, 2019 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. $\endgroup$
    – Mithridates the Great
    Commented Dec 11, 2019 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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