Consider the following double sum:

$$Q(n)=\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}\left [ \partial _{ij}lnf\left ( x \right ) \right ]^2$$ where $\partial_{ij}$ is the partial second order derivative (bounded for all indices), the function $f$ is n-dimensional. The problem is then the convergence of this sum to zero. Clearly the boundedness condition is not sufficient for the sum to converge to zero. Hence, the sum has to grow slightly slower than $n^2$, i.e. $Q(n)=o(n^2)$. One of my questions would be: what the condition $Q(n)=o(n^2)$ implies about the function $f$ itself? Or nothing specific could be deduced about the behaviour of the function?

In addition, I have the following inequality: $$\left | \frac{1}{n}\sum_{i=1}^{n}\left [ \partial _{ii}lnf+\left ( \partial_ilnf \right )^2 \right ] \right |<n^{-\frac{1}{8}}$$ But I do not think that the convergence of $Q(n)$ to zero could be deduced from this inequality. Or maybe I'm wrong?

Don't have much experience dealing with this kind of sums and their convergences. Therefore, I'm hoping for some help from the community. This is a research question that popped up while working on the convergence of particular infinitesimal generator on a stochastic process. The function $f$ is in fact a probability density function.

EDIT I observed that $Q(n)$ could be expressed as a trace of squared Hessian of $ln f(x)$: $$Q(n)=\frac{1}{n^2} trace(H^2)$$ Trace inequality $trace(AB)^2\leqslant \sum_{i=1 }^{n}\lambda_i^4$ implies that $$0\leqslant Q(n)\leqslant \frac{1}{n^2}\sum_{i=1 }^{n}\lambda_i^4\rightarrow 0$$ if all eigenvalues of the Hessian are bounded. Trace inequality is taken from this paper.

In conclusion, it looks like one in fact does not need that additional assumption $Q(n)=o(n^2)$, i.e. $Q(n)\rightarrow 0$ if eigenvalues of the Hessian of the function $ln f(x)$ are bounded.

Or have I missed something?