I have the following function $$H(\mathbf{x}) = a_{1}|x_{1}-A_{1}|+a_{2}|x_{2}-A_{2}|+\cdots+a_{n}|x_{n}-A_{n}| +b_{1}|x_{1}-B_{1}|+b_{2}|x_{2}-B_{2}|+\cdots+a_{n}|x_{n}-B_{n}| +c_{1}|x_{1}-x_{0}|+c_{2}|x_{2}-x_{1}|+\cdots+c_{n}|x_{n}-x_{n-1}|+c_{n+1}|x_{n+1}-x_{n}|$$ where all coefficients $a_{1}, a_{2}, \cdots, b_{1}, b_{2}, \cdots, A_{1}, A_{2}, \cdots, B_{1}, B_{2}, \cdots, c_{1}, \cdots$, $x_{0}$ and $x_{n+1}$ are positive constants.

For two dimensional case, I can plot $$H(x_{1},x_{2}) = a_{1}|x_{1}-A_{1}|+a_{2}|x_{2}-A_{2}| +b_{1}|x_{1}-B_{1}|+b_{2}|x_{2}-B_{2}| +c_{1}|x_{1}-x_{0}|+c_{2}|x_{2}-x_{1}|+c_{3}|x_{3}-x_{2}|$$ and it indeed has only one minimum,

enter image description here

However, I cannot prove it in a more general case. Does anyone once had encountered such problems?

  • $\begingroup$ If $H$ is a function of $x_1,x_2$, where does the $x_3$ (or the $x_{n+1}$) comes from? $\endgroup$
    – 49328481
    Jan 23 at 3:11
  • $\begingroup$ x0 and x_n+1 are constants. $\endgroup$ Jan 23 at 3:17
  • $\begingroup$ $a_1,...,a_n,b_1,...b_n,c_1,...,c_{n+1}$ are all non-negative? $\endgroup$
    – 49328481
    Jan 23 at 3:23
  • $\begingroup$ Yes, these are all positive real numbers. $\endgroup$ Jan 23 at 3:24
  • 1
    $\begingroup$ This is a convex function with compact sublevel sets that is bounded by from below (trivially since everything is nonnegative), so you can deduce the existence of a global minimizer by the Bolzano-Weierstrass theorem. $\endgroup$
    – VHarisop
    Jan 23 at 3:34

Your Answer

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

Browse other questions tagged or ask your own question.