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

Question

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

Answer 1

$N=6$, $K=4$, $P=\{1,2,4,5\}$, $x=(1,1,1,1)$, $l=1$ 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.