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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.

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