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Let $\Delta:=\lbrace p\in(0,1)^{n}\ \colon\ p_{1}+\dots+p_{n}=1 \rbrace$ and $H:=\lbrace h\in\mathbb{R}^{n}\ \colon\ h_{1}+\dots+h_{n}=0 \rbrace$. Now let $A\in\mathbb{R}^{n\times n}$ be a conditionally negative definite matrix, i.e. \begin{equation} h\cdot Ah<0,\quad \text{for all }h\in H. \end{equation} It is well-known that for such a matrix there exists $\lambda>0$ such that \begin{equation} h\cdot Ah\le -\lambda\|h\|^{2},\quad \text{for all }h\in H. \end{equation} Next, consider the matrix field $D(p)$, where for $p\in\Delta$ \begin{equation} D_{ij}(p):=\frac{\delta_{ij}}{p_{i}}-\frac{1}{np_{j}}. \end{equation} Clearly, $D$ is conditionally positive definite. In fact it is even 'conditionally elliptic' in that \begin{equation} h\cdot D(p)h\ge \|h\|^{2},\quad \text{for all }h\in H \text{ and }p\in\Delta. \end{equation} It it straight-forward to see that for all $p\in\Delta$ the product $AD(p)$ is again conditionally negative definite. The question which is bothering me is whether we can also get a uniform bound for the quadratic form, i.e. does there exist some $\rho>0$ for which \begin{equation} h\cdot AD(p)h\le -\rho\|h\|^{2},\quad \text{for all }h\in H\text{ and }p\in\Delta? \end{equation} I appreciate any help.

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