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It seems that the following claim is true, but I did not manage to prove it neither to find a reference.

Claim Let $f:\mathbb R^p\to\mathbb R$ be a three times differentiable function such that its Hessian is Lipschitz continuous: $$ \|\nabla^2 f(x)-\nabla^2 f(y)\|\le M\|x-y\|\qquad\forall x,y\in\mathbb R^p, $$ where the matrix norm is the largest singular value while the vector norm is the usual Euclidean norm. Then, for every $x\in\mathbb R^p$ we have $$ \|\boldsymbol\Delta [\nabla f(x)]\|^2=\sum_{i=1}^p \bigg( \sum_{j=1}^p \frac{\partial^3 f}{\partial x_i\partial x_j^2} (x)\bigg)^2 \le pM^2. $$

I can prove such an inequality with $p^2M^2$ instead of $pM^2$, but do not see how one can remove one power of $p$.

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2 Answers 2

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An equivalent formulation of the claim is the following. If the Hessian of a 3 times differentiable fnction $f$ is Lipschitz continuous with constant $M$, then the Laplacian of $f$ is Lipschitz continuous with the constant $M\sqrt{p}$.

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The claim is false. The counter example is the function $f(x) = \varphi(\|x\|_2)$ with an increasing and infinitely differentiable function $\varphi:[0,\infty)\to\mathbb R$ satisfying $$ \varphi(t) = \begin{cases} t^4,& \text{if } t\le 1,\\ 2, &\text{otherwise.} \end{cases} $$ There is also the example with $$ \varphi(t) = \begin{cases} 31t^2 ,& \text{if } t\le 1,\\ 3t^6-20.8t^5+51t^4-56t^3+56t^2-2.2, &\text{if } t\in[1,2],\\ 96t-75.8&\text{otherwise.} \end{cases} $$ The resulting function $f$ is then convex, since the function $\varphi$ is convex and increasing.

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    $\begingroup$ Your first $\varphi$ is discontinuous. $\endgroup$ Commented Oct 13, 2017 at 17:04

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