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I am considering the convexity of multivariate polynomials in the form of, say, $Px^4$, where tensor $P$ is a 4-order n-dimensional positive definite (that is, $\forall x\neq 0, Px^4>0$) symmetric tensor, $x\in \mathbb{R}^n$, and $Px^4$ is the abbreviation of $P_{ijkl}x_ix_jx_kx_l$ in Einstein sum notation.

My attempt: $Px^4$ is a polynomial function and has any order of derivative. The second-order convexity determination says that a function is strictly convex if its Hessian matrix is positive definite at any point in its domain. Simple calculation yields that $[12P_{ijkl}x_kx_l]_{ij}$ is its Hessian matrix, and all I need is to verify its positive definiteness. That means I need to determine whether $Px^2y^2$ is always positive, for any $x,y$.

My question: I don't think $Px^4$ is necessarily a convex function, but I can't find a counterexample. Is it truely convex or not? If it is convex, how is it proved, and can this conclusion be extended onto any even order tensor polynomials? If not, how can we find a counterexample?

Thanks anyone that may give me a hint.

Rodrigo de Azevedo
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dhliu
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My answer: $Px^4$ is not a convex function. Consider the simplest 2-dimensional case. Take the tensor $P$ as: $P_{1111}=P_{2222}=4, P_{1212}=P_{1221}=P_{1122}=P_{2211}=P_{2112}=P_{2121}=-1$, and all 0 at other indices. This $P$ is indeed a positive definite tensor, and this is because if $x=(a,b)^T\neq 0$, $$Px^4=4a^4-6a^2b^2+4b^4.$$ WLOG we set $b\neq 0$, and $Px^4=(4(a/b)^4-6(a/b)^2+4)b^4$, and the univariate polynomial $4x^4-6x^2+4=(2x^2-\sqrt{14}x+2)(2x^2+\sqrt{14}x+2)>0$.

And now we set $x=(x_1,x_2)^T, y=(y_1,y_2)^T$, then $$Px^2y^2 = 4x_1^2x_2^2+4x_2^2y_2^2-4x_1x_2y_1y_2-x_1^2y_2^2-x_2^2y_1^2 \\ = (2x_1y_1-2x_2y_2)^2-(x_1y_2-x_2y_1)^2+2x_1x_2y_1y_2.$$ Now take $x_2=0$, we have $Px^2y^2=x_1^2(4y_1^2-y_2^2).$ So we can simply choose $y_2$ that $|y_2|>2|y_1|$, then $Px^2y^2<0$, thus $Px^4$ is not convex.

dhliu
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