Let $a:\mathbb{R}^d\to S_d^{++}$ where $S_d^{++}$ is the set of $d\times d$ positive definite matrices. Suppose that $a$ is $C^1$. Then by a theorem of Phillips and Sarason (Rogers and Williams Theorem 12.12., p.134), we know that $\sigma(x) = a(x)^{1/2}$ is locally Lipschitz.
Question: Is the vector function $$b(x) = \frac{1}{\sqrt{\det a(x)}} \nabla \cdot (\sqrt{\det a(x)} a(x)),$$ locally Lipschitz too?
Here $\nabla \cdot \vec{f}$ is the Euclidean divergence of the vector function $f$. We extend this to matrix functions $A(x)$ by applying it row-wise.
A little attempt: We know that continuously differentiable real-valued functions are locally Lipschitz. Indeed, continuous functions are locally bounded, so if the derivatives are continuous, then they must be locally bounded too. We can use the mean value theorem to show this implies local Lipschitz continuity (and obviously local Lipschitz continuity implies the derivatives are locally bounded).
Can we extend this logic to this higher dimensional setting? Is it that easy?
Note: I do not care about counterexamples where globally Lipschitz continuity fails--only locally. This is because this question comes from studying solutions to SDEs and it is a famous result of Ito (or a refinement of it) that states locally Lipschitz coefficients $\mu(t, x)$ and $\sigma(t,x)$ are enough to guarantee unique strong solutions to the SDE $dX_t = \mu(t, X_t)dt+\sigma(t, X_t)dB_t$. See Theorem 12.1 pg 132 of Rogers and Williams's text.
The property already fails in $d = 1$. For instance, the function $a \colon \mathbb{R} \to S_1^{++} =(0,\infty)$ given by $a(x) = 1+|x|^{3/2}$ is $C^1$, while $b(x) = \frac{1}{\sqrt{a(x)}}(\sqrt{a(x)} a(x))'= \frac{3}{2} a'(x) = \left(\frac{3}{2}\right)^2\frac{x}{|x|^{1/2}}$ is not locally Lipschitz at $x=0$ (it is only $1/2$-Hölder).