Find conditions on the parameters such that a polynomial is globally nonnegative

Question

Let

\begin{align}%\label{eqn: S0} \notag f(x,y,z)=&z^2+\left(\beta+2\,c_3-4\right)\,y\,z+\left((\beta+1)\,c_3+c_6-4\right)\,y^2+z\left(\frac{1}{2}\left(10-c_3\right)x^2+c_6\,x\right) \\[2mm] \notag &+y\left(\left(\frac{1}{2}\,(\beta +1)\left(10-c_3\right)+3\,c_5+2\,c_6-3\right)\,x^2+\left(\beta\,c_6+2\,c_8\right)\,x\right) \\[2mm]%\notag &+c_5\,x^4+\left(\beta\,c_5+3\,c_8\right)\,x^3 +c_8\,(\beta -1)\,x^2. \end{align}

be a polynomial with $c_3,c_5,c_6,c_8\in\mathbb{R}$ and $\beta>0$. Could we find $c_3,c_5,c_6,c_8$ and $\beta$ such that $f(x,y,z)\ge 0$ for all $x,y,z\in\mathbb{R}$?

Any reference, suggestion, idea, or comment is welcome. Thank you!

Answer 1

I am unable It is impossible to find any $c_3, c_5, c_6, c_8$ and positive $\beta$ to make $f(x,y,z)$ globally non-negative (see update below). However, there There is a set of $14$ inequalities for the coefficients. If the coefficients satisfy all of them, then $f(x,y,z)$ will be globally non-negative.

Let $A$ be the $4\times 4$ matrix: $$\begin{bmatrix} 2 & \beta+2 c_3-4 & \frac{10-c_3}{2} & c_6\\ \beta+2c_3-4 & 2( c_3( \beta+1) + c_6-4) & \frac{(10-c_3)(\beta+1)+4 c_6+6c_5-6}{2} & c_6 \beta + 2 c_8\\ \frac{10-c_3}{2} & \frac{(10-c_3)(\beta+1)+4c_6+6c_5-6}{2} & 2c_5 & c_5\beta + 3c_8\\ c_6 & c_6\beta+2c_8 & c_5 \beta + 3c_8 & 2c_8(\beta-1) \end{bmatrix}$$ For $p = (x,y,z) \in \mathbb{R}^3$, let $U_p$ be the $4\times 1$ column vector $[z,y,x^2,x]^T$. In terms of $A$ and $U_p$, the function at hand equals to

$$f(x,y,z) = \frac12 U^T_p A U_p$$

When $A$ is positive semi-definite, it is easy to see $f(x,y,z)$ is globally non-negative.

If $A$ isn't positive semi-definite, then one can find a non-zero $U = [u_1,u_2,u_3,u_4]^T$ such that $U^T A U$ is negative. If one perturb $u_3,u_4$ for sufficient small amount, $U^T A U$ remains to be negative. WOLOG, we can assume $u_3, u_4 \ne 0$. For such a $U$, we find

$$f\left(\frac{u_3}{u_4},\frac{u_2u_3}{u_4^2},\frac{u_1u_3}{u_4^2}\right) = \frac{u_3^2}{2u_4^4} U^T A U < 0$$

This means $f(x,y,z)$ is globally non-negative when and only when $A$ is positive semi-definite.

By Sylvester's criterion for positive semidefinite matrices, $f(x,y,z)$ will be globally non-negative when all principal minors of $A$ have non-negative determinant.

Since $A$ is a $4\times 4$ matrix, it has $15 = 2^4-1$ principal minors. $4$ of them comes form diagonal elements of $A$. The inequality come from the first diagonal element is trivial. The other $3$ inequalities are

• $c_3(\beta+1) + c_6 - 4 \ge 0$,
• $c_5 \ge 0$,
• $c_8(\beta - 1) \ge 0$

There are $11 = 6 + 4 + 1$ more inequalities coming from principal minors obtained by removing $2$, $1$ or no rows/columns (the last one is simply $\det(A) \ge 0$). I'm not giving to list all of them here. Please compute them yourselves using a CAS.

Update

We are going to show $f(x,y,z)$ cannot be globally non-negative.

First, let use consider the case $c_5 > 0$.

Change variables to $(\beta, t,u,v,w)$ such that $(c_3,c_5,c_6,c_8) = (10-2ut,t^2, vt,wt^2)$. We will assume $t > 0$.

Let $A_{\overline{ij}}$ be the determinant of principal minor of $A$ with only $i^{th}/j^{th}$ rows/columns are kept.
With help of a CAS, one can verify

$$ \begin{align} A_{\overline{12}} = &\; 4tv-(4tu-15)^2-(1-\beta)(7-\beta)\\ A_{\overline{34}} = &\; t^2 \overbrace{(-4tv-12(\beta+1)tu - 8t^2+40\beta+36)}^{P}\\ &\;- (2tv+(\beta+1)tu+t^2-3)^2 \end{align}$$

When $A$ is positive semidefinite, $A_{\overline{12}}, A_{\overline{34}} \ge 0$. Since the factor $P$ in $A_{\overline{34}}$ can be expressed as a sum of $A_{\overline{34}}$ as square followed by division of $t^2$, we have $P \ge 0$. Adding a few more squares and simplify using an CAS, we obtain

$$5(\beta^2 + 6\beta - 11) = 4(A_{\overline{12}} + P) + (8tu+3\beta-18)^2 + 32t^2 \ge 0$$ This forces $$|\beta + 3| \ge \sqrt{20} \implies \beta \ge \sqrt{20} - 3$$

With help of CAS again, one find

$$A_{\overline{34}} = -t^4( 9w^2 + (2\beta+4)w + \beta^2)$$ Treat this as a quadratic polynomial in $w$. Notice the coefficient for $w^2$ and $w^0$ are negative. In order for it to have a chance to be non-negative, its discriminant need to be non-negative. This leads to

$$(2\beta+4)^2 - 36\beta^2 = 16(1-\beta)(1 + 2\beta) \ge 0\quad \implies\quad \beta \le 1$$ This contradicts with above result condition $\beta \ge \sqrt{20}-3 > 1$.

What this means is when $c_5 > 0$, it is impossible for $A_{\overline{12}}, A_{\overline{23}}, A_{\overline{34}}$ to be non-negative at the same time and hence $A$ cannot be positive semidefinite.

For the remaining case $c_5 = 0$, $A$ has a zero at the $3^{rd}$ diagonal element. In order for $A$ to be positive semidefinite, all entries in $3^{th}$ row/column need to vansih. This implies $c_3 = 10$, $c_6 = \frac32$ and $c_8 = 0$. When $c_8 = 0$, $A$ has a zero at the $4^{th}$ diagonal element. However the $4^{th}$ row/column has a non-zero entry and hennce $A$ cannot be positive semidefinite.

Combine these, we can conclude there are no $c_3, c_5, c_6, c_8$ and positive $\beta$ to make $A$ positive semidefinite and hence $f(x,y,z)$ is never globally non-negative.

Answer 2

Write it out as a homogeneos polynomial, i.e. in $x^2$, $x y$, and so on. You are asking for that to be positive definite...