# How to prove that homometric sets lead to same result in this problem? [closed]

First let me define Difference multiset for a set of integers $$P=\{p_1,p_2, \dots,p_K\} ,\quad p_i \in\{1,2,\dots,N\},\quad p_i\ne p_j$$ as below: $$D = \{p_i-p_j \mod N ,\quad i \ne j\}$$ I know that the minimum of $f(x)$ is same for all $P$'s having same difference multiset (homometric $P$'s) & and also the optimal $x$ (minimizing $f$) is same for all of them up to permutation of elements, (also for all homometric $P$'s same $l$ is the inner maximizer) , where $f(x)$ is a real function which is defined as:

$$\large f(x) = \max_{1 \leq l \leq N-1} {\sum_{j=1}^K \sum_{k=1}^K x_j x_k e^{\frac{i2\pi l(p_j-p_k)}N} \over \left( \sum_{i=1}^K x_i\right)^2 } = \max_{1 \leq l \leq N-1} { \sum_{i=1}^K x_i^2 + \sum_{j,k} x_j x_k cos(\frac{2\pi l(p_j-p_k)}N) \over \left( \sum_{i=1}^K x_i\right)^2 }$$ $x_i$'s are positive variables

I achieved this result from simulations. I'm looking for a proof or even a justification which helps me prove it.

Let me give you an example of my simulations if it helps:

suppose $(N,K)= (6, 4)$

$$P = \{1,2,3,5\} \Rightarrow D = \{1,1,2,2,2,3,3,4,4,4,5,5\}$$ minimizing $f(x)$ with $p_i$'s being members of $P$, led to this $x=(4,5,4,4)$

and for $$P' = \{1,2,4,6\} \Rightarrow D' = \{1,1,2,2,2,3,3,4,4,4,5,5\} = D$$ minimizing $f(x)$ with $p_i$'s being members of $P'$, led to this $x=(5,4,4,4)$

Also note that $f(x)$ is independent of $||x||_2$ and is just function of angle of vector $x$.

I've also asked this on ME

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## closed as off-topic by Gerry Myerson, Andrés Caicedo, Andrey Rekalo, David White, Scott Morrison♦Jul 25 '13 at 2:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Andrés Caicedo, Scott Morrison
If this question can be reworded to fit the rules in the help center, please edit the question.

May I suggest replacing the title with a more informative and descriptive one? That may attract more people who may be able to give you an answer. – Ricardo Andrade Jul 23 '13 at 8:19
crossposted: math.stackexchange.com/questions/450065/… – Greg Martin Jul 23 '13 at 8:26
@Mahdi: I apologize for being unclear before. My suggestion was to briefly describe your particular problem (i.e what you want to show) in the title. That could give people an idea of what the question is about, possibly prompting them to read the question. – Ricardo Andrade Jul 23 '13 at 8:47
This question isn't about Mathematics, it's about guessing what the question is. – Gerry Myerson Jul 23 '13 at 23:08
Best guess is that OP "knows" (from simulations) that "the minimum of $f(x)$ is the same...", but doesn't know it in the sense of possessing a rigorous proof, and that's what OP wants. Agreed that the question is a little jumbled. – Todd Trimble Jul 24 '13 at 2:59

$N=6$, $K=4$, $P=\{1,2,4,5\}$, $x=(1,1,1,1)$, $l=1$ seems to give $f(x)=0$.
There are values of the parameters for which $f$ has no minimum. Take $N=5$, $K=4$, $P=\{1,2,3,4\}$, $x=(a,b,b,a)$, $l=1$. Let $$\theta=\sum x_je^{2\pi il/N}=a(\beta+\beta^4)+b(\beta^2+\beta^3)$$ where $\beta=e^{2\pi i/5}$. Now, $\beta+\beta^4=(\sqrt5-1)/2$, $\beta^2+\beta^3=-(\sqrt5+1)/2$, and these two numbers are linearly independent over the rationals, so there are arbitrarily large $a$ and $b$ for which $|\theta|$ is arbitrarily small. So $$f(x)={|\theta|^2\over4(a+b)^2}$$ can be made arbitrarily close to zero. It can't be exactly zero, since $\beta,\dots,\beta^4$ are linearly independent over the rationals.