investigating positivity/negativity of a function [closed]

Question

I'm investigating if and how the positivity or negativity of a multivariable function can be proved. Consider $y_{1},y_{2},y_{3}\in\mathbb{R}$ and the following function $$f\left(y_{1},y_{2},y_{3}\right)=y_{1}\mbox{artanh}(y_{1}-y_{3})+y_{2}\mbox{artanh}(y_{2}-y_{1})+y_{3}\mbox{artanh}(y_{3}-y_{2})$$ where $\mbox{artanh}(\cdot)$ is the inverse hyperbolic tangent function. How can I prove whether $f(\cdot)$ is positive or negative? I have tried to find the critical points of this function by computing its gradient given by $$\Delta{f}\left(y_{1},y_{2},y_{3}\right)=\begin{bmatrix}\mbox{artanh}(y_{1}-y_{3})-\frac{y_{1}}{(y_{1}-y_{3})^2-1}+\frac{y_{2}}{(y_{1}-y_{2})^2-1}\\\frac{y_{3}}{(y_{2}-y_{3})^2-1}-\frac{y_{2}}{(y_{1}-y_{2})^2-1}-\mbox{artanh}(y_{1}-y_{2})\\ \frac{y_{1}}{(y_{1}-y_{3})^2-1}-\mbox{artanh}(y_{2}-y_{3})-\frac{y_{3}}{(y_{2}-y_{3})^2-1} \end{bmatrix}^{T}$$ which becomes equal to zero when $y_{1}=y_{2}=y_{3}=y_{\ast}$ where $y_{\ast}\in\mathbb{R}$ is an arbitrary real number. The Hessian of the function computed at these points turned out to be a positive semi definite matrix. However, I'm not sure whether this solution is unique. Is this a good way to approach the problem or is there another way to show whether the function $f(\cdot)$ is positive or negative? Thank you.

Kind regards,

Answer 1

This function can be either positive or negative, even if $y_1,y_2,y_3$ are close enough that the hyperbolic tangents are real (see Robert Israel's comment). Fix $\eta_1,\eta_2,\eta_3$ such that $$ \phi(\eta_1,\eta_2,\eta_3) := {\mathop{\rm arctanh}} (\eta_1 - \eta_3) + {\mathop{\rm arctanh}} (\eta_2 - \eta_1) + {\mathop{\rm arctanh}} (\eta_3 - \eta_2) \neq 0. $$ (Almost any choice works as long as $0 < |\eta_i-\eta_j| < 1$ for all $i \neq j$.) Then $$ f(t+\eta_1,t+\eta_2,t+\eta_3) = \phi(\eta_1,\eta_2,\eta_3) t + f(\eta_1,\eta_2,\eta_3) $$ has one sign for $t \rightarrow +\infty$ and the opposite sign for $t \rightarrow -\infty$.

[added later $-$ Note that this used almost nothing about the arctanh function: the same argument works for any function of the form $$ f(y_1,y_2,y_3) = y_1 \, a(y_1-y_3) + y_2 \, a(y_2-y_1) + y_3 \, a(y_3-y_2) $$ as long as the real-valued function $a$ is not linear.]

Answer 2

Basically you are trying to do global optimization of a function on $\mathbb R^n$.

The bad news is that in principle this is undecidable, even if $n=1$ and for a rather small class of functions. According to Richardson's theorem, if $E$ is the smallest class of functions containing $\pi$, $\log(2)$, $e^x$ and $\sin(x)$ and closed under addition, subtraction, multiplication and composition, then the question "is $A(x) \ge 0$ for all $x \in \mathbb R$?" for functions $A \in E$ is undecidable.

The good news is that in many cases it is possible to solve, if you can characterize the behaviour "at infinity".