Let $\mu : \mathbb{R}^d \rightarrow \mathbb{R}^d $ and $\sigma: \mathbb{R}^{d} \rightarrow \mathbb{R}^{d \times d} $ be smooth and Lipschitz continuous. Furthermore, let $\varphi : \mathbb{R}^d \rightarrow \mathbb{R}$ be continuous and at most polynomially growing.
The equation \begin{gather} 0 = \frac{1}{2} \text{Trace}[\sigma(x) \sigma(x) (\text{Hessian}_x (u))(\theta)] + \langle \mu (x) , (\nabla_x u) (\theta) \rangle - \frac{\partial u}{\partial t} (\theta) \\ \\ := F(\theta, u(\theta) , \nabla u (\theta), \text{Hessian} \ (u) (\theta)), \\ \\ u(0,x) = \varphi(x), \end{gather} with $$ F : \mathbb{R}^{d+1} \times \mathbb{R} \times \mathbb{R}^{d+1} \times \mathbb{R}^{d+1 \times d+1} \rightarrow \mathbb{R} $$ and $$ \theta = (t,x_1,...,x_d) $$ is called Kolmogorov partial differential equation.
A function $$H :\mathbb{R}^{d+1} \times \mathbb{R} \times \mathbb{R}^{d+1} \times \mathbb{R}^{d+1 \times d+1} \rightarrow \mathbb{R}$$ is called degenerate elliptic if, for any two symmetric matrices $A,B$ for which $B-A$ is positive definite, we have $$ H(x,s,v, A) \geq H(x,u,v,B) $$ for all $x, v \in \mathbb{R}^{d+1}$ and $s \in \mathbb{R}$.
Is $F$ degenerate elliptic? How could I prove this?