2
$\begingroup$

assuming we have two smooth function ${f_1},{f_2}:{R^N} \to R$,

  1. under what condition, we have

${f_1}\left( {{{\bf{x}}_1}} \right) \ge {f_1}\left( {{{\bf{x}}_2}} \right) \leftrightarrow {f_2}\left( {{{\bf{x}}_1}} \right) \ge {f_2}\left( {{{\bf{x}}_2}} \right), \forall \bf{x}_1\neq \bf{x}_2\in{R^N}$

  1. what property is called in mathematical terminology about $f_1,f_2$?

we can assume both of the function are convex if needed. I know that monotone transformation preserve the above property. but I am not sure if it is necessary to be able to define a monotone transformation between these two functions or not?

$\endgroup$
3
  • 2
    $\begingroup$ I've seen this property called comonotone $\endgroup$
    – Suvrit
    Apr 1, 2015 at 18:00
  • $\begingroup$ comonotone is for random variables i guess, its not for convex functions! if i am wrong please correct me. $\endgroup$ Apr 1, 2015 at 19:53
  • $\begingroup$ well, the property that you describe has nothing special to do with convex functions. "Varying in the same direction in a monotone fashion" can be quite easily called "comonotone" in general (without being tied down to convexity or random variables etc.) $\endgroup$
    – Suvrit
    Apr 1, 2015 at 21:49

1 Answer 1

2
$\begingroup$

Let Property A be the property described above.

Let Property B be the property that $f_2=g \circ f_1$ for some strictly monotonically increasing function $g:R\rightarrow R$.

Then Property A and Property B are equivalent.

To see that Property B implies Property A, suppose that $f_1(x_1)\geq f_1(x_2)$. Since $g$ is monotonic, if we apply $g$ we get $f_2(x_1)\geq f_2(x_2)$. Since $g$ strictly monotonic implies that $g^{-1}$ is also strictly monotonic, we get the other direction.

To see that Property A implies Property B, assume that Property A holds. Then $f_1(x_1)=f_1(x_2)$ iff $f_2(x_1)=f_2(x_2)$. Therefore, we can define $g$ on $Im(f_1)$ by setting $g(f_1(x))=f_2(x)$; by the preceding observation, $g$ is well-defined, and $f_2=g\circ f_1$.

$\endgroup$

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.