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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.

Nap D. Lover
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  • This is false, even for scalar functions. If $f \in C^1(\mathbb{R})$, then $f'$ is continuous but need not be Lipschitz. Think of $f(x) = |x|^{3/2}$ at $x = 0$. – Ignatius Oct 01 '23 at 01:01
  • @Ignatius If $f$ is $C^1$ then its derivative is continuous hence locally bounded, hence locally Lipschitz, right? I do not need or care about globally Lipschitz (because this is from theory of solutions to SDEs--locally Lipschitz coefficients are enough to guarantee strong solutions to SDEs). I think my parentheses in my attempt might have been misleading, sorry, I will edit. – Nap D. Lover Oct 01 '23 at 19:10
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    Not all continuous functions are locally Lipschitz. While $C^1$ functions are locally Lipschitz, their derivatives need not have this property. In my example, local continuity of $f'$ fails at $x=0$. In your question, you know that $a$ is locally Lipschitz and you are asking whether $b$ (which is essentially some derivative of $a$) is locally Lipschitz, correct? This cannot be true in general! – Ignatius Oct 01 '23 at 22:12
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    Obviously, I meant that local Lipschitz continuity of $f'$ fails at $x=0$ (not continuity). – Ignatius Oct 01 '23 at 22:25
  • @Ignatius Ah I see now! Thank you for being patient and clarifying. If you wish to write this up as an answer, I will gladly accept it. – Nap D. Lover Oct 01 '23 at 22:52

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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).

Ignatius
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