# Inferring the Lipschitz constant of a vector Laplacian, given Lipschitz constants of the second and third derivative

I have a function $$f: \mathbb{R}^d \to \mathbb{R}$$, which is three times continuously differentiable. I know that the Lipschitz constants for the second and third derivative of this function are $$\mu_2$$ and $$\mu_3$$, i.e. $$\| \nabla^2 f(x) - \nabla^2 f(y) \|_{\text{op}} \le \mu_2 \| x -y \|_2,$$ $$\| \nabla^3 f(x) - \nabla^3 f(y) \|_{\text{op}} \le \mu_3 \| x -y \|_2,$$ where the operator norms are defined to be $$\| \nabla^2 f(x) \|_{\text{op}} = \sup_{ \| u \|\le 1} \| \nabla^2 f(x)u \|_2,$$ $$\| \nabla^3 f(x) \|_{\text{op}} = \sup_{ \| u \|, \| v \| \le 1} \left\| \left(v^\top \nabla^2 \frac{\partial f}{\partial x_1} u, \dots, v^\top \nabla^2 \frac{\partial f}{\partial x_d} u \right) \right\|_2.$$ Now, I'm interested in the Lipschitz constant of the vector Laplacian of the gradient of $$f$$, i.e. $$\vec{\Delta}(\nabla f)$$, where each component is $$\vec{\Delta}(\nabla f)_i = \Delta \left(\frac{\partial f}{\partial x_i}\right)$$.

I'm guessing the constant is $$d \mu_3$$. Is there a way to formally show this?

It's trivial to show $$d^{3/2} \mu_3$$ is a valid Lipschitz constant. However, whether this is the best we can do seems non-trivial.